On 11/15/2024 12:06 PM, Jim Burns wrote:
On 11/15/2024 1:05 PM, Ross Finlayson wrote:
Non-standard models of integers exist.
Yes, and,
when we discuss non.standard models,
we can assemble claim.sequences which
start with a description of a non.standard model.
And, when we do that,
augmenting true.or.not.false claims
will be true about the described non.standard models.
However,
non.standard models of integers are not
standard models of integers.
Ah, yet according to Mirimanoff,
there do not exist standard models of integers,
that Russell has fooled you with a contradictory statement,
and there are only fragments, if unbounded,
and extensions, sublime.
On 11/15/2024 02:37 PM, Jim Burns wrote:
On 11/15/2024 4:32 PM, Ross Finlayson wrote:
Ah, yet according to Mirimanoff,
there do not exist standard models of integers,
If it is true that
our domain of discourse is a model of ST+PQ
then it is true that
our domain of discourse holds a standard integer.model.
What is Mirimanoff's argument that
it doesn't exist?
Mirimanoff's? Russell's Paradox.
That
"If it is true that our domain of discourse
is a model of ST+PQ then it is true that our
domain of discourse holds a standard integer.model"
is a pretty long axiom - why not just say
"infinity", that's the usual approach.
On 11/16/2024 08:58 AM, Ross Finlayson wrote:
On 11/16/2024 02:22 AM, Jim Burns wrote:
On 11/15/2024 9:52 PM, Ross Finlayson wrote:
On 11/15/2024 02:37 PM, Jim Burns wrote:
On 11/15/2024 4:32 PM, Ross Finlayson wrote:
Ah, yet according to Mirimanoff,
there do not exist standard models of integers,
If it is true that
our domain of discourse is a model of ST+PQ
then it is true that
our domain of discourse holds a standard integer.model.
What is Mirimanoff's argument that
it doesn't exist?
Mirimanoff's? Russell's Paradox.
ST+PQ does not suffer from claiming
that the set of all non.self.membered sets
is self.membered or claiming it isn't.
I don't say "infinity" is an axiom
primarily because
"infinity" is not an axiom of ST+PQ
ST+PQ:
⎛ set {} exists
⎜ set x∪{y} exists
⎜ set.extensionality
⎜ plurality ⦃z:P(z)⦄ exists
⎝ plurality.extensionality
"Infinity exists" ==
"the minimal inductive plurality exists"
is a theorem of those axioms.
^- Fragment
Let's recall an example geometrically of what's
so inductively and not so in the limit.
Take a circle and draw a diameter, then bisect
the diameter resulting diameters of common circles,
all sharing a common diameter, vertical, say.
Then, notice the length of the circle, is
same, as the sum of the lengths of the half-diameter
circles, their sum.
So, repeat his dividing ad infinitum. In the limit,
the length is that of the diameter, not the perimeter,
while inductively, it's the diameter.
Thusly, a clear example "not.first.false" being
"ultimately.untrue".
Then, with regards to your fragment,
congratulations,
you have ignored Russell his paradox and so on
and quite fully revived Frege and given yourself
a complete theory and consistent as it may be, and
can entirely ignore all of 20'th century mathematics.
It's small, .... Fragment
On 11/16/2024 12:29 PM, Jim Burns wrote:
On 11/16/2024 12:07 PM, Ross Finlayson wrote:
On 11/16/2024 08:58 AM, Ross Finlayson wrote:
On 11/16/2024 02:22 AM, Jim Burns wrote:
On 11/15/2024 9:52 PM, Ross Finlayson wrote:
On 11/15/2024 02:37 PM, Jim Burns wrote:
On 11/15/2024 4:32 PM, Ross Finlayson wrote:
Ah, yet according to Mirimanoff,
there do not exist standard models of integers,
If it is true that
our domain of discourse is a model of ST+PQ
then it is true that
our domain of discourse holds a standard integer.model.
What is Mirimanoff's argument that
it doesn't exist?
Mirimanoff's? Russell's Paradox.
ST+PQ does not suffer from claiming
that the set of all non.self.membered sets
is self.membered or claiming it isn't.
I don't say "infinity" is an axiom
primarily because
"infinity" is not an axiom of ST+PQ
ST+PQ:
⎛ set {} exists
⎜ set x∪{y} exists
⎜ set.extensionality
⎜ plurality ⦃z:P(z)⦄ exists
⎝ plurality.extensionality
"Infinity exists" ==
"the minimal inductive plurality exists"
is a theorem of those axioms.
^- Fragment
No.
The minimal inductive plurality is
a standard model of the integers.
Let's recall an example geometrically of what's
so inductively and not so in the limit.
Take a circle and draw a diameter, then bisect
the diameter resulting diameters of common circles,
all sharing a common diameter, vertical, say.
Then, notice the length of the circle, is
same, as the sum of the lengths of the half-diameter
circles, their sum.
So, repeat his dividing ad infinitum. In the limit,
the length is that of the diameter, not the perimeter,
while inductively, it's the diameter.
Thusly, a clear example "not.first.false" being
"ultimately.untrue".
A finite sequence of claims, each claim of which
is true.or.not.first.false is
a finite sequence of claims, each claim of which
is true.
The reason that that's true is that
THE SEQUENCE OF CLAIMS is finite.
Whatever those CLAIMS refer to,
none of those CLAIMS are first.false.
(They're each not.first.false.)
Since none of those CLAIMS are first.false,
none of those CLAIMS are false.
(That sequence is finite.)
What those claims are ABOUT doesn't affect that.
For example,
being ABOUT an indefinite one of infinitely.many
doesn't affect that.
Discovering
a finite sequence of claims, each claim of which
is true.or.not.first.false
in which there IS an untrue claim
is akin to
counting the eggs in a carton and
discovering that, there, in that carton,
7 is NOT between 6 and 8.
There is a problem, but not with mathematics.
Then, with regards to your fragment,
...the minimal inductive plurality...
congratulations,
Thank you.
you have ignored Russell his paradox and so on
No.
Selecting axioms which do not suffer from claiming
that the set of all non.self.membered sets
is self.membered or claiming it isn't
is not ignoring Russel,
it is responding to Russell.
Russell points out that
_we do not want_ to claim
that the set of all non.self.membered sets
is self.membered or claiming it isn't.
We respond: Okay, we'll stop doing that.
and quite fully revived Frege and given yourself
a complete theory and consistent as it may be, and
can entirely ignore all of 20'th century mathematics.
It's small, .... Fragment
The minimal inductive plurality.
Big or small, that's the thing,
the whole thing, and nothing but the thing.
Bzzt, flake-out.
It's not pretty the act of making lies.
On 11/16/2024 02:46 PM, Jim Burns wrote:
On 11/16/2024 5:31 PM, Ross Finlayson wrote:
On 11/16/2024 12:29 PM, Jim Burns wrote:
On 11/16/2024 12:07 PM, Ross Finlayson wrote:
On 11/16/2024 08:58 AM, Ross Finlayson wrote:
On 11/16/2024 02:22 AM, Jim Burns wrote:
On 11/15/2024 9:52 PM, Ross Finlayson wrote:
On 11/15/2024 02:37 PM, Jim Burns wrote:
On 11/15/2024 4:32 PM, Ross Finlayson wrote:
Ah, yet according to Mirimanoff,
there do not exist standard models of integers,
If it is true that
our domain of discourse is a model of ST+PQ
then it is true that
our domain of discourse holds a standard integer.model.
What is Mirimanoff's argument that
it doesn't exist?
Mirimanoff's? Russell's Paradox.
ST+PQ does not suffer from claiming
that the set of all non.self.membered sets
is self.membered or claiming it isn't.
I don't say "infinity" is an axiom
primarily because
"infinity" is not an axiom of ST+PQ
ST+PQ:
⎛ set {} exists
⎜ set x∪{y} exists
⎜ set.extensionality
⎜ plurality ⦃z:P(z)⦄ exists
⎝ plurality.extensionality
"Infinity exists" ==
"the minimal inductive plurality exists"
is a theorem of those axioms.
^- Fragment
No.
The minimal inductive plurality is
a standard model of the integers.
Let's recall an example geometrically of what's
so inductively and not so in the limit.
Take a circle and draw a diameter, then bisect
the diameter resulting diameters of common circles,
all sharing a common diameter, vertical, say.
Then, notice the length of the circle, is
same, as the sum of the lengths of the half-diameter
circles, their sum.
So, repeat his dividing ad infinitum. In the limit,
the length is that of the diameter, not the perimeter,
while inductively, it's the diameter.
Thusly, a clear example "not.first.false" being
"ultimately.untrue".
A finite sequence of claims, each claim of which
is true.or.not.first.false is
a finite sequence of claims, each claim of which
is true.
The reason that that's true is that
THE SEQUENCE OF CLAIMS is finite.
Whatever those CLAIMS refer to,
none of those CLAIMS are first.false.
(They're each not.first.false.)
Since none of those CLAIMS are first.false,
none of those CLAIMS are false.
(That sequence is finite.)
What those claims are ABOUT doesn't affect that.
For example,
being ABOUT an indefinite one of infinitely.many
doesn't affect that.
Discovering
a finite sequence of claims, each claim of which
is true.or.not.first.false
in which there IS an untrue claim
is akin to
counting the eggs in a carton and
discovering that, there, in that carton,
7 is NOT between 6 and 8.
There is a problem, but not with mathematics.
Then, with regards to your fragment,
...the minimal inductive plurality...
congratulations,
Thank you.
you have ignored Russell his paradox and so on
No.
Selecting axioms which do not suffer from claiming
that the set of all non.self.membered sets
is self.membered or claiming it isn't
is not ignoring Russel,
it is responding to Russell.
Russell points out that
_we do not want_ to claim
that the set of all non.self.membered sets
is self.membered or claiming it isn't.
We respond: Okay, we'll stop doing that.
and quite fully revived Frege and given yourself
a complete theory and consistent as it may be, and
can entirely ignore all of 20'th century mathematics.
It's small, .... Fragment
The minimal inductive plurality.
Big or small, that's the thing,
the whole thing, and nothing but the thing.
Bzzt, flake-out.
It's not pretty the act of making lies.
Tell me what you think is a lie:
On 11/16/2024 08:18 PM, Jim Burns wrote:
On 11/16/2024 7:17 PM, Ross Finlayson wrote:
On 11/16/2024 02:46 PM, Jim Burns wrote:
On 11/16/2024 5:31 PM, Ross Finlayson wrote:
On 11/16/2024 12:29 PM, Jim Burns wrote:
On 11/16/2024 12:07 PM, Ross Finlayson wrote:
On 11/16/2024 08:58 AM, Ross Finlayson wrote:
On 11/16/2024 02:22 AM, Jim Burns wrote:
On 11/15/2024 9:52 PM, Ross Finlayson wrote:
On 11/15/2024 02:37 PM, Jim Burns wrote:
On 11/15/2024 4:32 PM, Ross Finlayson wrote:
Ah, yet according to Mirimanoff,
there do not exist standard models of integers,
If it is true that
our domain of discourse is a model of ST+PQ
then it is true that
our domain of discourse holds a standard integer.model.
What is Mirimanoff's argument that
it doesn't exist?
Mirimanoff's? Russell's Paradox.
ST+PQ does not suffer from claiming
that the set of all non.self.membered sets
is self.membered or claiming it isn't.
I don't say "infinity" is an axiom
primarily because
"infinity" is not an axiom of ST+PQ
ST+PQ:
⎛ set {} exists
⎜ set x∪{y} exists
⎜ set.extensionality
⎜ plurality ⦃z:P(z)⦄ exists
⎝ plurality.extensionality
"Infinity exists" ==
"the minimal inductive plurality exists"
is a theorem of those axioms.
^- Fragment
No.
The minimal inductive plurality is
a standard model of the integers.
Let's recall an example geometrically of what's
so inductively and not so in the limit.
Take a circle and draw a diameter, then bisect
the diameter resulting diameters of common circles,
all sharing a common diameter, vertical, say.
Then, notice the length of the circle, is
same, as the sum of the lengths of the half-diameter
circles, their sum.
So, repeat his dividing ad infinitum. In the limit,
the length is that of the diameter, not the perimeter,
while inductively, it's the diameter.
Thusly, a clear example "not.first.false" being
"ultimately.untrue".
A finite sequence of claims, each claim of which
is true.or.not.first.false is
a finite sequence of claims, each claim of which
is true.
The reason that that's true is that
THE SEQUENCE OF CLAIMS is finite.
Whatever those CLAIMS refer to,
none of those CLAIMS are first.false.
(They're each not.first.false.)
Since none of those CLAIMS are first.false,
none of those CLAIMS are false.
(That sequence is finite.)
What those claims are ABOUT doesn't affect that.
For example,
being ABOUT an indefinite one of infinitely.many
doesn't affect that.
Discovering
a finite sequence of claims, each claim of which
is true.or.not.first.false
in which there IS an untrue claim
is akin to
counting the eggs in a carton and
discovering that, there, in that carton,
7 is NOT between 6 and 8.
There is a problem, but not with mathematics.
Then, with regards to your fragment,
...the minimal inductive plurality...
congratulations,
Thank you.
you have ignored Russell his paradox and so on
No.
Selecting axioms which do not suffer from claiming
that the set of all non.self.membered sets
is self.membered or claiming it isn't
is not ignoring Russel,
it is responding to Russell.
Russell points out that
_we do not want_ to claim
that the set of all non.self.membered sets
is self.membered or claiming it isn't.
We respond: Okay, we'll stop doing that.
and quite fully revived Frege and given yourself
a complete theory and consistent as it may be, and
can entirely ignore all of 20'th century mathematics.
It's small, .... Fragment
The minimal inductive plurality.
Big or small, that's the thing,
the whole thing, and nothing but the thing.
Bzzt, flake-out.
It's not pretty the act of making lies.
Tell me what you think is a lie:
Quote what I wrote which you think is a lie:
Well, it's among what you clipped because it was un-answerable,
On 11/16/2024 08:57 PM, Jim Burns wrote:
On 11/16/2024 11:35 PM, Ross Finlayson wrote:
On 11/16/2024 08:18 PM, Jim Burns wrote:
On 11/16/2024 7:17 PM, Ross Finlayson wrote:
On 11/16/2024 02:46 PM, Jim Burns wrote:
On 11/16/2024 5:31 PM, Ross Finlayson wrote:
On 11/16/2024 12:29 PM, Jim Burns wrote:
On 11/16/2024 12:07 PM, Ross Finlayson wrote:
you have ignored Russell his paradox and so on
No.
Selecting axioms which do not suffer from claiming
that the set of all non.self.membered sets
is self.membered or claiming it isn't
is not ignoring Russel,
it is responding to Russell.
Russell points out that
_we do not want_ to claim
that the set of all non.self.membered sets
is self.membered or claiming it isn't.
We respond: Okay, we'll stop doing that.
Bzzt, flake-out.
It's not pretty the act of making lies.
Quote what I wrote which you think is a lie.
"We respond"
On 11/16/2024 02:22 AM, Jim Burns wrote:
On 11/15/2024 9:52 PM, Ross Finlayson wrote:
On 11/15/2024 02:37 PM, Jim Burns wrote:
On 11/15/2024 4:32 PM, Ross Finlayson wrote:
Ah, yet according to Mirimanoff,
there do not exist standard models of integers,
If it is true that
our domain of discourse is a model of ST+PQ
then it is true that
our domain of discourse holds a standard integer.model.
What is Mirimanoff's argument that
it doesn't exist?
Mirimanoff's? Russell's Paradox.
ST+PQ does not suffer from claiming
that the set of all non.self.membered sets
is self.membered or claiming it isn't.
While I am at it,
ZFC does not suffer from claiming
that the set of all non.self.membered sets
is self.membered or claiming it isn't,
and
ordinal.theory=Well.Order
does not suffer from claiming
that the set of all non.self.membered sets
is self.membered or claiming it isn't.
On 11/16/2024 12:07 PM, Ross Finlayson wrote:
you have ignored Russell his paradox and so on
On 11/16/2024 10:11 PM, Ross Finlayson wrote:
On 11/16/2024 09:59 PM, Ross Finlayson wrote:
On 11/16/2024 09:56 PM, Jim Burns wrote:
⎛ The modern study of set theory was initiated by
⎜ Georg Cantor and Richard Dedekind in the 1870s.
⎜ However,
⎜ the discovery of paradoxes in naive set theory,
⎜ such as Russell's paradox,
⎜ led to the desire for
⎜ a more rigorous form of set theory
⎝ that was free of these paradoxes.
https://en.wikipedia.org/wiki/Zermelo%E2%80%93Fraenkel_set_theory
How about Finsler and Boffa?
On 11/17/2024 10:35 AM, Jim Burns wrote:
how about Finsler or Boffa or Mirimanoff or
non.well.founded set theories?
Do they show that ST+PQ or ZFC or ordinals
suffer from Russell's {S:S∉S}?
No. They do not show that.
The less.interesting reason that they don't
is that
they are different domains of discourse.
⎛ 0 < 1/2 < 1 does not show that
⎜ there is an integer between 0 and 1
⎝ because 1/2 isn't an integer.
That less.interesting reason seems to
lie close to the heart of your objection.
You (RF) seem to not.believe that
things can be not.referred to.
In that respect,
I don't see what I can do for you.
I will continue to not.refer to
what I choose to not.refer to.
You (RF) seem to argue that
☠⎛ they cannot not.refer to Russell's {S:S∉S}
☠⎜ and therefore they ARE talking gibberish
☠⎝ and a standard model of the integers not.exists.
☠( and anyone who disagrees with that is a liar.
It is so that
that is what I argue, for, yes.
On 11/18/2024 07:46 AM, Jim Burns wrote:
I plan to turn to your argument
once we have finished with
⎛ A FINITE SEQUENCE OF CLAIMS, each claim of which
⎜ is true.or.not.first.false is
⎜ a FINITE SEQUENCE OF CLAIMS, each claim of which
⎝ is true.
What do you have to say about that, Ross?
I already did you keep clipping it.
Why don't you look back about the last three posts
and see an example where an inductive argument FAILS
and is nowhere finitely "not.first.false",
that it yet FAILS.
So, it's a counterexample,
and illustrates why what's not.first.false must
also be not.ultimately.untrue to not FAIL.
Then about Russell's retro-thesis and
On 11/18/2024 05:45 PM, Ross Finlayson wrote:
On 11/18/2024 04:56 PM, Jim Burns wrote:
On 11/18/2024 12:59 PM, Ross Finlayson wrote:
On 11/18/2024 07:46 AM, Jim Burns wrote:
I plan to turn to your argument
once we have finished with
⎛ A FINITE SEQUENCE OF CLAIMS, each claim of which
⎜ is true.or.not.first.false is
⎜ a FINITE SEQUENCE OF CLAIMS, each claim of which
⎝ is true.
What do you have to say about that, Ross?
I already did you keep clipping it.
Why don't you look back about the last three posts
and see an example where an inductive argument FAILS
and is nowhere finitely "not.first.false",
that it yet FAILS.
So, it's a counterexample,
and illustrates why what's not.first.false must
also be not.ultimately.untrue to not FAIL.
Then about Russell's retro-thesis and
First things first.
Are there non.well.ordered finite sequences, Ross?
Giving an ordinal assignment as "being", cardinals,
results then there are those among CH and not CH,
that would result contradictory models,
and not even necessarily considering Cohen's forcing,
and that he takes an ordinal out of a well-ordering,
since you asked for example of out-of-order ordinals.
Giving an assignment of reals
On 11/19/2024 05:15 AM, Jim Burns wrote:
On 11/18/2024 11:39 PM, Ross Finlayson wrote:
On 11/18/2024 05:45 PM, Ross Finlayson wrote:
On 11/18/2024 04:56 PM, Jim Burns wrote:
First things first.
Are there non.well.ordered finite sequences, Ross?
Of course I won't address your response.
You're insisting on speaking a language
which sounds misleadingly like mine,
misleadingly like, for example, Cohen's, too.
Why would you say, "misleadingly"?
On 11/19/2024 09:50 AM, Ross Finlayson wrote:
On 11/19/2024 09:41 AM, Jim Burns wrote:
On 11/19/2024 8:30 AM, Ross Finlayson wrote:
On 11/19/2024 05:15 AM, Jim Burns wrote:
On 11/18/2024 11:39 PM, Ross Finlayson wrote:
On 11/18/2024 05:45 PM, Ross Finlayson wrote:
On 11/18/2024 04:56 PM, Jim Burns wrote:
First things first.
Are there non.well.ordered finite sequences, Ross?
Of course I won't address your response.
You're insisting on speaking a language
which sounds misleadingly like mine,
misleadingly like, for example, Cohen's, too.
Why would you say, "misleadingly"?
Are there non.well.ordered finite sequences, Ross?
How about a stoplight?
In the temporal, and modal, and finite, thus fixed,
there's a well-ordering of that.
In the quasi-modal of whatever variety: there is not.
What I'm saying is that you can take back
that "not.first.false" guarantees a claim
for inference, when it doesn't.
"Broadly" speaking, ....
Earlier you disputed that "not.first.false"
and "not.ultimately.untrue" were any different.
Now you don't.
The "bait-and-switch" and "back-slide"
don't go well together.
In either order, ....
On 11/19/2024 11:56 AM, Jim Burns wrote:
On 11/19/2024 12:52 PM, Ross Finlayson wrote:
The "bait-and-switch" and "back-slide"
don't go well together.
In either order, ....
⎛ Necessary and sufficient conditions for finiteness
⎜
⎜ 3. (Paul Stäckel)
⎜ S can be given a total ordering which is
⎜ well-ordered both forwards and backwards.
⎜ That is, every non-empty subset of S has both
⎝ a least and a greatest element in the subset.
https://en.wikipedia.org/wiki/Finite_set
Yeah we looked at that before also,
and I wrote another, different, definition of finite.
Jim Burns was thinking very hard :
On 11/19/2024 4:38 PM, Ross Finlayson wrote:
On 11/19/2024 11:56 AM, Jim Burns wrote:
On 11/19/2024 12:52 PM, Ross Finlayson wrote:
The "bait-and-switch" and "back-slide"
don't go well together.
In either order, ....
⎛ Necessary and sufficient conditions for finiteness
⎜
⎜ 3. (Paul Stäckel)
⎜ S can be given a total ordering which is
⎜ well-ordered both forwards and backwards.
⎜ That is, every non-empty subset of S has both
⎝ a least and a greatest element in the subset.
https://en.wikipedia.org/wiki/Finite_set
Yeah we looked at that before also,
and I wrote another, different, definition of finite.
Thank you for admitting that.
However, you (RF) might NOT be bait.and.switch.ing
if the definitions are equivalent.
What was the definition you wrote before?
I didn't see it in the rest of your post.
Didn't he use not.ultimately.untrue
instead of not.first.false?
Is that just an inversion?
It seems to imply an ultimate or last
instead of a first or least.
Just like WM
when he inverted the naturals
to unit fractions.
On 11/19/2024 02:18 PM, Jim Burns wrote:
On 11/19/2024 4:38 PM, Ross Finlayson wrote:
On 11/19/2024 11:56 AM, Jim Burns wrote:
On 11/19/2024 12:52 PM, Ross Finlayson wrote:
The "bait-and-switch" and "back-slide"
don't go well together.
In either order, ....
⎛ Necessary and sufficient conditions for finiteness
⎜
⎜ 3. (Paul Stäckel)
⎜ S can be given a total ordering which is
⎜ well-ordered both forwards and backwards.
⎜ That is, every non-empty subset of S has both
⎝ a least and a greatest element in the subset.
https://en.wikipedia.org/wiki/Finite_set
Yeah we looked at that before also,
and I wrote another, different, definition of finite.
Thank you for admitting that.
However, you (RF) might NOT be bait.and.switch.ing
if the definitions are equivalent.
What was the definition you wrote before?
I didn't see it in the rest of your post.
Oh, no, it was a definition of finite good for both.
On 11/19/2024 4:38 PM, Ross Finlayson wrote:
On 11/19/2024 11:56 AM, Jim Burns wrote:
⎛ Necessary and sufficient conditions for finiteness
⎜
⎜ 3. (Paul Stäckel)
⎜ S can be given a total ordering which is
⎜ well-ordered both forwards and backwards.
⎜ That is, every non-empty subset of S has both
⎝ a least and a greatest element in the subset.
https://en.wikipedia.org/wiki/Finite_set
Yeah we looked at that before also, and
I wrote another, different, definition of finite.
On 11/19/2024 07:45 PM, Ross Finlayson wrote:
On 11/19/2024 02:38 PM, FromTheRafters wrote:
[...]
In
"Replacement of Cardinality (infinite middle)", 8/19 2024,
this was:
I mean it's a great definition that finite has that
there exists a normal ordering that's a well-ordering
and that all the orderings of the set are well-orderings.
That's a great definition of finite and now it stands
for itself in enduring mathematical definition in defense.
Why is it you think that Stackel's definition of finite
and "not Dedekind's definition of countably infinite"
don't agree?
On 11/20/2024 11:19 AM, Jim Burns wrote:
I am satisfied that using the other definition
which you mentioned isn't bait.and.switch.ing.
Dedekind.finite with countable.choice is
equivalent to Stäckel.finite.
⎛ Countable.choice:
⎜ ∃S: ℕ→Collection: ∀k∈ℕ:S(k)≠{} ⇒
⎝ ∃ch: ℕ→⋃Collection: ∀k∈ℕ:ch(k)∈S(k)
Therefore,
if
P is a finite sequence of claims, each claim of which
is true.or.not.first.false,
then
P is a finite sequence of claims, each claim of which
is true.
That conclusion is the telescope which
finite beings use to observe the infinite,
because
each claim in finite.length.P is true whether.or.not
it is a claim referring to one of infinitely.many.
I am satisfied that using the other definition
which you mentioned isn't bait.and.switch.ing.
Are you, though?
With regards to choice and countable choice,
the weaker form that goes without saying anyways,
the existence of a choice function being [implies?]
a bijection [between?] any given set, and
an [at least one?] ordinal's elements lesser ordinals,
making a well-ordering of the set,
has that,
well-foundedness
and
well-ordering
sort of result dis-agreement.
On 11/21/2024 09:57 AM, Jim Burns wrote:
[...]
(the existence of a choice function,
i.e. a bijection between any set and some ordinal)
Yeah, Well-Ordering and Choice (the existence of
a choice function, i.e. a bijection between any
set and some ordinal) are same.
Countable-choice is weak and trivial.
On 11/27/2024 6:04 AM, WM wrote:
However,
one makes a quantifier shift, unreliable,
to go from that to
⛔⎛ there is an end segment such that
⛔⎜ for each number (finite cardinal)
⛔⎝ the number isn't in the end segment.
Each end.segment is infinite.
Their intersection of all is empty.
These claims do not conflict.
In an infinite endsegment
numbers are remaining.
In many infinite endsegments infinitely many numbers are the same.
And the intersection of all,
which isn't any end.segment,
is empty.
On 27.11.2024 16:57, Jim Burns wrote:Yes.
On 11/27/2024 6:04 AM, WM wrote:
However,
one makes a quantifier shift, unreliable, to go from that to ⛔⎛ there
is an end segment such that ⛔⎜ for each number (finite cardinal) ⛔⎝ the
number isn't in the end segment.
If all endsegments are infinite then infinitely
many natbumbers remain in all endsegments.
Infinite endsegments with an empty intersection are excluded byNot by intersecting infinitely many segments.
inclusion monotony.
Because that would mean infinitely many differentWeird way to put it.
numbers in infinite endsegments.
That is true.Each end.segment is infinite.That means it has infinitely many numbers in common with every other
infinite endsegment. If not, then there is an infinite endsegment with infinitely many numbers but not with infinitely many numbers in common
with other infinite endsegments. Contradiction by inclusion monotony.
They are infinite, they don't need to gain elements.Their intersection of all is empty. These claims do not conflict.It conflicts with the fact, that the endsegments can lose elements but
never gain elements.
I.e. never.Wrong. Up to every endsegment the intersection is this endsegment. Up to every infinite endsegment the intersection is infinite. This cannotIn an infinite endsegment numbers are remaining.And the intersection of all, which isn't any end.segment, is empty.
In many infinite endsegments infinitely many numbers are the same.
change as long as infinite endsegments exist.
Am Wed, 27 Nov 2024 20:59:43 +0100 schrieb WM:
Infinite endsegments with an empty intersection are excluded byNot by intersecting infinitely many segments.
inclusion monotony.
That is true.Each end.segment is infinite.That means it has infinitely many numbers in common with every other
infinite endsegment. If not, then there is an infinite endsegment with
infinitely many numbers but not with infinitely many numbers in common
with other infinite endsegments. Contradiction by inclusion monotony.
They are infinite, they don't need to gain elements.Their intersection of all is empty. These claims do not conflict.It conflicts with the fact, that the endsegments can lose elements but
never gain elements.
I.e. never.Wrong. Up to every endsegment the intersection is this endsegment. Up toIn an infinite endsegment numbers are remaining.And the intersection of all, which isn't any end.segment, is empty.
In many infinite endsegments infinitely many numbers are the same.
every infinite endsegment the intersection is infinite. This cannot
change as long as infinite endsegments exist.
On 27.11.2024 16:57, Jim Burns wrote:
On 11/27/2024 6:04 AM, WM wrote:
However,
one makes a quantifier shift, unreliable,
to go from that to
⛔⎛ there is an end segment such that
⛔⎜ for each number (finite cardinal)
⛔⎝ the number isn't in the end segment.
Don't blather nonsense.
If all endsegments are infinite
then infinitely many natbumbers
remain in all endsegments.
Consider the sequence of claims.
⎛⎛ [∀∃] for each end.segment
⎜⎜ there is an infinite set such that
⎜⎝ the infinite set subsets the end.segment
⎜
⎜⎛ [∃∀] there is an infinite set such that
⎜⎜ for each end.segment
⎝⎝ the infinite set subsets the end.segment
We cannot SEE,
just by looking at the claims,
that, after [∀∃], [∃∀] is not.first.false.
Am Thu, 28 Nov 2024 11:39:05 +0100 schrieb WM:
A simpler arguments is this: All endsegments are in a decreasingThere is no decrease, they are all infinite.
sequence.
Before the decrease has reached finite endsegments, all areAll segments are infinite. Nothing can come "afterwards".
infinite and share an infinite contents from E(1) = ℕ on. They have not
yet had the chance to reduce their infinite subset below infinity.
On 28.11.2024 09:34, Jim Burns wrote:Naturally.
Consider the sequence of claims.and its predecessors!
⎛⎛ [∀∃] for each end.segment ⎜⎜ there is an infinite set such that ⎜⎝
the infinite set subsets the end.segment
If each endsegment is infinite, then this is valid for each endsegmentThat is what "every" means.
with no exception. because all are predecessors of an infinite
endsegment. That means it is valid for all endsegments.
The trick here is that the infinite set has no specified natural number (because all fall out at some endsegment) but it is infinite without any other specification.Yes. You can call it omega or N.
Uh, that is wrong.⎜⎛ [∃∀] there is an infinite set such that ⎜⎜ for each end.segment ⎝⎝
the infinite set subsets the end.segment
We cannot SEE,
just by looking at the claims,
that, after [∀∃], [∃∀] is not.first.false.
I have proved above that [∃∀] is true for all infinite endsegments.
A simpler arguments is this: All endsegments are in a decreasingThere is no decrease, they are all infinite.
sequence.
Before the decrease has reached finite endsegments, all areAll segments are infinite. Nothing can come "afterwards".
infinite and share an infinite contents from E(1) = ℕ on. They have not
yet had the chance to reduce their infinite subset below infinity.
Am Thu, 28 Nov 2024 18:09:16 +0100 schrieb WM:
On 28.11.2024 17:45, joes wrote:It is called a subset. It is still infinite
Am Thu, 28 Nov 2024 11:39:05 +0100 schrieb WM:Every endsegment has one number less than its predecessor.
A simpler arguments is this: All endsegments are in a decreasingThere is no decrease, they are all infinite.
sequence.
That is called decrease.
No finite intersection anyway.Then the intersection is never empty.Before the decrease has reached finite endsegments, all are infiniteAll segments are infinite. Nothing can come "afterwards".
and share an infinite contents from E(1) = ℕ on. They have not yet had >>>> the chance to reduce their infinite subset below infinity.
On 28.11.2024 17:45, joes wrote:It is called a subset. It is still infinite
Am Thu, 28 Nov 2024 11:39:05 +0100 schrieb WM:
Every endsegment has one number less than its predecessor.A simpler arguments is this: All endsegments are in a decreasingThere is no decrease, they are all infinite.
sequence.
That is called decrease.
No finite intersection anyway.Then the intersection is never empty.Before the decrease has reached finite endsegments, all are infiniteAll segments are infinite. Nothing can come "afterwards".
and share an infinite contents from E(1) = ℕ on. They have not yet had >>> the chance to reduce their infinite subset below infinity.
On 28.11.2024 18:36, joes wrote:Trademark ambiguous phrasing.
Am Thu, 28 Nov 2024 18:09:16 +0100 schrieb WM:Yes this decrease produces subsets. All infinite subsets produce
On 28.11.2024 17:45, joes wrote:It is called a subset. It is still infinite
Am Thu, 28 Nov 2024 11:39:05 +0100 schrieb WM:Every endsegment has one number less than its predecessor.
A simpler arguments is this: All endsegments are in a decreasingThere is no decrease, they are all infinite.
sequence.
That is called decrease.
infinite intersections.
The infinite intersection is empty.No intersection of infinite endsegments is finite.No finite intersection anyway.Then the intersection is never empty.Before the decrease has reached finite endsegments, all are infinite >>>>> and share an infinite contents from E(1) = ℕ on. They have not yet >>>>> had the chance to reduce their infinite subset below infinity.All segments are infinite. Nothing can come "afterwards".
Every infinite endsegments has an infinite intersection with all its predecessors.Itself, yes.
If all endsegments are infinite, then this holds for allIt does. But it does not for the union of all of them - N itself,
endsegments. They simply had not the chance to lose these numbers.
WM used his keyboard to write :
On 28.11.2024 17:45, joes wrote:
Am Thu, 28 Nov 2024 11:39:05 +0100 schrieb WM:
A simpler arguments is this: All endsegments are in a decreasingThere is no decrease, they are all infinite.
sequence.
Every endsegment has one number less than its predecessor.
That is called decrease.
More like the subset relation. It is not a decrease in cardinality.
Am Thu, 28 Nov 2024 18:46:21 +0100 schrieb WM:
The intersection of all infinitely many segments is empty.
The infinite intersection is empty.
Every infinite endsegments has an infinite intersection with all itsItself, yes.
predecessors.
On 28.11.2024 20:28, FromTheRafters wrote:That long you haven't taken all segments yet.
WM used his keyboard to write :Of course not. Cardinality is nothing else than infinitely many.
On 28.11.2024 17:45, joes wrote:More like the subset relation. It is not a decrease in cardinality.
Am Thu, 28 Nov 2024 11:39:05 +0100 schrieb WM:Every endsegment has one number less than its predecessor.
A simpler arguments is this: All endsegments are in a decreasingThere is no decrease, they are all infinite.
sequence.
That is called decrease.
But as long as infinitely many natnumbers have not left the endsegments,
they stay inside all of them.
And many are the same for all endsegments.Only for every finite number.
Am Thu, 28 Nov 2024 21:20:10 +0100 schrieb WM:
On 28.11.2024 20:28, FromTheRafters wrote:That long you haven't taken all segments yet.
WM used his keyboard to write :Of course not. Cardinality is nothing else than infinitely many.
On 28.11.2024 17:45, joes wrote:More like the subset relation. It is not a decrease in cardinality.
Am Thu, 28 Nov 2024 11:39:05 +0100 schrieb WM:Every endsegment has one number less than its predecessor.
A simpler arguments is this: All endsegments are in a decreasingThere is no decrease, they are all infinite.
sequence.
That is called decrease.
But as long as infinitely many natnumbers have not left the endsegments,
they stay inside all of them.
And many are the same for all endsegments.Only for every finite number.
WM expressed precisely :
But as long as infinitely many natnumbers have not left the
endsegments, they stay inside all of them. And many are the same for
all endsegments. Therefore the intersection of infinite endsegments is
infinite.
Natural numbers don't "leave", sets don't change.
You don't 'run out of
indices' or elements to index.
On 28.11.2024 09:34, Jim Burns wrote:
Consider the sequence of claims.
⎛⎛ [∀∃] for each end.segment
⎜⎜ there is an infinite set such that
⎜⎝ the infinite set subsets the end.segment
and its predecessors!
If each endsegment is infinite,
then this is valid
for each endsegment with no exception
because all are
predecessors of an infinite endsegment.
That means it is valid for all endsegments.
The trick here is that
the infinite set has no specified natural number
(because all fall out at some endsegment)
but it is infinite
without any other specification.
WM was thinking very hard :
On 29.11.2024 09:54, FromTheRafters wrote:
WM expressed precisely :
But as long as infinitely many natnumbers have not left the
endsegments, they stay inside all of them. And many are the same for
all endsegments. Therefore the intersection of infinite endsegments
is infinite.
Natural numbers don't "leave", sets don't change.
Call it as you like. Fact is that the function of endsegments is
losing elements. The limit is the empty endsegment.
Your sequence of endsegments (which are each countably infinte) is
indeed losing an element of N with each iteration. Losing an element is
not the same as reducing an infinite set's size though.
Yes,
if each end.segment is infinite
then each end.segment is infinite.
because all are
predecessors of an infinite endsegment.
That means it is valid for all endsegments.
The trick here is that
the infinite set has no specified natural number (because all fall out
at some endsegment)
all fall out == empty
but it is infinite
infinite and empty
On 29.11.2024 19:53, Jim Burns wrote:
Yes,
if each end.segment is infinite
then each end.segment is infinite.
And their intersection is infinite.
because all are
predecessors of an infinite endsegment.
That means it is valid for all endsegments.
The trick here is that
the infinite set has no specified natural number
(because all fall out at some endsegment)
all fall out == empty
== empty endsegment.
but it is infinite
infinite and empty
According to set theory.
Set theory is wrong.
On 11/29/2024 2:41 PM, WM wrote:
all fall out == empty
== empty endsegment.
No.
Empty intersection of all and only infinite end.segments.
infinite and empty
According to set theory.
You imagine that's what set theory says.
It doesn't say that.
On 29.11.2024 21:37, Jim Burns wrote:
On 11/29/2024 2:41 PM, WM wrote:
On 29.11.2024 19:53, Jim Burns wrote:
all fall out == empty
== empty endsegment.
No.
Empty intersection of all and only infinite end.segments.
All natnumbers fall out of the endsegmets
but the endsegments keep infinitely many natnumbers.
Can a stronger nonsense exist?
infinite and empty
According to set theory.
You imagine that's what set theory says.
It doesn't say that.
It does say that
infinite endsegments have an empty intersection
because
all natnumbers fall out of the endsegments.
Where else could they fall out?
The end.segments are infinite.
Their intersection is empty.
Nothing is infinite and empty.
WM wrote on 11/29/2024 :
The size of the intersection remains infinite as long as all
endsegments remain infinite (= as long as only infinite endsegments
are considered).
Endsegments are defined as infinite,
all of them and
each and every one of them.
The intersection is empty.
WM explained :
On 29.11.2024 22:50, FromTheRafters wrote:
WM wrote on 11/29/2024 :
The size of the intersection remains infinite as long as all
endsegments remain infinite (= as long as only infinite endsegments
are considered).
Endsegments are defined as infinite,
Endsegments are defined as endsegments. They have been defined by
myself many years ago.
As what is left after not considering a finite initial segment in your
new set and considering only the tail of the sequence.
Almost all
elements are considered in the new set, which means all endsegments are infinite.
Try to understand inclusion monotony. The sequence of endsegments
decreases.
In what manner are they decreasing?
When you filter out the FISON, the
rest, the tail, as a set, stays the same size of aleph_zero.
As long as it has not decreased below ℵo elements, the intersection
has not decreased below ℵo elements.
It doesn't decrease in size at all.
Finite sets versus infinite sets. Finite ordered sets have a last
element which can be in the intersection of all previously considered
finite sets. Infinite ordered sets have no such last element.
On 30.11.2024 12:54, FromTheRafters wrote:For an intersection, the "smallest" set matters, which there isn't
Finite sets versus infinite sets. Finite ordered sets have a lastBut they have infinitely many elements which contribute to the
element which can be in the intersection of all previously considered
finite sets. Infinite ordered sets have no such last element.
intersection.
The intersection of the "finite initial segment" of endsegments isIt does for all finite k.
∩{E(1), E(2), ..., E(k)} = E(k)
is a function which remains infinite for all infinite endsegments. If
all endsegments remain infinite forever, then this function remains
infinite forever.
Am Sat, 30 Nov 2024 13:19:46 +0100 schrieb WM:
On 30.11.2024 12:54, FromTheRafters wrote:For an intersection, the "smallest" set matters, which there isn't
Finite sets versus infinite sets. Finite ordered sets have a lastBut they have infinitely many elements which contribute to the
element which can be in the intersection of all previously considered
finite sets. Infinite ordered sets have no such last element.
intersection.
in this infinite sequence, only a "biggest".
The intersection of the "finite initial segment" of endsegments isIt does for all finite k.
∩{E(1), E(2), ..., E(k)} = E(k)
is a function which remains infinite for all infinite endsegments. If
all endsegments remain infinite forever, then this function remains
infinite forever.
On 30.11.2024 16:58, joes wrote:True. All endsegments are infinite. But they form a chain of inclusion,
Am Sat, 30 Nov 2024 13:19:46 +0100 schrieb WM:If all sets are infinite, then there is no smaller set than an infinite
On 30.11.2024 12:54, FromTheRafters wrote:For an intersection, the "smallest" set matters, which there isn't in
Finite sets versus infinite sets. Finite ordered sets have a lastBut they have infinitely many elements which contribute to the
element which can be in the intersection of all previously considered
finite sets. Infinite ordered sets have no such last element.
intersection.
this infinite sequence, only a "biggest".
set.
All natural k are finite.Of course. Only for finite k the endsegments are infinite.The intersection of the "finite initial segment" of endsegments isIt does for all finite k.
∩{E(1), E(2), ..., E(k)} = E(k)
is a function which remains infinite for all infinite endsegments. If
all endsegments remain infinite forever, then this function remains
infinite forever.
Am Sat, 30 Nov 2024 18:20:51 +0100 schrieb WM:
True. All endsegments are infinite. But they form a chain of inclusion,For an intersection, the "smallest" set matters, which there isn't inIf all sets are infinite, then there is no smaller set than an infinite
this infinite sequence, only a "biggest".
set.
and there is no smallest set, because that chain is infinite.
All natural k are finite.Of course. Only for finite k the endsegments are infinite.The intersection of the "finite initial segment" of endsegments isIt does for all finite k.
∩{E(1), E(2), ..., E(k)} = E(k)
is a function which remains infinite for all infinite endsegments. If
all endsegments remain infinite forever, then this function remains
infinite forever.
On 30.11.2024 18:45, joes wrote:How surprising.
Am Sat, 30 Nov 2024 18:20:51 +0100 schrieb WM:
There is an infinite sequence of endsegments E(1), E(2), E(3), ... andTrue. All endsegments are infinite. But they form a chain of inclusion,For an intersection, the "smallest" set matters, which there isn't inIf all sets are infinite, then there is no smaller set than an
this infinite sequence, only a "biggest".
infinite set.
and there is no smallest set, because that chain is infinite.
an infinite sequence of their intersections E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ... .
Both are identical - from the first endsegment on until every existing endsegment.
...for every natural (which are finite), but not for the limit.Then all endsegments are infinite like their intersections.All natural k are finite.Of course. Only for finite k the endsegments are infinite.The intersection of the "finite initial segment" of endsegments isIt does for all finite k.
∩{E(1), E(2), ..., E(k)} = E(k)
is a function which remains infinite for all infinite endsegments.
If all endsegments remain infinite forever, then this function
remains infinite forever.
Am Sat, 30 Nov 2024 20:10:49 +0100 schrieb WM:
There is an infinite sequence of endsegments E(1), E(2), E(3), ... andHow surprising.
an infinite sequence of their intersections E(1), E(1)∩E(2),
E(1)∩E(2)∩E(3), ... .
Both are identical - from the first endsegment on until every existing
endsegment.
...for every natural (which are finite), but not for the limit.Then all endsegments are infinite like their intersections.Of course. Only for finite k the endsegments are infinite.All natural k are finite.
On 11/30/2024 3:12 AM, WM wrote:
Finite initial segment[s]: F(n) = {1, 2, 3, ..., n} (n e IN).
Endsegment[s]: E(n) = {n, n+1, n+2, ...} (n e IN).
The sequence of endsegments decreases. (WM)
In what manner [is it] decreasing?
The [endegmanets] are losing elements, one after the other:
∀k ∈ ℕ : E(k+1) = E(k) \ {k}
But each endsegment has only one element less than its predecessor.
When you filter out the FISON, the rest, the tail, as a set, stays
the same size of aleph_zero.
On 11/30/2024 3:12 AM, WM wrote:
Finite initial segment[s]: F(n) = {1, 2, 3, ..., n} (n e IN).
Endsegment[s]: E(n) = {n, n+1, n+2, ...} (n e IN).
The sequence of endsegments decreases. (WM)
In what manner [is it] decreasing?
The [endegmanets] are losing elements, one after the other:
∀k ∈ ℕ : E(k+1) = E(k) \ {k}
But each endsegment has only one element less than its predecessor.
When you filter out the FISON, the rest, the tail, as a set, stays
the same size of aleph_zero.
On 11/30/2024 3:12 AM, WM wrote:
Finite initial segment[s]: F(n) = {1, 2, 3, ..., n} (n e IN)
Finite?
On 11/30/2024 3:12 AM, WM wrote:
Finite initial segment[s]: F(n) = {1, 2, 3, ..., n} (n e IN)
Finite?
Am 02.12.2024 um 00:11 schrieb Chris M. Thomasson:
On 11/30/2024 3:12 AM, WM wrote:
FISON(s)
Finite initial segment[s]: F(n) = {1, 2, 3, ..., n} (n e IN)
You see it's a DEFINITION:
F(n) = {1, 2, 3, ..., n} (n e IN) .
This "means" (implies): F(1) (i.e. {1}) is a FISON, F(2) (i.e. {1, 2}) is a FISON,
F(3) (i.e. {1, 2, 3}) is a FISION, and so on (ad infinitum). F(1), F(2), F(3), ... are FISONs.
Finite?
Yeah, finite. For each and eveer n e IN F(n) (i.e. {1, 2, 3, ..., n}) is finite (i.e. a finite set).
Huh? The natural numbers don't stop at n! WTF!!!
No one (except possibly Mückenheim) said they did.
Hint: There are _infinitely many_ finite initial segments (one for each
and every natural number n). :-)
Am 02.12.2024 um 00:11 schrieb Chris M. Thomasson:
On 11/30/2024 3:12 AM, WM wrote:
FISON(s)
Finite initial segment[s]: F(n) = {1, 2, 3, ..., n} (n e IN)
You see it's a DEFINITION:
F(n) = {1, 2, 3, ..., n} (n e IN) .
This "means" (implies): F(1) is a FISON (i.e. {1}), F(2) is a FISON
(i.e. {1, 2}), F(3) is a FISION (i.e. {1, 2, 3}), and so on (ad
infinitum). F(1), F(2), F(3), ... are FISONs.
Finite?
Yeah, finite. For each and eveer n e IN F(n) (i.e. {1, 2, 3, ..., n}) is finite (i.e. a finite set).
Huh? The natural numbers don't stop at n! WTF!!!
No one (except possibly Mückenheim) said they did.
Hint: There are _infinitely many_ finite initial segments (one for each
and every natural number n). :-)
On 02.12.2024 12:53, FromTheRafters wrote:They ARE infinite sets.
Infinite endsegments contain an infinite set eachThis is his definition of endsegment, which as almost anyone can see,Endsegment E(n) = {n, n+1, n+2, ...}
has no last element, so yes it is infinite. He says 'infinite
endsegment' as if there were a choice, only to add confusion.
infinitely many elements of which are in the intersection.The intersection of all infinite segments?
An empty intersection cannotWhich happens only in the limit.
come before an empty endsegment has been produced by losing one element
at every step.
E(1), E(2), E(3), ...What do you reckon the limit is?
and E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
are identical for every n and in the limit because E(1)∩E(2)∩...∩E(n) = E(n).
Endsegment E(n) = {n, n+1, n+2, ...}
This is his definition of endsegment, which as almost anyone can see,
has no last element, so yes it is infinite. He says 'infinite
endsegment' as if there were a choice, only to add confusion.
WM wrote :
E(1), E(2), E(3), ...
and
E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
are identical for every n and in the limit because
E(1)∩E(2)∩...∩E(n) = E(n).
Non sequitur. That which is true for finite sequences is not necessarily
true for infinite sequences.
Am Mon, 02 Dec 2024 15:28:30 +0100 schrieb WM:
On 02.12.2024 12:53, FromTheRafters wrote:They ARE infinite sets.
Infinite endsegments contain an infinite set eachThis is his definition of endsegment, which as almost anyone can see,Endsegment E(n) = {n, n+1, n+2, ...}
has no last element, so yes it is infinite. He says 'infinite
endsegment' as if there were a choice, only to add confusion.
infinitely many elements of which are in the intersection.The intersection of all infinite segments?
An empty intersection cannotWhich happens only in the limit.
come before an empty endsegment has been produced by losing one element
at every step.
E(1), E(2), E(3), ...
and E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
are identical for every n and in the limit because E(1)∩E(2)∩...∩E(n) =
E(n).
What do you reckon the limit is?
On 02.12.2024 16:46, FromTheRafters wrote:What is the limit?
WM wrote :
As easily can be obtaied from the above it is necessarily true that upE(1), E(2), E(3), ...Non sequitur. That which is true for finite sequences is not
and E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
are identical for every n and in the limit because E(1)∩E(2)∩...∩E(n) >>> = E(n).
necessarily true for infinite sequences.
to every term and therefore also in the limit the sequences of
endsegments and of intersections are identical. Every contrary opinion
is matheology, outside of mathematics.
On 02.12.2024 15:52, joes wrote:No, they contain numbers.
Am Mon, 02 Dec 2024 15:28:30 +0100 schrieb WM:Nevertheless they also contain an infinite set
On 02.12.2024 12:53, FromTheRafters wrote:They ARE infinite sets.
Infinite endsegments contain an infinite set eachThis is his definition of endsegment, which as almost anyone can see,Endsegment E(n) = {n, n+1, n+2, ...}
has no last element, so yes it is infinite. He says 'infinite
endsegment' as if there were a choice, only to add confusion.
Ah, no. The intersection is infinite only for a finite number of segments.So it is.infinitely many elements of which are in the intersection.The intersection of all infinite segments?
It is the empty set.Whatever the limit is, it is the same for the sequence of endsegmentsAn empty intersection cannot come before an empty endsegment has beenWhich happens only in the limit.
produced by losing one element at every step.
E(1), E(2), E(3), ...What do you reckon the limit is?
and E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
are identical for every n and in the limit because E(1)∩E(2)∩...∩E(n) >>> = E(n).
and the sequence of intersections.
Chris M. Thomasson brought next idea :
On 11/30/2024 3:12 AM, WM wrote:
Endsegment E(n) = {n, n+1, n+2, ...}
This is his definition of endsegment, which
as almost anyone can see,
has no last element, so
yes it is infinite.
He says 'infinite endsegment' as if
there were a choice,
only to add confusion.
On 12/1/2024 9:50 PM, Moebius wrote:
Am 02.12.2024 um 00:11 schrieb Chris M. Thomasson:[...]
On 11/30/2024 3:12 AM, WM wrote:
Finite initial segment[s]: F(n) = {1, 2, 3, ..., n} (n e IN).
When WM writes:
{1, 2, 3, ..., n}
I think he might mean that n is somehow a largest natural number?
On 12/2/2024 3:56 PM, Moebius wrote:
Am 03.12.2024 um 00:51 schrieb Chris M. Thomasson:
On 12/1/2024 9:50 PM, Moebius wrote:
Am 02.12.2024 um 00:11 schrieb Chris M. Thomasson:[...]
On 11/30/2024 3:12 AM, WM wrote:
Finite initial segment[s]: F(n) = {1, 2, 3, ..., n} (n e IN).
When WM writes:
{1, 2, 3, ..., n}
I think he might mean that n is somehow a largest natural number?
Nope, he just means some n e IN.
So if n = 5, the FISON is:
{ 1, 2, 3, 4, 5 }
n = 3
{ 1, 2, 3 }
Right?
On 02.12.2024 12:53, FromTheRafters wrote:
[...]
Infinite endsegments contain an infinite set each,
infinitely many elements of which
are in the intersection.
An empty intersection cannot come before
an empty endsegment has been produced by
losing one element at every step.
E(1), E(2), E(3), ...
and
E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
are identical for every n and in the limit
because
E(1)∩E(2)∩...∩E(n) = E(n).
An empty intersection cannot come before
an empty endsegment has been produced by
losing one element at every step.
On 12/1/2024 9:50 PM, Moebius wrote:
Am 02.12.2024 um 00:11 schrieb Chris M. Thomasson:
On 11/30/2024 3:12 AM, WM wrote:
[...]Finite initial segment[s]:
F(n) = {1, 2, 3, ..., n} (n e IN).
When WM writes:
{1, 2, 3, ..., n}
I think he might mean that
n is somehow a largest natural number?
On 12/2/2024 5:02 PM, Jim Burns wrote:
That theory implies that, on some level of his mind,
WM knows he is trolling us.
For some reason
a thought crossed my mind several times before.
Is WM trolling us?
He knows better and is not stupid at all!
;^o ? Humm...
Just a thought. I hope I am right. :^D
On 12/02/2024 04:32 PM, Jim Burns wrote:
On 12/2/2024 9:28 AM, WM wrote:
Infinite endsegments contain an infinite set each,
infinitely many elements of which
are in the intersection.
Yes to:
⎛ regarding finite.cardinals,
⎜ for each end.segment E(k)
⎜ there is a subset S such that
⎝ for each finite cardinal j, j < |S| ≤ |E(k)|
No to:
⛔⎛ regarding finite.cardinals,
⛔⎜ ⮣ there is a subset S such that ⮧
⛔⎜ ⮤ for each end.segment E(k) ⮠
⛔⎝ for each finite cardinal j, j < |S| ≤ |E(k)|
A quantifier shift tells you (WM) what you (WM) _expect_
but a quantifier shift is untrustworthy.
An empty intersection cannot come before
an empty endsegment has been produced by
losing one element at every step.
No.
Because see below. [redacted by JB]
The usual idea of wrestling with a pig is
that you both get dirty, and the pig likes it.
Quit letting that pig dirty things.
For some reason a thought crossed my mind several times before. Is WM trolling us? He knows better and is not stupid at all! ;^o ? Humm...
Just a thought. I hope I am right. :^D
On 12/2/2024 4:00 PM, Chris M. Thomasson wrote:
On 12/2/2024 3:59 PM, Moebius wrote:
Am 03.12.2024 um 00:58 schrieb Chris M. Thomasson:
On 12/2/2024 3:56 PM, Moebius wrote:
Am 03.12.2024 um 00:51 schrieb Chris M. Thomasson:
On 12/1/2024 9:50 PM, Moebius wrote:
Am 02.12.2024 um 00:11 schrieb Chris M. Thomasson:
On 11/30/2024 3:12 AM, WM wrote:
Finite initial segment[s]: F(n) = {1, 2, 3, ..., n} (n e IN). >>>>>> [...]
When WM writes:
{1, 2, 3, ..., n}
I think he might mean that n is somehow a largest natural number?
Nope, n here may be any element in IN.
So if n = 5, the FISON is:
{ 1, 2, 3, 4, 5 }
[If} n = 3
{ 1, 2, 3 }
Right?
Right.
Thank you Moebius. :^)
So, if n = all_of_the_naturals, then
{ 1, 2, 3, ... }
What about {1, 2, 3, ..., n}, where n is taken to infinity? No limit?
What about {1, 2, 3, ..., n}, where n is taken to infinity? No limit?
{1, 2, 3, ..., n} when n is taken to infinity = the set of all natural numbers?
Am 03.12.2024 um 06:34 schrieb Chris M. Thomasson:
Then the following can be shown:
lim_(n->oo) {1, 2, 3, ..., n} = {1, 2, 3, ...} .
lim_(n->oo) {n, n+1, n+2, ...} = {} .
Hope this helps. :-P
.
.
.
Sometimes I like to think of the set of all natural numbers as an n-ary
tree, binary here, wrt zero as a main root, so to speak:
0
/ \
/ \
/ \
/ \
1 2
/ \ / \
/ \ / \
3 4 5 6
.........................
On and on. A lot of math can be applied to it.
What about {1, 2, 3, ..., n}, where n is taken to infinity? No limit?
0
/ \
/ \
/ \
/ \
1 2
/ \ / \
/ \ / \
3 4 5 6
.........................
1
/ \
/ \
/ \
/ \
10 11
/ \ / \
/ \ / \
100 101 110 111
.........................
Am 03.12.2024 um 07:17 schrieb Chris M. Thomasson:
0
/ \
/ \
/ \
/ \
1 2
/ \ / \
/ \ / \
3 4 5 6
.........................
Though we may take 1 for the root too. This way we would get (using
binary representation):
1
/ \
/ \
/ \
/ \
10 11
/ \ / \
/ \ / \
100 101 110 111
.........................
I guess you get the pattern. :-P
*
/ \
/ \
/ \
/ \
l r
/ \ / \
/ \ / \
ll lr rl rr
.........................
() [<<< the "empty l-r-sequence"]
/ \
/ \
/ \
/ \
(l) (r)
/ \ / \
/ \ / \
(l,l) (l,r) (r,l) (r,r)
.........................
What about {1, 2, 3, ..., n}, where n is taken to infinity? No limit?
{1, 2, 3, ..., n} when n is taken to infinity = the set of all natural numbers?
Am 03.12.2024 um 07:17 schrieb Chris M. Thomasson:
0
/ \
/ \
/ \
/ \
1 2
/ \ / \
/ \ / \
3 4 5 6
.........................
Though we may take 1 for the root too. This way we would get (using
binary representation):
1
/ \
/ \
/ \
/ \
10 11
/ \ / \
/ \ / \
100 101 110 111
.........................
I guess you get the pattern. :-P
*
/ \
/ \
/ \
/ \
l r
/ \ / \
/ \ / \
ll lr rl rr
.........................
() [<<< the "empty l-r-sequence"]
/ \
/ \
/ \
/ \
(l) (r)
/ \ / \
/ \ / \
(l,l) (l,r) (r,l) (r,r)
.........................
Am Mon, 02 Dec 2024 17:51:22 +0100 schrieb WM:
On 02.12.2024 16:46, FromTheRafters wrote:What is the limit?
WM wrote :As easily can be obtaied from the above it is necessarily true that up
E(1), E(2), E(3), ...Non sequitur. That which is true for finite sequences is not
and E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
are identical for every n and in the limit because E(1)∩E(2)∩...∩E(n)
= E(n).
necessarily true for infinite sequences.
to every term and therefore also in the limit the sequences of
endsegments and of intersections are identical. Every contrary opinion
is matheology, outside of mathematics.
On 12/2/2024 4:00 PM, Chris M. Thomasson wrote:
On 12/2/2024 3:59 PM, Moebius wrote:
Am 03.12.2024 um 00:58 schrieb Chris M. Thomasson:Thank you Moebius. :^)
On 12/2/2024 3:56 PM, Moebius wrote:
Am 03.12.2024 um 00:51 schrieb Chris M. Thomasson:
On 12/1/2024 9:50 PM, Moebius wrote:
Am 02.12.2024 um 00:11 schrieb Chris M. Thomasson:
On 11/30/2024 3:12 AM, WM wrote:
Finite initial segment[s]: F(n) = {1, 2, 3, ..., n} (n e IN). >>>>>> [...]
When WM writes:
{1, 2, 3, ..., n}
I think he might mean that n is somehow a largest natural number?
Nope, he just means some n e IN.
So if n = 5, the FISON is:
{ 1, 2, 3, 4, 5 }
n = 3
{ 1, 2, 3 }
Right?
Right.
So, i n = all_of_the_naturals, then
{ 1, 2, 3, ... }
Aka, there is no largest natural number and they are not limited. Aka, no limit?
Right?
WM formulated on Monday :
Yes, your nth term is the term common to all previous sets as members of
the sequence. This final 'n' is always a member of the naturals. For
infinite sets of naturals, there is no last element to be common to all previous sets, so it, the intersection, is empty.
and therefore also in the limit the sequences of endsegments and of
intersections are identical.
Says you, but you can't prove the conjecture.
Am Mon, 02 Dec 2024 17:46:04 +0100 schrieb WM:
On 02.12.2024 15:52, joes wrote:No, they contain numbers.
Am Mon, 02 Dec 2024 15:28:30 +0100 schrieb WM:Nevertheless they also contain an infinite set
On 02.12.2024 12:53, FromTheRafters wrote:They ARE infinite sets.
Infinite endsegments contain an infinite set eachThis is his definition of endsegment, which as almost anyone can see, >>>>> has no last element, so yes it is infinite. He says 'infiniteEndsegment E(n) = {n, n+1, n+2, ...}
endsegment' as if there were a choice, only to add confusion.
Ah, no. The intersection is infinite only for a finite number of segments.So it is.infinitely many elements of which are in the intersection.The intersection of all infinite segments?
It is the empty set.Whatever the limit is, it is the same for the sequence of endsegmentsAn empty intersection cannot come before an empty endsegment has beenWhich happens only in the limit.
produced by losing one element at every step.
E(1), E(2), E(3), ...What do you reckon the limit is?
and E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
are identical for every n and in the limit because E(1)∩E(2)∩...∩E(n)
= E(n).
and the sequence of intersections.
On 12/2/2024 9:28 AM, WM wrote:
A quantifier shift tells you (WM) what you (WM) _expect_
but a quantifier shift is untrustworthy.
E(1), E(2), E(3), ...
and
E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
are identical for every n and in the limit
because
E(1)∩E(2)∩...∩E(n) = E(n).
No.
For the set of finite cardinals,
EVEN IF NO END.SEGMENT IS EMPTY,
the intersection of all end segments is empty.
⎜ EVEN IF NO END.SEGMENT IS EMPTY,
⎝ the intersection of all end segments is empty.
WM used his keyboard to write :
Identical sequences have the same limit.
Running with buffaloes does not make one a buffalo.
Am 03.12.2024 um 06:47 schrieb Moebius:
Am 03.12.2024 um 06:34 schrieb Chris M. Thomasson:
Then the following can be shown:
lim_(n->oo) {1, 2, 3, ..., n} = {1, 2, 3, ...} .
While this seems to be intuitively clear, the following is less clear
(I'd say):
lim_(n->oo) {n, n+1, n+2, ...} = {} .
On 03.12.2024 01:32, Jim Burns wrote:This is an equality:
On 12/2/2024 9:28 AM, WM wrote:
A quantifier shift tells you (WM) what you (WM) _expect_Here is no quantifier shift but an identity:
but a quantifier shift is untrustworthy.
That limit being the empty set.E(1), E(2), E(3), ...
and E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
are identical for every n and in the limit because E(1)∩E(2)∩...∩E(n) >>> = E(n).
No.I cannot read or understand the above. The following is gibberish.
For the set of finite cardinals,
EVEN IF NO END.SEGMENT IS EMPTY,
the intersection of all end segments is empty.
⎜ EVEN IF NO END.SEGMENT IS EMPTY,E(1)∩E(2)∩...∩E(n) = E(n).
⎝ the intersection of all end segments is empty.
Sequences which are identical in every term have identical limits.
The sequence of FISONs has a limit. Indeed that's one way to define N
as the least upper bound of the sequence
{1}, {1, 2}, {1, 2, 3}, ...
Am Tue, 03 Dec 2024 14:02:05 +0100 schrieb WM:
On 03.12.2024 01:32, Jim Burns wrote:This is an equality:
On 12/2/2024 9:28 AM, WM wrote:Here is no quantifier shift but an identity:
A quantifier shift tells you (WM) what you (WM) _expect_
but a quantifier shift is untrustworthy.
E(1), E(2), E(3), ...
and E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
are identical for every n and in the limit because E(1)∩E(2)∩...∩E(n)
= E(n).
That limit being the empty set.
No.I cannot read or understand the above. The following is gibberish.
For the set of finite cardinals,
EVEN IF NO END.SEGMENT IS EMPTY,
the intersection of all end segments is empty.
⎜ EVEN IF NO END.SEGMENT IS EMPTY,E(1)∩E(2)∩...∩E(n) = E(n).
⎝ the intersection of all end segments is empty.
Sequences which are identical in every term have identical limits.
On 03.12.2024 01:32, Jim Burns wrote:
On 12/2/2024 9:28 AM, WM wrote:
A quantifier shift tells you (WM) what you (WM) _expect_
but a quantifier shift is untrustworthy.
Here is no quantifier shift but an identity:
E(1), E(2), E(3), ...
and
E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
are identical for every n and in the limit
because
E(1)∩E(2)∩...∩E(n) = E(n).
No.
For the set of finite cardinals,
EVEN IF NO END.SEGMENT IS EMPTY,
the intersection of all end segments is empty.
You cannot read or understand the above.
E(1)∩E(2)∩...∩E(n) = E(n).
Sequences which are identical in every term
have identical limits.
On 12/3/2024 12:29 AM, Moebius wrote:
Am 03.12.2024 um 07:24 schrieb Moebius:
Am 03.12.2024 um 07:17 schrieb Chris M. Thomasson:
0
/ \
/ \
/ \
/ \
1 2
/ \ / \
/ \ / \
3 4 5 6
.........................
Though we may take 1 for the root too. This way we would get (using
binary representation):
; 1
; / \
; / \
; / \
; / \
; 10 11
; / \ / \
; / \ / \
; 100 101 110 111
; .........................
I guess you get the pattern. :-P
Another one (without numbers) but /left/ (l), /right/ (r):
; *
; / \
; / \
; / \
; / \
; l r
; / \ / \
; / \ / \
; ll lr rl rr
; .........................
This way we may even "identify" each node in the tree with a (finite)
l- r-sequence:
; () [<<< the "empty l-r-sequence"] >>> >>> / \
; / \
; / \
; / \
; (l) (r)
; / \ / \
; / \ / \
; (l,l) (l,r) (r,l) (r,r)
; .........................
:-P
So each node actually "is" (or represents) the path leading to it. :-P
(l, l) means take two left branches from the root.
(r, l) means take one right and one left from the root.
I see the pattern. It makes me think some more about my original tree
with node, say, 6.
It's parent is (6-2)/2 = 2 that is a right (-2) wrt 6 has a parent at 2.
At node 2 take a right to get at node 6
Now, lets try 5. It's parent is (5-1)/2 = 2. That is a left from 2 wrt
(-1). There is a pattern here as well.
At node 2 take a left to get at node 5.
Make any sense to you? Thanks.
On 12/3/2024 3:35 AM, Ben Bacarisse wrote:
"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
On 12/2/2024 4:00 PM, Chris M. Thomasson wrote:
On 12/2/2024 3:59 PM, Moebius wrote:
Am 03.12.2024 um 00:58 schrieb Chris M. Thomasson:
On 12/2/2024 3:56 PM, Moebius wrote:
Am 03.12.2024 um 00:51 schrieb Chris M. Thomasson:So if n = 5, the FISON is:
On 12/1/2024 9:50 PM, Moebius wrote:Nope, n may be any natural number [actually, "n" is a variable here]. >>>>>>
Am 02.12.2024 um 00:11 schrieb Chris M. Thomasson:
On 11/30/2024 3:12 AM, WM wrote:
Finite initial segment[s]: F(n) = {1, 2, 3, ..., n} (n e IN). >>>>>>>> [...]
When WM writes:
{1, 2, 3, ..., n}
I think he might mean that n is somehow a largest natural number? >>>>>>>
{ 1, 2, 3, 4, 5 }
[If] n = 3
{ 1, 2, 3 }
Right?
Right.
Thank you Moebius. :^)
So, i n = all_of_the_naturals, then
You are in danger of falling into one of WM's traps here. Above, you
had n = 3 and n = 5. 3 and 5 are naturals. Switching to "n =
all_of_the_naturals" is something else. It's not wrong because there are >> models of the naturals in which they are all sets, but it's open to
confusing interpretations and being unclear about definition is the key
to WM's endless posts.
{ 1, 2, 3, ... }
Aka, there is no largest natural number and they are not limited.
Aka, no limit?
The sequence of FISONs has a limit [namely a "set-theoretical limit" --Moebius]. Indeed that's one way to define N
as the least upper bound of the sequence
{1}, {1, 2}, {1, 2, 3}, ...
although all terms involved need to be carefully defined.
Right?
The numerical sequence 1, 2, 3, ... has no conventional numerical limit,
but, again, if the symbols 1, 2, 3 etc stand for sets (as in, say, Von
Neumann's model for the naturals) then the set sequence
1, 2, 3, ...
does have a set-theoretical limit: N.
However, there is no largest natural number, when I think of that I see
no limit to the naturals. I must be missing something here? ;^o
Am 03.12.2024 um 22:59 schrieb Chris M. Thomasson:
However, there is no largest natural number, when I think of that I
see no limit to the naturals.
I must be missing something here? ;^o
On 12/3/2024 3:35 AM, Ben Bacarisse wrote:
"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
On 12/2/2024 4:00 PM, Chris M. Thomasson wrote:
On 12/2/2024 3:59 PM, Moebius wrote:
Am 03.12.2024 um 00:58 schrieb Chris M. Thomasson:
On 12/2/2024 3:56 PM, Moebius wrote:
Am 03.12.2024 um 00:51 schrieb Chris M. Thomasson:So if n = 5, the FISON is:
On 12/1/2024 9:50 PM, Moebius wrote:Nope, n may be any natural number [actually, "n" is a variable here]. >>>>>>
Am 02.12.2024 um 00:11 schrieb Chris M. Thomasson:
On 11/30/2024 3:12 AM, WM wrote:
Finite initial segment[s]: F(n) = {1, 2, 3, ..., n} (n e IN). >>>>>>>> [...]
When WM writes:
{1, 2, 3, ..., n}
I think he might mean that n is somehow a largest natural number? >>>>>>>
{ 1, 2, 3, 4, 5 }
[If] n = 3
{ 1, 2, 3 }
Right?
Right.
Thank you Moebius. :^)
So, if n = all_of_the_naturals, then
You are in danger of falling into one of WM's traps here. Above, you
had n = 3 and n = 5. 3 and 5 are naturals. Switching to "n =
all_of_the_naturals" is something else. It's not wrong because there are >> models of the naturals in which they are all sets, but it's open to
confusing interpretations and being unclear about definition is the key
to WM's endless posts.
{ 1, 2, 3, ... }
Aka, there is no largest natural number and they are not limited.
Aka, no limit?
The sequence of FISONs has a limit [namely a "set-theoretical limit" --Moebius]. Indeed that's one way to define N
as the least upper bound of the sequence
{1}, {1, 2}, {1, 2, 3}, ...
although all terms involved need to be carefully defined.
Right?
The numerical sequence 1, 2, 3, ... has no conventional numerical limit,
but, again, if the symbols 1, 2, 3 etc stand for sets (as in, say, Von
Neumann's model for the naturals) then the set sequence
1, 2, 3, ...
does have a set-theoretical limit: N.
However, there is no largest natural number, when I think of that I see
no limit to the naturals. I must be missing something here? ;^o
On 12/3/2024 2:32 PM, Moebius wrote:
Am 03.12.2024 um 23:16 schrieb Moebius:
Am 03.12.2024 um 22:59 schrieb Chris M. Thomasson:
However, there is no largest natural number, when I think of that I
see no limit to the naturals.
Right. No "coventional" limit. Actually,
"lim_(n->oo) n"
does not exist.
In the sense of as n tends to infinity there is no limit that can be
reached as in a so-called largest natural number type of shit? Fair enough?
On 12/02/2024 05:22 PM, Jim Burns wrote:
[...]
You mean like "do you pick?".
Remember "do you pick?".
See, in mathematics, all of which are mathematical
objects, in one theory called mathematics, there's
the anti-diagonal argument, which here used to be
called the diagonal argument which is the wrong name,
has after an only-diagonal argument, what results
that you either get both or none, though that the
only-diagonal itself is constructive for itself
while the anti-diagonal is a non-constructive argument
when you look at it that way.
So, "do you pick?".
I think us long-term readers can generously,
generously, aver the "Burse's memories", are,
at best, regularly erased.
Many of which elicited a spark of thought
then out-went-the-lights. Most of which
went starkers bat-shit.
So, do you even remember?
Or did you just get told again?
Am 04.12.2024 um 01:47 schrieb Chris M. Thomasson:
On 12/3/2024 2:32 PM, Moebius wrote:
Am 03.12.2024 um 23:16 schrieb Moebius:
Am 03.12.2024 um 22:59 schrieb Chris M. Thomasson:
However, there is no largest natural number, when I think of that I
see no limit to the naturals.
Right. No "coventional" limit. Actually,
"lim_(n->oo) n"
does not exist.
In the sense of as n tends to infinity there is no limit that can be
reached [...]?
Exactly.
We say, n is "growing beyond all bounds". :-P
0 = {}, 1 = {0}, 2 = {0, 1}, ...
On 12/3/2024 8:02 AM, WM wrote:
E(1)∩E(2)∩...∩E(n) = E(n).
Sequences which are identical in every term
have identical limits.
An empty intersection does not require
an empty end.segment.
"Set-theoretical limit" and "coventional limit" (as defined in
real analysis) are different notions.
Maybe it would be helful to write "LIM_(n->oo) ..." (instead of
"lim_(n->oo) ...") for the former ...
On 12/3/2024 3:35 AM, Ben Bacarisse wrote:
"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
On 12/2/2024 4:00 PM, Chris M. Thomasson wrote:You are in danger of falling into one of WM's traps here. Above, you
On 12/2/2024 3:59 PM, Moebius wrote:
Am 03.12.2024 um 00:58 schrieb Chris M. Thomasson:Thank you Moebius. :^)
On 12/2/2024 3:56 PM, Moebius wrote:
Am 03.12.2024 um 00:51 schrieb Chris M. Thomasson:
On 12/1/2024 9:50 PM, Moebius wrote:Nope, he just means some n e IN.
Am 02.12.2024 um 00:11 schrieb Chris M. Thomasson:
On 11/30/2024 3:12 AM, WM wrote:
Finite initial segment[s]: F(n) = {1, 2, 3, ..., n} (n e IN). >>>>>>>> [...]
When WM writes:
{1, 2, 3, ..., n}
I think he might mean that n is somehow a largest natural number? >>>>>>>
So if n = 5, the FISON is:
{ 1, 2, 3, 4, 5 }
n = 3
{ 1, 2, 3 }
Right?
Right.
So, i n = all_of_the_naturals, then
had n = 3 and n = 5. 3 and 5 are naturals. Switching to n =
all_of_the_naturals is something else. It's not wrong because there are
models of the naturals in which they are all sets, but it's open to
confusing interpretations and being unclear about definition is the key
to WM's endless posts.
{ 1, 2, 3, ... }The sequence of FISONs has a limit. Indeed that's one way to define N
Aka, there is no largest natural number and they are not limited. Aka, no >>> limit?
as the least upper bound of the sequence
{1}, {1, 2}, {1, 2, 3}, ...
although the all terms involved need to be carefully defined.
Right?The numerical sequence 1, 2, 3, ... has no conventional numerical limit,
but, again, if the symbols 1, 2, 3 etc stand for sets (as in, say, Von
Neumann's model for the naturals) then the set sequence
1, 2, 3, ...
does have a set-theoretical limit: N.
However, there is no largest natural number,
when I think of that I see no
limit to the naturals. I must be missing something here? ;^o
FromTheRafters <FTR@nomail.afraid.org> writes:
Moebius expressed precisely :
Am 04.12.2024 um 02:02 schrieb Moebius:
Am 04.12.2024 um 01:47 schrieb Chris M. Thomasson:
On 12/3/2024 2:32 PM, Moebius wrote:Exactly.
Am 03.12.2024 um 23:16 schrieb Moebius:
Am 03.12.2024 um 22:59 schrieb Chris M. Thomasson:
However, there is no largest natural number, when I think of that I >>>>>>>> see no limit to the naturals.
Right. No "coventional" limit. Actually,
"lim_(n->oo) n"
does not exist.
In the sense of as n tends to infinity there is no limit that can be >>>>> reached [...]?
We say, n is "growing beyond all bounds". :-P
On the other hand, if we focus on the fact that the natural numbers are
sets _in the context of set theory_, namely
0 = {}, 1 = {{}}, 2 = {{}, {{}}}, ...
0 = {}, 1 = {0}, 2 = {0, 1}, ...
(due to von Neumann)
then we may conisider the "set-theoretic limit" of the sequence
(0, 1, 2, ...) = ({}, {0}, {0, 1}, ...).
This way we get:
LIM_(n->oo) n = {0, 1, 2, ...} = IN. :-P
I'd like to mention that "lim_(n->oo) n" is "old math" (oldies but
goldies) while "LIM_(n->oo) n" is "new math" (only possible after the
invention of set theory (->Cantor) and later developments (->axiomatic
set theory, natural numbers due to von Neumann, etc.).
If you say so, but I haven't seen this written anywhere.
It's usually framed in terms of least upper bounds, so that might be why
you are not recalling it.
Ironically, there is a very common example of a "set theoretic limit"
which is the point-wise limit of a sequence of functions. Since
functions are just sets of pairs, these long-known limits are just the
limits of sequences of sets. It's ironic because WM categorically
denies that /any/ non-constant sequence of sets has a limit, yet the
basic mathematics textbook he wrote includes the definition of the
point-wise limit, as well as stating that functions are just sets of
pairs. He includes examples of something he categorically denies!
Moebius expressed precisely :
Am 04.12.2024 um 02:02 schrieb Moebius:
Am 04.12.2024 um 01:47 schrieb Chris M. Thomasson:
On 12/3/2024 2:32 PM, Moebius wrote:Exactly.
Am 03.12.2024 um 23:16 schrieb Moebius:
Am 03.12.2024 um 22:59 schrieb Chris M. Thomasson:
However, there is no largest natural number, when I think of that I >>>>>>> see no limit to the naturals.
Right. No "coventional" limit. Actually,
"lim_(n->oo) n"
does not exist.
In the sense of as n tends to infinity there is no limit that can be
reached [...]?
We say, n is "growing beyond all bounds". :-P
On the other hand, if we focus on the fact that the natural numbers are
sets _in the context of set theory_, namely
0 = {}, 1 = {{}}, 2 = {{}, {{}}, ...
Typo, needs another closing curly bracket.
0 = {}, 1 = {0}, 2 = {0, 1}, ...
(due to von Neumann)
then we may conisider the "set-theoretic limit" of the sequence
(0, 1, 2, ...) = ({}, {0}, {0, 1}, ...).
This way we get:
LIM_(n->oo) n = {0, 1, 2, ...} = IN. :-P
I'd like to mention that "lim_(n->oo) n" is "old math" (oldies but
goldies) while "LIM_(n->oo) n" is "new math" (only possible after the
invention of set theory (->Cantor) and later developments (->axiomatic
set theory, natural numbers due to von Neumann, etc.).
If you say so, but I haven't seen this written anywhere.
WM formulated the question :
On 03.12.2024 21:34, Jim Burns wrote:
On 12/3/2024 8:02 AM, WM wrote:
E(1)∩E(2)∩...∩E(n) = E(n).
Sequences which are identical in every term
have identical limits.
An empty intersection does not require
an empty end.segment.
A set of non-empty endsegments has a non-empty intersection. The
reason is inclusion-monotony.
Conclusion not supported by facts.
Am 04.12.2024 um 12:26 schrieb Ben Bacarisse:
FromTheRafters <FTR@nomail.afraid.org> writes:
Moebius expressed precisely :
Am 04.12.2024 um 02:02 schrieb Moebius:
Am 04.12.2024 um 01:47 schrieb Chris M. Thomasson:
On 12/3/2024 2:32 PM, Moebius wrote:Exactly.
Am 03.12.2024 um 23:16 schrieb Moebius:
Am 03.12.2024 um 22:59 schrieb Chris M. Thomasson:
However, there is no largest natural number, when I think of that I >>>>>>>>> see no limit to the naturals.
Right. No "coventional" limit. Actually,
"lim_(n->oo) n"
does not exist.
In the sense of as n tends to infinity there is no limit that can be >>>>>> reached [...]?
We say, n is "growing beyond all bounds". :-P
On the other hand, if we focus on the fact that the natural numbers are >>>> sets _in the context of set theory_, namely
0 = {}, 1 = {{}}, 2 = {{}, {{}}}, ...
0 = {}, 1 = {0}, 2 = {0, 1}, ...
(due to von Neumann)
then we may conisider the "set-theoretic limit" of the sequence
(0, 1, 2, ...) = ({}, {0}, {0, 1}, ...).
This way we get:
LIM_(n->oo) n = {0, 1, 2, ...} = IN. :-P
I'd like to mention that "lim_(n->oo) n" is "old math" (oldies but
goldies) while "LIM_(n->oo) n" is "new math" (only possible after the
invention of set theory (->Cantor) and later developments (->axiomatic >>>> set theory, natural numbers due to von Neumann, etc.).
If you say so, but I haven't seen this written anywhere.
@FromTheAfter: https://en.wikipedia.org/wiki/Set-theoretic_limit
It's usually framed in terms of least upper bounds, so that might be why
you are not recalling it.
Ironically, there is a very common example of a "set theoretic limit"
which is the point-wise limit of a sequence of functions. Since
functions are just sets of pairs, these long-known limits are just the
limits of sequences of sets. It's ironic because WM categorically
denies that /any/ non-constant sequence of sets has a limit, yet the
basic mathematics textbook he wrote includes the definition of the
point-wise limit, as well as stating that functions are just sets of
pairs. He includes examples of something he categorically denies!
Same with the notions of /bijections/. Explained in his book but denied by
WM these days.
On 04.12.2024 11:33, FromTheRafters wrote:Please expand how this works in the infinite case.
WM formulated the question :
On 03.12.2024 21:34, Jim Burns wrote:
On 12/3/2024 8:02 AM, WM wrote:
E(1)∩E(2)∩...∩E(n) = E(n).
Sequences which are identical in every term have identical limits.
An empty intersection does not require
an empty end.segment.
A set of non-empty endsegments has a non-empty intersection. The
reason is inclusion-monotony.
Conclusion not supported by facts.
In two sets A and B which are non-empty both but have an empty
intersection, there must be at least two elements a and b which are in
one endsegment but not in the other:
a ∈ A but a ∉ B and b ∉ A but b ∈ B.
Same with a set of endsegments. It can be divided into two sets for both
of which the same is required.
On 03.12.2024 21:34, Jim Burns wrote:*finite
On 12/3/2024 8:02 AM, WM wrote:
A* set of non-empty endsegments has a non-empty intersection. The reasonE(1)∩E(2)∩...∩E(n) = E(n).An empty intersection does not require an empty end.segment.
Sequences which are identical in every term have identical limits.
is inclusion-monotony.
Am Wed, 04 Dec 2024 14:31:12 +0100 schrieb WM:
In two sets A and B which are non-empty both but have an emptyPlease expand how this works in the infinite case.
intersection, there must be at least two elements a and b which are in
one endsegment but not in the other:
a ∈ A but a ∉ B and b ∉ A but b ∈ B.
Same with a set of endsegments. It can be divided into two sets for both
of which the same is required.
Am Wed, 04 Dec 2024 10:00:08 +0100 schrieb WM:
A* set of non-empty endsegments has a non-empty intersection. The reason*finite
is inclusion-monotony.
On 04.12.2024 11:33, FromTheRafters wrote:
WM formulated the question :
On 03.12.2024 21:34, Jim Burns wrote:
On 12/3/2024 8:02 AM, WM wrote:
E(1)∩E(2)∩...∩E(n) = E(n).
Sequences which are identical in every term
have identical limits.
An empty intersection does not require
an empty end.segment.
A set of non-empty endsegments has
a non-empty intersection.
The reason is inclusion-monotony.
Conclusion not supported by facts.
In two sets A and B which
are non-empty both
but have an empty intersection,
there must be at least
two elements a and b which are
in one endsegment but not in the other:
a ∈ A but a ∉ B and b ∉ A but b ∈ B.
Same with a set of endsegments.
It can be divided into two sets
for both of which the same is required.
On 12/4/2024 8:31 AM, WM wrote:
Below: two is finite.
In two sets A and B which
are non-empty both
but have an empty intersection,
there must be at least
two elements a and b which are
in one endsegment but not in the other:
a ∈ A but a ∉ B and b ∉ A but b ∈ B.
Same with a set of endsegments.
It can be divided into two sets
for both of which the same is required.
No.
Not all sets of end.segments
can be subdivided into two FINITE sets.
GREATERS is inclusion.monotonic and {}.free.
⋂GREATERS = {}
On 04.12.2024 16:33, joes wrote:What makes you think that?
Am Wed, 04 Dec 2024 10:00:08 +0100 schrieb WM:
No. Why should the intersection of endsegments get empty before the endsegments?A* set of non-empty endsegments has a non-empty intersection. The*finite
reason is inclusion-monotony.
On 12/03/2024 04:16 PM, Jim Burns wrote:
[...]
it was very brave of you when you admitted that
"not.first.false"
is not so much justifying itself and
not un-justifying itself,
with regards to the "yin-yang ad infinitum",
which inductively is a constant
yet in its completion is different,
with regards to the "yin-yang ad infinitum",
which inductively is a constant
yet in its completion is different,
Yet, I think that I've always been
both forthcoming and forthright
in providing answers, and context, in
this loooong conversation [...]
On 12/4/2024 6:11 AM, Ben Bacarisse wrote:
Moebius <invalid@example.invalid> writes:
Am 04.12.2024 um 12:26 schrieb Ben Bacarisse:Ah, I had not seen him deny the notion of a bijection. Do you have a
FromTheRafters <FTR@nomail.afraid.org> writes:
Moebius expressed precisely :
Am 04.12.2024 um 02:02 schrieb Moebius:
Am 04.12.2024 um 01:47 schrieb Chris M. Thomasson:
On 12/3/2024 2:32 PM, Moebius wrote:Exactly.
Am 03.12.2024 um 23:16 schrieb Moebius:
Am 03.12.2024 um 22:59 schrieb Chris M. Thomasson:
However, there is no largest natural number, when I think of that I >>>>>>>>>>> see no limit to the naturals.
Right. No "coventional" limit. Actually,
"lim_(n->oo) n"
does not exist.
In the sense of as n tends to infinity there is no limit that can be >>>>>>>> reached [...]?
We say, n is "growing beyond all bounds". :-P
On the other hand, if we focus on the fact that the natural numbers are >>>>>> sets _in the context of set theory_, namely
0 = {}, 1 = {{}}, 2 = {{}, {{}}}, ...
0 = {}, 1 = {0}, 2 = {0, 1}, ...
(due to von Neumann)
then we may conisider the "set-theoretic limit" of the sequence
(0, 1, 2, ...) = ({}, {0}, {0, 1}, ...).
This way we get:
LIM_(n->oo) n = {0, 1, 2, ...} = IN. :-P
I'd like to mention that "lim_(n->oo) n" is "old math" (oldies but >>>>>> goldies) while "LIM_(n->oo) n" is "new math" (only possible after the >>>>>> invention of set theory (->Cantor) and later developments (->axiomatic >>>>>> set theory, natural numbers due to von Neumann, etc.).
If you say so, but I haven't seen this written anywhere.
@FromTheAfter: https://en.wikipedia.org/wiki/Set-theoretic_limit
It's usually framed in terms of least upper bounds, so that might be why >>>> you are not recalling it.
Ironically, there is a very common example of a "set theoretic limit"
which is the point-wise limit of a sequence of functions. Since
functions are just sets of pairs, these long-known limits are just the >>>> limits of sequences of sets. It's ironic because WM categorically
denies that /any/ non-constant sequence of sets has a limit, yet the
basic mathematics textbook he wrote includes the definition of the
point-wise limit, as well as stating that functions are just sets of
pairs. He includes examples of something he categorically denies!
Same with the notions of /bijections/. Explained in his book but denied by >>> WM these days.
message ID? I used to collect explicit statements, though I don't post
enough to make it really worth while anymore.
I think he has said that Cantor Pairing does not work with "certain"
natural numbers?
On 12/4/2024 2:56 AM, Ben Bacarisse wrote:
"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
On 12/3/2024 3:35 AM, Ben Bacarisse wrote:Yes, there is no largest natural. Let's not loose sight of that.
"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
On 12/2/2024 4:00 PM, Chris M. Thomasson wrote:You are in danger of falling into one of WM's traps here. Above, you
On 12/2/2024 3:59 PM, Moebius wrote:
Am 03.12.2024 um 00:58 schrieb Chris M. Thomasson:Thank you Moebius. :^)
On 12/2/2024 3:56 PM, Moebius wrote:
Am 03.12.2024 um 00:51 schrieb Chris M. Thomasson:
On 12/1/2024 9:50 PM, Moebius wrote:Nope, he just means some n e IN.
Am 02.12.2024 um 00:11 schrieb Chris M. Thomasson:
On 11/30/2024 3:12 AM, WM wrote:
Finite initial segment[s]: F(n) = {1, 2, 3, ..., n} (n e IN). >>>>>>>>>> [...]
When WM writes:
{1, 2, 3, ..., n}
I think he might mean that n is somehow a largest natural number? >>>>>>>>>
So if n = 5, the FISON is:
{ 1, 2, 3, 4, 5 }
n = 3
{ 1, 2, 3 }
Right?
Right.
So, i n = all_of_the_naturals, then
had n = 3 and n = 5. 3 and 5 are naturals. Switching to n =
all_of_the_naturals is something else. It's not wrong because there are >>>> models of the naturals in which they are all sets, but it's open to
confusing interpretations and being unclear about definition is the key >>>> to WM's endless posts.
{ 1, 2, 3, ... }The sequence of FISONs has a limit. Indeed that's one way to define N >>>> as the least upper bound of the sequence
Aka, there is no largest natural number and they are not limited. Aka, no >>>>> limit?
{1}, {1, 2}, {1, 2, 3}, ...
although the all terms involved need to be carefully defined.
Right?The numerical sequence 1, 2, 3, ... has no conventional numerical limit, >>>> but, again, if the symbols 1, 2, 3 etc stand for sets (as in, say, Von >>>> Neumann's model for the naturals) then the set sequence
1, 2, 3, ...
does have a set-theoretical limit: N.
However, there is no largest natural number,
when I think of that I see noIt's just that there are lots of kinds of limit, and a limit is not
limit to the naturals. I must be missing something here? ;^o
always in the set in question. Very often, limits take us outside of
the set in question. R (the reals) can be defined as the "smallest" set
closed under the taking of certain limits -- the limits of Cauchy
sequences, the elements of which are simply rationals.
So, the limit of the natural numbers is _outside_ of the set of all natural numbers?
Even if we don't consider FISONs, we can define a limit (technically a
least upper bound) for the sequence 1, 2, 3, ... It won't be a natural
number. We will have to expand our ideas of "number" and "size" to get
the smallest "thing", larger than all naturals. This is how the study
of infinite ordinals starts.
Okay... I need to ponder on that. Can I just, sort of, make up a "symbol"
and say this is larger than any natural number, however it is definitely _not_ a natural number in and of itself?
Damn. Still missing something here. Thanks for your patience... ;^o
On 04.12.2024 18:16, Jim Burns wrote:
On 12/4/2024 8:31 AM, WM wrote:
[...]
Below: two is finite.
No, they may be finite or infinite.
In two sets A and B which
are non-empty both
but have an empty intersection,
there must be at least
two elements a and b which are
in one endsegment but not in the other:
a ∈ A but a ∉ B and b ∉ A but b ∈ B.
Same with a set of endsegments.
It can be divided into two sets
for both of which the same is required.
No.
Not all sets of end.segments
can be subdivided into two FINITE sets.
The set of all endsegments
can be subdivided into two sets,
one of which is finite and the other is infinite.
GREATERS is inclusion.monotonic and {}.free.
Then
⋂GREATERS = {}
is wrong.
E(1), E(2), E(3), ...
and
E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
are identical for every n and
in the limit because
E(1)∩E(2)∩...∩E(n) = E(n).
For very naive readers I recommend the bathtub.
All its states have a non-empty intersection
unless one of the states is the empty state.
On 12/04/2024 11:37 AM, Jim Burns wrote:
On 12/3/2024 8:09 PM, Ross Finlayson wrote:
[...]
⎜ not.first.false claim [1]:
⎜ not.first.false claim [2]:
Yet, I think that I've always been
both forthcoming and forthright
in providing answers, and context,
in this loooong conversation [...]
Please continue being forthcoming and forthright
by confirming or correcting my impression that
"yin-yang ad infinitum"
refers to how, up to ω, claim [1] is true,
about immediate [predecessors],
but, from ω onward, it's negation is true.
The thing is,
'not.first.false' is not used to describe ordinals,
in the way that 'yin.yang.ad.infinitum'
is used to describe ordinals.
'Not.first.false' is used to describe
_claims about ordinals_ of which we are
here only concerned with finitely.many claims.
There is no 'ad infinitum' for 'not.first.false'.
It is in part the absence of 'ad infinitum'
which justifies claims such as [1] and [2]
A linearly.ordered _finite_ set must be well.ordered.
If all claims are true.or.not.first.false,
there is no first false claim.
Because well.ordered,
if there is no first false,
then there is no false,
and all those not.first.false claims are justified.
The natural numbers are not finitely.many.
But that isn't a problem for this argument,
because it isn't the finiteness of the _numbers_
which it depends upon,
but the finiteness of the claim.sequence.
So, not.first.false,
So, not.first.false, is only
after some pair-wise comprehension,
because, there are ready example that
(which you have not yet done)
in "super-task comprehension",
with regards to the "yin-yang ad infinitum",
which inductively is a constant
yet in its completion is different,
Consider this finite sequence of claims
⎛⎛ By 'ordinals', we mean those which
⎜⎜ have only sets.with.minimums and {}
⎜⎝ ('well.ordered')
⎜
⎜⎛ By 'natural numbers', we mean those which
⎜⎜ have a successor,
⎜⎜ are a successor or 0, and
⎜⎝ are an ordinal.
⎜
⎜ (not.first.false claim)
⎜
⎜ (not.first.false claim)
⎜
⎜ (not.first.false claim)
⎜
⎜ ...
⎜
⎜ not.first.false claim [1]:
⎜⎛ Each non.zero natural number
⎜⎜ has,
⎜⎜ for it and for each of its non.zero priors,
⎜⎝ an immediate ordinal.predecessor.
⎜
⎜ not.first.false claim [2]:
⎜⎛ The first transfinite ordinal, which we name 'ω',
⎜⎜ and each of its ordinal.followers
⎜⎜ does not have,
⎜⎜ for it and for each of its non.zero priors,
⎜⎜ an immediate ordinal.predecessor.
⎜⎜ That is,
⎜⎜ there is a non.zero prior without
⎝⎝ an immediate ordinal.predecessor.
I like to look at it as {0,1,2,...} has a larger 'scope' of natural
numbers than {1,2,3,...} while retaining the same set size. It does this
by not being finite.
Not sure why WM thinks that Cantor Pairing does not work with any
natural number... I think I am not misunderstanding WM here.
Am Wed, 04 Dec 2024 18:20:45 +0100 schrieb WM:
Why should the intersection of endsegments get empty before theWhat makes you think that?
endsegments?
WM laid this down on his screen :
In two sets A and B which are non-empty both but have an empty
intersection, there must be at least two elements a and b which are in
one endsegment but not in the other:
a ∈ A but a ∉ B and b ∉ A but b ∈ B.
Finite thinking.
Same with a set of endsegments.
No, because they are infinite and have no last element to be in every participating endsegment.
On 12/4/2024 12:29 PM, WM wrote:
The set of all endsegments
can be subdivided into two sets,
one of which is finite and the other is infinite.
The intersection of the infinite one is empty.
E(1)∩E(2)∩...∩E(n) = E(n).
No finite.cardinal is in the intersection of
all end.segments.
No intersection of
more.than.finitely.many end.segments
of the finite.cardinals
holds a finite.cardinal, or
is non.empty.
On 04.12.2024 20:59, Chris M. Thomasson wrote:Yes, of course they can? Why shouldn't they? What does it mean to you?
Not sure why WM thinks that Cantor Pairing does not work with anyTake any natnumber you can. Almost all natnumbers are following.
natural number... I think I am not misunderstanding WM here.
Infinitely many of them cannot be "taken" or "given" and cannot be
proven to be in any mapping. But Cantor claims that all without any
exception can be taken.
Am Thu, 05 Dec 2024 09:54:11 +0100 schrieb WM:
On 04.12.2024 20:59, Chris M. Thomasson wrote:Yes, of course they can? Why shouldn't they? What does it mean to you?
Not sure why WM thinks that Cantor Pairing does not work with anyTake any natnumber you can. Almost all natnumbers are following.
natural number... I think I am not misunderstanding WM here.
Infinitely many of them cannot be "taken" or "given" and cannot be
proven to be in any mapping. But Cantor claims that all without any
exception can be taken.
On 05.12.2024 11:53, joes wrote:
Am Thu, 05 Dec 2024 09:54:11 +0100 schrieb WM:All means all with no exception. But every number you can take belongs
On 04.12.2024 20:59, Chris M. Thomasson wrote:Yes, of course they can? Why shouldn't they? What does it mean to you?
Not sure why WM thinks that Cantor Pairing does not work with anyTake any natnumber you can. Almost all natnumbers are following.
natural number... I think I am not misunderstanding WM here.
Infinitely many of them cannot be "taken" or "given" and cannot be
proven to be in any mapping. But Cantor claims that all without any
exception can be taken.
to a vanishing subset of ℕ.
Regards, WM
On 04.12.2024 20:59, Chris M. Thomasson wrote:
Not sure why WM thinks that Cantor Pairing does not work with any
natural number... I think I am not misunderstanding WM here.
Take any natnumber you can. Almost all natnumbers are following.
Infinitely many of them cannot be "taken" or "given" and cannot be
proven to be in any mapping. But Cantor claims that all without any
exception can be taken.
Regards, WM
On 05.12.2024 11:53, joes wrote:What does that have to do with the ability to be "chosen"?
Am Thu, 05 Dec 2024 09:54:11 +0100 schrieb WM:But every number you can take belongs to a vanishing subset of ℕ.
On 04.12.2024 20:59, Chris M. Thomasson wrote:Yes, of course they can? Why shouldn't they? What does it mean to you?
Not sure why WM thinks that Cantor Pairing does not work with anyTake any natnumber you can. Almost all natnumbers are following.
natural number... I think I am not misunderstanding WM here.
Infinitely many of them cannot be "taken" or "given" and cannot be
proven to be in any mapping. But Cantor claims that all without any
exception can be taken.
Am Thu, 05 Dec 2024 12:42:17 +0100 schrieb WM:
But every number you can take belongs to a vanishing subset of ℕ.What does that have to do with the ability to be "chosen"?
Which ones can not be "taken" or "given".
WM used his keyboard to write :
All means all with no exception. But every number you can take belongs
to a vanishing subset of ℕ.
What do you mean by vanishing?
On 04.12.2024 21:36, Jim Burns wrote:
On 12/4/2024 12:29 PM, WM wrote:
[...]
No intersection of
more.than.finitely.many end.segments
of the finite.cardinals
holds a finite.cardinal, or
is non.empty.
Small wonder.
More than finitely many endsegments
require
infinitely many indices, i.e., all indices.
No natnumbers are remaining in the contents.
The set of all endsegments
can be subdivided into two sets,
one of which is finite and the other is infinite.
The intersection of the infinite one is empty.
The intersection of the two sets is not empty,
wherever the cut is made.
E(1)∩E(2)∩...∩E(n) = E(n).
No finite.cardinal is in the intersection of
all end.segments.
No finite cardinal is in all endsegments.
The sequences of E(1)∩E(2)∩...∩E(n) and
of E(n) both have an empty limit.
No intersection of
more.than.finitely.many end.segments
of the finite.cardinals
holds a finite.cardinal, or
is non.empty.
Small wonder.
More than finitely many endsegments
require
infinitely many indices, i.e., all indices.
No natnumbers are remaining in the contents.
On 12/04/2024 02:12 PM, Jim Burns wrote:
On 12/4/2024 4:39 PM, Ross Finlayson wrote:
On 12/04/2024 11:37 AM, Jim Burns wrote:
On 12/3/2024 8:09 PM, Ross Finlayson wrote:
Yet, I think that I've always been
both forthcoming and forthright
in providing answers, and context,
in this loooong conversation [...]
Please continue being forthcoming and forthright
by confirming or correcting my impression that
"yin-yang ad infinitum"
refers to how, up to ω, claim [1] is true,
about immediate [predecessors],
but, from ω onward, it's negation is true.
Thank you in advance for confirming or correcting
my impression of what you mean
(something you have not yet done),
in furtherance of your
forthcoming and forthright posting history.
The thing is,
'not.first.false' is not used to describe ordinals,
in the way that 'yin.yang.ad.infinitum'
is used to describe ordinals.
'Not.first.false' is used to describe
_claims about ordinals_ of which we are
here only concerned with finitely.many claims.
There is no 'ad infinitum' for 'not.first.false'.
It is in part the absence of 'ad infinitum'
which justifies claims such as [1] and [2]
A linearly.ordered _finite_ set must be well.ordered.
If all claims are true.or.not.first.false,
there is no first false claim.
Because well.ordered,
if there is no first false,
then there is no false,
and all those not.first.false claims are justified.
The natural numbers are not finitely.many.
But that isn't a problem for this argument,
because it isn't the finiteness of the _numbers_
which it depends upon,
but the finiteness of the claim.sequence.
About your posited point of detail, or question,
about this yin-yang infinitum,
which is non-inductive, and
a neat also graphical example of the non-inductive,
a counter-example to the naively inductive,
as with regards to whether it's not so
at some finite or not ultimately untrue,
I'd aver that it introduces a notion of "arrival"
at "the trans-finite case",
Anyways your point stands that
"not.first.false" is not necessarily
"not.ultimately.untrue",
and so does _not_ decide the outcome.
On 12/5/2024 4:00 AM, WM wrote:
On 04.12.2024 21:36, Jim Burns wrote:
On 12/4/2024 12:29 PM, WM wrote:
[...]
No intersection of
more.than.finitely.many end.segments
of the finite.cardinals
holds a finite.cardinal, or
is non.empty.
Small wonder.
More than finitely many endsegments
require
infinitely many indices, i.e., all indices.
No natnumbers are remaining in the contents.
⎛ That's the intersection.
⎜
Small wonder.
More than finitely many endsegments
require
infinitely many indices, i.e., all indices.
No natnumbers are remaining in the contents.
Yes.
WM pretended :
On 05.12.2024 14:40, FromTheRafters wrote:
WM used his keyboard to write :
All means all with no exception. But every number you can take
belongs to a vanishing subset of ℕ.
What do you mean by vanishing?
The subset of numbers that can be taken is ℕ_def. It is smaller than
|ℕ|/k for every k ∈ ℕ.
Have you meanwhile found an example for sequences with terms a_n = b_n
for every n but different limit?
Why should I?
On 05.12.2024 18:12, Jim Burns wrote:There is no empty segment.
On 12/5/2024 4:00 AM, WM wrote:And it is the empty endsegment.
On 04.12.2024 21:36, Jim Burns wrote:⎛ That's the intersection.
On 12/4/2024 12:29 PM, WM wrote:
No intersection of more.than.finitely.many end.segments of theSmall wonder.
finite.cardinals holds a finite.cardinal, or is non.empty.
More than finitely many endsegments require infinitely many indices,
i.e., all indices. No natnumbers are remaining in the contents.
The contents cannot disappear "in theSame thing. Every finite number is "lost" in some segment. No natural
limit". It has to be lost one by one if ∀k ∈ ℕ : E(k+1) = E(k) \ {k} is really true for all natnumbers.
I really don't understand this connection. First, this also makesMore than finitely many endsegments require infinitely many indices,
i.e., all indices. No natnumbers are remaining in the contents.
Am Thu, 05 Dec 2024 20:30:22 +0100 schrieb WM:
On 05.12.2024 18:12, Jim Burns wrote:There is no empty segment.
On 12/5/2024 4:00 AM, WM wrote:And it is the empty endsegment.
On 04.12.2024 21:36, Jim Burns wrote:⎛ That's the intersection.
On 12/4/2024 12:29 PM, WM wrote:
No intersection of more.than.finitely.many end.segments of theSmall wonder.
finite.cardinals holds a finite.cardinal, or is non.empty.
More than finitely many endsegments require infinitely many indices,
i.e., all indices. No natnumbers are remaining in the contents.
The contents cannot disappear "in theSame thing. Every finite number is "lost" in some segment.
limit". It has to be lost one by one if ∀k ∈ ℕ : E(k+1) = E(k) \ {k} is
really true for all natnumbers.
All segments are infinite.
I really don't understand this connection. First, this also makesMore than finitely many endsegments require infinitely many indices,
i.e., all indices. No natnumbers are remaining in the contents.
every segment infinite. The set of all indices is the infinite N.
On 05.12.2024 18:12, Jim Burns wrote:
On 12/5/2024 4:00 AM, WM wrote:
On 04.12.2024 21:36, Jim Burns wrote:
No intersection of
more.than.finitely.many end.segments
of the finite.cardinals
holds a finite.cardinal, or
is non.empty.
Small wonder.
More than finitely many endsegments
require
infinitely many indices, i.e., all indices.
No natnumbers are remaining in the contents.
⎛ That's the intersection.
And it is the empty endsegment.
On 05.12.2024 13:26, Richard Damon wrote:
Which ones can not be "taken" or "given".
Those with less than infinitely many successors. Cantor claims that all numbers are in his bijections. No successors remaining.
Regards, WM
On 12/5/2024 8:08 AM, WM wrote:
On 05.12.2024 13:26, Richard Damon wrote:
Which ones can not be "taken" or "given".
Those with less than infinitely many successors.
Do you even know how to take any natural number, create a unique pair
and then get back to the original number from said pair?
On 12/5/2024 5:40 AM, FromTheRafters wrote:
WM used his keyboard to write :
On 05.12.2024 11:53, joes wrote:
Am Thu, 05 Dec 2024 09:54:11 +0100 schrieb WM:All means all with no exception. But every number you can take
On 04.12.2024 20:59, Chris M. Thomasson wrote:Yes, of course they can? Why shouldn't they? What does it mean to you? >>>>
Not sure why WM thinks that Cantor Pairing does not work with anyTake any natnumber you can. Almost all natnumbers are following.
natural number... I think I am not misunderstanding WM here.
Infinitely many of them cannot be "taken" or "given" and cannot be
proven to be in any mapping. But Cantor claims that all without any
exception can be taken.
belongs to a vanishing subset of ℕ.
What do you mean by vanishing?
Perhaps his brain in a toilet after a couple of flushes? Ugggh...
On 12/5/2024 2:30 PM, WM wrote:
And it is the empty endsegment.
Depending upon how 'end.segment' is defined,
{} either is or isn't an end.segment.
Consider the options.
⎛ With {} as an end.segment,
⎜ there are more.than.finite.many end.segments,
⎜ too many for any finite.cardinal to be
⎜ in common with all end segments.
⎜ And therefore,
⎜ the intersection of all holds no finite cardinal.
⎜
⎜ With {} NOT as an end.segment,
⎜ there STILL are more.than.finite.many end.segments,
⎜ too many for any finite.cardinal to be
⎜ in common with all end segments.
⎜ And therefore,
⎝ the intersection of all STILL holds no finite cardinal.
The intersection of all non.empty.end.segments
of the finite.cardinals,
which are each infinite non.empty.end.segments,
is empty.
Because,
⎛ for each finite.cardinal,
⎜ there are fewer finite.cardinals before it
⎝ than there are finite.cardinals after it.
Not because
( an end.segment is empty.
WM presented the following explanation :
Your answer to Identical sequences have the same limit: "Running with
buffaloes does not make one a buffalo" appeared to doubt my claim.
Your example ignores the step-by-step dwindling aspect of the
intersections of your infinite sequence of endsegments.
On 12/5/24 11:08 AM, WM wrote:
On 05.12.2024 13:26, Richard Damon wrote:
Which ones can not be "taken" or "given".
Those with less than infinitely many successors. Cantor claims that
all numbers are in his bijections. No successors remaining.
Which since such numbers don't exist,
On 06.12.2024 00:48, Richard Damon wrote:Just because you can't picture infinity?
On 12/5/24 11:08 AM, WM wrote:If so, then infinity cannot be used completely.
On 05.12.2024 13:26, Richard Damon wrote:Which since such numbers don't exist,
Which ones can not be "taken" or "given".Those with less than infinitely many successors. Cantor claims that
all numbers are in his bijections. No successors remaining.
"The infinite sequence thus defined has the peculiar property to containAnd each of them has infinitely many successors, for which the same goes,
the positive rational numbers completely, and each of them only once at
a determined place." [G. Cantor, letter to R. Lipschitz (19 Nov 1883)]
None is missing, let alone a natural number or infinitely many
successors.
On 05.12.2024 21:11, joes wrote:This happens only in the limit. And then there are no more numbers.
Am Thu, 05 Dec 2024 20:30:22 +0100 schrieb WM:If all natnumbers have been lost, then nothing remains. If there are infinitely many endsegments, then all contents has become indices.
On 05.12.2024 18:12, Jim Burns wrote:There is no empty segment.
On 12/5/2024 4:00 AM, WM wrote:And it is the empty endsegment.
On 04.12.2024 21:36, Jim Burns wrote:⎛ That's the intersection.
On 12/4/2024 12:29 PM, WM wrote:
No intersection of more.than.finitely.many end.segments of theSmall wonder.
finite.cardinals holds a finite.cardinal, or is non.empty.
More than finitely many endsegments require infinitely many indices, >>>>> i.e., all indices. No natnumbers are remaining in the contents.
Only as long as you only consider a finite number of them.Two identical sequences have the same limit.The contents cannot disappear "in the limit". It has to be lost one bySame thing. Every finite number is "lost" in some segment.
one if ∀k ∈ ℕ : E(k+1) = E(k) \ {k} is really true for all natnumbers.
All segments are infinite.
E(1)∩E(2)∩...∩E(n) = E(n).
As long as all endsegments are infinite so is their intersection.
There is no segment "after" all the others.Yes. It is E(1) having all natnumbers as its content.I really don't understand this connection. First, this also makes everyMore than finitely many endsegments require infinitely many indices, >>>>> i.e., all indices. No natnumbers are remaining in the contents.
segment infinite. The set of all indices is the infinite N.
Am Fri, 06 Dec 2024 10:01:43 +0100 schrieb WM:
"The infinite sequence thus defined has the peculiar property to containAnd each of them has infinitely many successors,
the positive rational numbers completely, and each of them only once at
a determined place." [G. Cantor, letter to R. Lipschitz (19 Nov 1883)]
None is missing, let alone a natural number or infinitely many
successors.
Am Thu, 05 Dec 2024 22:31:01 +0100 schrieb WM:
If all natnumbers have been lost, then nothing remains. If there areThis happens only in the limit.
infinitely many endsegments, then all contents has become indices.
Two identical sequences have the same limit.Only as long as you only consider a finite number of them.
E(1)∩E(2)∩...∩E(n) = E(n).
As long as all endsegments are infinite so is their intersection.
On 05.12.2024 13:56, joes wrote:Duh. There are no other numbers.
Am Thu, 05 Dec 2024 12:42:17 +0100 schrieb WM:
It is impossible to choose a number outside of a tiny subset.But every number you can take belongs to a vanishing subset of ℕ.What does that have to do with the ability to be "chosen"?
On 06.12.2024 00:48, Richard Damon wrote:
On 12/5/24 11:08 AM, WM wrote:
On 05.12.2024 13:26, Richard Damon wrote:
Which ones can not be "taken" or "given".
Those with less than infinitely many successors. Cantor claims that
all numbers are in his bijections. No successors remaining.
Which since such numbers don't exist,
If so, then infinity cannot be used completely.
"The infinite sequence thus defined has the peculiar property to contain
the positive rational numbers completely, and each of them only once at
a determined place." [G. Cantor, letter to R. Lipschitz (19 Nov 1883)]
None is missing, let alone a natural number or infinitely many successors.
Regards, WM
On 06.12.2024 11:32, joes wrote:
Am Fri, 06 Dec 2024 10:01:43 +0100 schrieb WM:
"The infinite sequence thus defined has the peculiar property to contain >>> the positive rational numbers completely, and each of them only once atAnd each of them has infinitely many successors,
a determined place." [G. Cantor, letter to R. Lipschitz (19 Nov 1883)]
None is missing, let alone a natural number or infinitely many
successors.
The sequence has no successors. Therefore all numbers including their successors are contained in the sequence.
Regards, WM
On 06.12.2024 13:23, joes wrote:"The end" is not a natural number.
Am Thu, 05 Dec 2024 22:31:01 +0100 schrieb WM:
It happens in the end. ∀k ∈ ℕ : E(k+1) = E(k) \ {k}.If all natnumbers have been lost, then nothing remains. If there areThis happens only in the limit. And then there are no more numbers.
infinitely many endsegments, then all contents has become indices.
It's your fault of not considering all infinitely many segments at once.As long as all endsegments are infinite, there is no infinite number ofTwo identical sequences have the same limit. E(1)∩E(2)∩...∩E(n) =Only as long as you only consider a finite number of them.
E(n).
As long as all endsegments are infinite so is their intersection.
them.
On 12/05/2024 10:14 AM, Jim Burns wrote:
On 12/4/2024 5:44 PM, Ross Finlayson wrote:
About your posited point of detail, or question,
about this yin-yang infinitum,
which is non-inductive, and
a neat also graphical example of the non-inductive,
a counter-example to the naively inductive,
as with regards to whether it's not so
at some finite or not ultimately untrue,
I'd aver that it introduces a notion of "arrival"
at "the trans-finite case",
Anyways your point stands that
"not.first.false" is not necessarily
"not.ultimately.untrue",
and so does _not_ decide the outcome.
Thank you for what seems to be
a response to my request.
You seem to have clarified that
your use of
'not.ultimately.untrue' and 'yin-yang ad infinitum'
is utterly divorced from
my use of
'not.first.false'.
A couple thousand years ago,
the Pythagoreans developed a good argument
that √2 is irrational.
⎛ The arithmetical case was made that,
⎜ for each rational expression of √2
⎜ that expression is not.first.√2
⎜
⎜ But that can only be true if
⎜ there _aren't any_ rational expressions of √2
⎜
⎜ So, there aren't any,
⎝ and √2 is irrational.
Mathematicians,
ever loath to let a good argument go to waste,
took that and applied it (joyously, I imagine)
in a host of other domains.
Applied, for example, in the domain of claims.
In the domain of claims,
there are claims.
There are claims about rational.numbers,
irrational.numbers, sets, functions, classes, et al.
An argument over the domain of claims
makes claims about claims,
claims about claims about rational numbers, et al.
We narrow our focus to
claims meeting certain conditions,
that they are in a finite sequence of claims,
each claim of which is true.or.not.first.false.
What is NOT a condition on the claims is that
the claims are about only finitely.many, or
are independently verifiable, or,
in some way, leave the infinite unconsidered.
We narrow our focus, and then,
for those claims,
we know that none of them are false.
We know it by an argument echoing
a thousands.years.old argument.
⎛ There is no first (rational√2, false.claim),
⎝ thus, there is no (rational√2, false.claim).
You seem to have clarified that
your use of
'not.ultimately.untrue' and 'yin-yang ad infinitum'
is utterly divorced from
my use of
'not.first.false'.
No, I say "not.ultimately.untrue" is
_more_ than "not.first.false".
On 05.12.2024 23:20, Jim Burns wrote:
Depending upon how 'end.segment' is defined,
{} either is or isn't an end.segment.
Consider the options.
⎛ With {} as an end.segment,
⎜ there are more.than.finite.many end.segments,
⎜ too many for any finite.cardinal to be
⎜ in common with all end segments.
⎜ And therefore,
⎜ the intersection of all holds no finite cardinal.
Yes.
⎜ With {} NOT as an end.segment,
all endsegments hold content.
⎜ there STILL are
⎜ more.than.finite.many end.segments,
Not actually infinitely many however.
If all endsegments have content,
then not all natnumbers are indices,
then the indices have an upper bound.
⎜ too many for any finite.cardinal to be
⎜ in common with all end segments.
Too many for all definable natnumbers.
But by assumption of content
not all natnumbers have become indices.
⎜ And therefore,
⎜ the intersection of all
⎝ STILL holds no finite cardinal.
No definable finite cardinal.
The intersection of all non.empty.end.segments
of the finite.cardinals,
which are each infinite non.empty.end.segments,
is empty.
That is a wrong conclusion because
inclusion monotony prevents
an empty intersection of non-empty endsegments:
Because,
⎛ for each finite.cardinal,
⎜ there are fewer finite.cardinals before it
⎝ than there are finite.cardinals after it.
That is true only for definable or accessible cardinals.
Not because
( an end.segment is empty.
This argument is wrong if
infinite bijections are assumed to exist.
On 12/06/2024 10:51 AM, Jim Burns wrote:
On 12/5/2024 9:25 PM, Ross Finlayson wrote:
On 12/05/2024 10:14 AM, Jim Burns wrote:
On 12/4/2024 5:44 PM, Ross Finlayson wrote:
About your posited point of detail, or question,
about this yin-yang infinitum,
which is non-inductive, and
a neat also graphical example of the non-inductive,
a counter-example to the naively inductive,
as with regards to whether it's not so
at some finite or not ultimately untrue,
I'd aver that it introduces a notion of "arrival"
at "the trans-finite case",
Anyways your point stands that
"not.first.false" is not necessarily
"not.ultimately.untrue",
and so does _not_ decide the outcome.
Thank you for what seems to be
a response to my request.
You seem to have clarified that
your use of
'not.ultimately.untrue' and 'yin-yang ad infinitum'
is utterly divorced from
my use of
'not.first.false'.
A couple thousand years ago,
the Pythagoreans developed a good argument
that √2 is irrational.
⎛ The arithmetical case was made that,
⎜ for each rational expression of √2
⎜ that expression is not.first.√2
⎜
⎜ But that can only be true if
⎜ there _aren't any_ rational expressions of √2
⎜
⎜ So, there aren't any,
⎝ and √2 is irrational.
Mathematicians,
ever loath to let a good argument go to waste,
took that and applied it (joyously, I imagine)
in a host of other domains.
Applied, for example, in the domain of claims.
In the domain of claims,
there are claims.
There are claims about rational.numbers,
irrational.numbers, sets, functions, classes, et al.
An argument over the domain of claims
makes claims about claims,
claims about claims about rational numbers, et al.
We narrow our focus to
claims meeting certain conditions,
that they are in a finite sequence of claims,
each claim of which is true.or.not.first.false.
What is NOT a condition on the claims is that
the claims are about only finitely.many, or
are independently verifiable, or,
in some way, leave the infinite unconsidered.
We narrow our focus, and then,
for those claims,
we know that none of them are false.
We know it by an argument echoing
a thousands.years.old argument.
⎛ There is no first (rational√2, false.claim),
⎝ thus, there is no (rational√2, false.claim).
----
You seem to have clarified that
your use of
'not.ultimately.untrue' and 'yin-yang ad infinitum'
is utterly divorced from
my use of
'not.first.false'.
No, I say "not.ultimately.untrue" is
_more_ than "not.first.false".
Here is how to tell:
I have here in my hand a list of claims,
each claim true.or.not.first.false,
considering each point between a split of ℚ
(what I consider ℝ)
It is, of course, a finite list, since
I am not a god.like being (trust me on this).
If anything here is not.ultimately.untrue
_what_ is not.ultimately.untrue?
The points?
The claims, trustworthily true of the points?
Clams?
Where are the clams at/from?
Yet, I think that I've always been
both forthcoming and forthright
in providing answers, and context, in
this loooong conversation [...]
On 12/6/24 8:04 AM, WM wrote:
And each of them has infinitely many successors,
The sequence has no successors. Therefore all numbers including their
successors are contained in the sequence.
A sequence doesn't have successors,
Am Thu, 05 Dec 2024 17:04:58 +0100 schrieb WM:
On 05.12.2024 13:56, joes wrote:Duh. There are no other numbers.
Am Thu, 05 Dec 2024 12:42:17 +0100 schrieb WM:It is impossible to choose a number outside of a tiny subset.
But every number you can take belongs to a vanishing subset of ℕ.What does that have to do with the ability to be "chosen"?
Am Fri, 06 Dec 2024 14:10:10 +0100 schrieb WM:
On 06.12.2024 13:23, joes wrote:"The end" is not a natural number.
Am Thu, 05 Dec 2024 22:31:01 +0100 schrieb WM:It happens in the end. ∀k ∈ ℕ : E(k+1) = E(k) \ {k}.
If all natnumbers have been lost, then nothing remains. If there areThis happens only in the limit. And then there are no more numbers.
infinitely many endsegments, then all contents has become indices.
It's your fault of not considering all infinitely many segments at once.As long as all endsegments are infinite, there is no infinite number ofTwo identical sequences have the same limit. E(1)∩E(2)∩...∩E(n) = >>>> E(n).Only as long as you only consider a finite number of them.
As long as all endsegments are infinite so is their intersection.
them.
On 12/6/2024 3:19 AM, WM wrote:
⎜ With {} NOT as an end.segment,
all endsegments hold content.
But no common.to.all finite.cardinals.
⎜ there STILL are
⎜ more.than.finite.many end.segments,
Not actually infinitely many however.
More.than.finitely.many are enough to
break the rules we devise for finitely.many.
For each finite.cardinal,
up.to.that.cardinal are finitely.many.
A rule for finitely.many holds.
All.the.finite.cardinals are more.than.finitely.many.
A rule for more.than.finitely.many holds.
If all endsegments have content,
then not all natnumbers are indices,
That seems to be based on the idea that
no finite.cardinal is both index and content.
Elsewhere, considering one set, that's true.
No element is both
index(minimum) and content(non.minimum).
However,
here, we're considering all the end segments.
Each content is index in a later set.
Each non.zero index is content in an earlier set.
Each content is index in a later set.
⎜ And therefore,
⎜ the intersection of all
⎝ STILL holds no finite cardinal.
No definable finite cardinal.
Wasn't there a time when you (WM)
thought 'undefinable finite.cardinal'
was contradictory?
Round up the usual suspects
and label them 'definable'.
The intersection of all non.empty.end.segments
of the definable finite.cardinals,
which are each infinite non.empty.end.segments,
is empty.
Generalizing,
the intersection of all non.empty end.segments is empty.
It is an argument considering finites,
of which there are more.than.finitely.many.
On 06.12.2024 14:30, joes wrote:This is so dumb.
Am Thu, 05 Dec 2024 17:04:58 +0100 schrieb WM:May be. My statements however concern completed infinity. The subset is
On 05.12.2024 13:56, joes wrote:Duh. There are no other numbers.
Am Thu, 05 Dec 2024 12:42:17 +0100 schrieb WM:It is impossible to choose a number outside of a tiny subset.
But every number you can take belongs to a vanishing subset of ℕ.What does that have to do with the ability to be "chosen"?
never complete.
On 06.12.2024 14:30, joes wrote:
Am Thu, 05 Dec 2024 17:04:58 +0100 schrieb WM:
On 05.12.2024 13:56, joes wrote:Duh. There are no other numbers.
Am Thu, 05 Dec 2024 12:42:17 +0100 schrieb WM:It is impossible to choose a number outside of a tiny subset.
But every number you can take belongs to a vanishing subset of ℕ.What does that have to do with the ability to be "chosen"?
May be. My statements however concern completed infinity. The subset is
never complete.
Regards, WM
Le 07/12/2024 à 12:09, crank Wolfgang Mückenheim from Hochschule
Augsburg, aka WM a écrit :
[W. Mückenheim: "Evidence for Dark Numbers", ELIVA Press, Chisinau
2024. ISBN 978-99993-2-218-8, in press]
https://beallslist.net/vanity-press/
List of vanity press
Here we list the known vanity press outlets. Please be cautious about
sending them any of your articles or theses.
...
Eliva Press
...
On 06.12.2024 16:16, Richard Damon wrote:
On 12/6/24 8:04 AM, WM wrote:
And each of them has infinitely many successors,
The sequence has no successors. Therefore all numbers including their
successors are contained in the sequence.
A sequence doesn't have successors,
Therefore it includes all terms. So it is possible to use all terms. But
it is impossible to use them as individuals.
Regards, WM
On 06.12.2024 19:17, Jim Burns wrote:Not two, but infinitely many.
On 12/6/2024 3:19 AM, WM wrote:
Show two endsegments which do not hold common content.But no common.to.all finite.cardinals.⎜ With {} NOT as an end.segment,all endsegments hold content.
No. Larger than any finite number = infinite.More than finitely many are finitely many,More.than.finitely.many are enough to break the rules we devise for⎜ there STILL are ⎜ more.than.finite.many end.segments,Not actually infinitely many however.
finitely.many.
This right here is you cardinal (heh) mistake. Infinite means thereFor each finite.cardinal,All the fite cardinals are actually infinitely many. That is impossible
up.to.that.cardinal are finitely.many.
A rule for finitely.many holds.
All.the.finite.cardinals are more.than.finitely.many.
A rule for more.than.finitely.many holds.
as long as an upper bound rests in the contentents of endsegments.
And since the "content" is infinite, there are inf.many indices.By an unfortunate definition (made by myself) there is always oneIf all endsegments have content, then not all natnumbers are indices,That seems to be based on the idea that no finite.cardinal is both
index and content.
cardinal content and index: E(2) = {2, 3, 4, ...}. But that is not
really a problem.
There are no steps. Something that holds for every finite partElsewhere, considering one set, that's true."All at once" is the seductive attempt of tricksters. All that happens
No element is both index(minimum) and content(non.minimum).
However,
here, we're considering all the end segments.
Each content is index in a later set.
Each non.zero index is content in an earlier set.
in a sequence can be investigated at every desired step.
It is possible in the limit.Each content is index in a later set.Only if all content is lost. That is not possible for visible
endsegments.
They all are infinite and therefore are finitely many.Ridiculous. The "contents" "become" indices.
[citation needed]Yes, until about six years ago.Wasn't there a time when you (WM)⎜ And therefore,No definable finite cardinal.
⎜ the intersection of all ⎝ STILL holds no finite cardinal.
thought 'undefinable finite.cardinal'
was contradictory?
Although I was a strong opponent of Cantor's
actual infinity, an internet discussion in 2018 [1] has changed my mind
in that without actual infinity the real axis would have gaps.
It doesn't "come down", only in the infinite.Round up the usual suspects and label them 'definable'.∀k ∈ ℕ : E(k+1) = E(k) \ {k} cannot come down to the empty set in definable numbers. No other way however is accessible.
No, there are infinitely many.The intersection of all non.empty.end.segments of the definableA clear selfcontradiction because of inclusion monotony.
finite.cardinals,
which are each infinite non.empty.end.segments,
is empty.
It's just not finite.Generalizing,It is violating mathematics and logic. Like Bob.
the intersection of all non.empty end.segments is empty.
It is an argument considering finites,
of which there are more.than.finitely.many.
Le 07/12/2024 à 12:09, crank Wolfgang Mückenheim from Hochschule
Augsburg, aka WM a écrit :
[snip nonsense]
[W. Mückenheim: "Evidence for Dark Numbers", ELIVA Press, Chisinau
2024. ISBN 978-99993-2-218-8, in press]
https://beallslist.net/vanity-press/
List of vanity press
Here we list the known vanity press outlets. Please be cautious about
sending them any of your articles or theses.
...
Eliva Press
...
On 05.12.2024 21:11, joes wrote:Yes, and then we don't need any more "contents".
Am Thu, 05 Dec 2024 20:30:22 +0100 schrieb WM:If all natnumbers have been lost, then nothing remains. If there are infinitely many endsegments, then all contents has become indices.
On 05.12.2024 18:12, Jim Burns wrote:There is no empty segment.
On 12/5/2024 4:00 AM, WM wrote:And it is the empty endsegment.
On 04.12.2024 21:36, Jim Burns wrote:⎛ That's the intersection.
On 12/4/2024 12:29 PM, WM wrote:
No intersection of more.than.finitely.many end.segments of theSmall wonder.
finite.cardinals holds a finite.cardinal, or is non.empty.
More than finitely many endsegments require infinitely many indices, >>>>> i.e., all indices. No natnumbers are remaining in the contents.
As long as you intersect only finitely many segments.Two identical sequences have the same limit.The contents cannot disappear "in the limit". It has to be lost one bySame thing. Every finite number is "lost" in some segment.
one if ∀k ∈ ℕ : E(k+1) = E(k) \ {k} is really true for all natnumbers.
All segments are infinite.
As long as all endsegments are infinite so is their intersection.
And there is no empty segment.Yes. It is E(1) having all natnumbers as its content.I really don't understand this connection. First, this also makes everyMore than finitely many endsegments require infinitely many indices, >>>>> i.e., all indices. No natnumbers are remaining in the contents.
segment infinite. The set of all indices is the infinite N.
On 12/5/2024 11:06 PM, Moebius wrote:
Am 06.12.2024 um 01:30 schrieb Chris M. Thomasson:
On 12/5/2024 8:08 AM, WM wrote:
On 05.12.2024 13:26, Richard Damon wrote:
Which ones can not be "taken" or "given".
Those with less than infinitely many successors.
Mückenheim, bei Dir sind wirklich ein paar Schrauben locker.
Indeed! Numbers which do not exist can not be "taken" or "given"
Mückenheim is completely right here! On the other hand, "those"?!
Do you even know how to take any natural number, create a unique pair
and then get back to the original number from said pair?
I certainly don't know. Please tell me!
Take any natural number and run it through Cantor Pairing to get a
unique pair. From this pair alone we can also get back to the original number.
On 06.12.2024 19:17, Jim Burns wrote:
On 12/6/2024 3:19 AM, WM wrote:
On 05.12.2024 23:20, Jim Burns wrote:
⎜ With {} NOT as an end.segment,
all endsegments hold content.
But no common.to.all finite.cardinals.
Show two endsegments which
do not hold common content.
⎜ there STILL are
⎜ more.than.finite.many end.segments,
Not actually infinitely many however.
More.than.finitely.many are enough to
break the rules we devise for finitely.many.
More than finitely many are finitely many,
More than finitely many are finitely many,
unless they are actually infinitely many.
Therefore they are no enough.
For each finite.cardinal,
up.to.that.cardinal are finitely.many.
A rule for finitely.many holds.
All.the.finite.cardinals are more.than.finitely.many.
A rule for more.than.finitely.many holds.
All the fite cardinals are actually infinitely many.
That is impossible as long as
an upper bound rests in the contentents of endsegments.
If all endsegments have content,
then not all natnumbers are indices,
That seems to be based on the idea that
no finite.cardinal is both index and content.
By an unfortunate definition (made by myself)
there is always one cardinal content and index:
E(2) = {2, 3, 4, ...}.
But that is not really a problem.
Elsewhere, considering one set, that's true.
No element is both
index(minimum) and content(non.minimum).
However,
here, we're considering all the end segments.
Each content is index in a later set.
Each non.zero index is content in an earlier set.
"All at once" is
the seductive attempt of tricksters.
All that happens in a sequence
can be investigated at every desired step.
Each content is index in a later set.
Only if all content is lost.
Round up the usual suspects
and label them 'definable'.
∀k ∈ ℕ : E(k+1) = E(k) \ {k}
cannot come down to the empty set
∀k ∈ ℕ : E(k+1) = E(k) \ {k}
cannot come down to the empty set
in definable numbers.
No other way however is accessible.
On 12/06/2024 08:34 PM, Jim Burns wrote:
On 12/6/2024 4:32 PM, Ross Finlayson wrote:
On 12/06/2024 10:51 AM, Jim Burns wrote:
On 12/5/2024 9:25 PM, Ross Finlayson wrote:
No, I say "not.ultimately.untrue" is
_more_ than "not.first.false".
Here is how to tell:
I have here in my hand a list of claims,
each claim true.or.not.first.false,
considering each point between a split of ℚ
(what I consider ℝ)
It is, of course, a finite list, since
I am not a god.like being (trust me on this).
If anything here is not.ultimately.untrue
_what_ is not.ultimately.untrue?
The points?
The claims, trustworthily true of the points?
Clams?
Where are the clams at/from?
s/clams/claims
If you need to know where the claims are at/from
in order to answer my question,
that also answers my question:
Your use of
'not.ultimately.untrue' and 'yin-yang ad infinitum'
is utterly divorced from
my use of
'not.first.false'.
No, the "clams", where do you address that
the "clams", inductive clams, must both close
and open to be fit for the soup?
Then, your "utterly divorced" implies a bad marriage.
<RF>
</RF>
Yet, I think that I've always been
both forthcoming and forthright
in providing answers, and context, in
this loooong conversation [...]
...For.Certain.Values.Of forthcoming and forthright.
On 07.12.2024 13:12, Python wrote:Source?
Le 07/12/2024 à 12:09, crank Wolfgang Mückenheim from HochschuleLiars write those lists and stupids believe them.
Augsburg, aka WM a écrit :
[snip nonsense]
[W. Mückenheim: "Evidence for Dark Numbers", ELIVA Press, Chisinau
2024. ISBN 978-99993-2-218-8, in press]
https://beallslist.net/vanity-press/
List of vanity press
Here we list the known vanity press outlets. Please be cautious about
sending them any of your articles or theses.
...
Eliva Press ...
The list starts: "WhatWhich books, the same one?
is vanity press? Vanity press is a type of publishing, where authors pay
to have their work published;"
Eliva publishes two of my books. I had not to pay a cent.
SSRN-Elsevier and De Gruyter published my books. I had not to pay a
cent. What is the difference?
If you ever have something to publish that passes Eliva's terms ofDifferent time, different person.
acceptance you will see that you have nothing to pay for that.
By the way, Cantor payed for publishing his works frequently.
On 12/7/2024 6:09 AM, WM wrote:
On 06.12.2024 19:17, Jim Burns wrote:
On 12/6/2024 3:19 AM, WM wrote:
On 05.12.2024 23:20, Jim Burns wrote:
⎜ With {} NOT as an end.segment,
all endsegments hold content.
But no common.to.all finite.cardinals.
Show two endsegments which
do not hold common content.
I will, after you
show me a more.than.finitely.many two.
⎜ there STILL are
⎜ more.than.finite.many end.segments,
Not actually infinitely many however.
More.than.finitely.many are enough to
break the rules we devise for finitely.many.
More than finitely many are finitely many,
You (WM) define it that way,
which turns your arguments into gibberish.
Each finite.cardinality
cannot be more.than.finitely.many
⎜ If n is a finite.cardinal, then
⎜ n is.less.than n+1, and n+1 is finite.
⎝ n cannot be more.than.finitely.many.
More than finitely many are finitely many,
unless they are actually infinitely many.
Therefore they are not enough.
Call it 'potential'.
None of that changes that
each finite.cardinal is followed by
a finite.cardinal,
Have you ever implemented a Cantor Pairing function that can go back and forth wrt the original number to unique pair and back to the original
number? They are pretty fun to play around with.
Am Sat, 07 Dec 2024 18:01:17 +0100 schrieb WM:
On 07.12.2024 13:12, Python wrote:Source?
Le 07/12/2024 à 12:09, crank Wolfgang Mückenheim from HochschuleLiars write those lists and stupids believe them.
Augsburg, aka WM a écrit :
[snip nonsense]
[W. Mückenheim: "Evidence for Dark Numbers", ELIVA Press, Chisinau
2024. ISBN 978-99993-2-218-8, in press]
https://beallslist.net/vanity-press/
List of vanity press
Here we list the known vanity press outlets. Please be cautious about
sending them any of your articles or theses.
...
Eliva Press ...
Am Thu, 05 Dec 2024 22:31:01 +0100 schrieb WM:
On 05.12.2024 21:11, joes wrote:Yes, and then we don't need any more "contents".
Am Thu, 05 Dec 2024 20:30:22 +0100 schrieb WM:If all natnumbers have been lost, then nothing remains. If there are
On 05.12.2024 18:12, Jim Burns wrote:There is no empty segment.
On 12/5/2024 4:00 AM, WM wrote:And it is the empty endsegment.
On 04.12.2024 21:36, Jim Burns wrote:⎛ That's the intersection.
On 12/4/2024 12:29 PM, WM wrote:
No intersection of more.than.finitely.many end.segments of theSmall wonder.
finite.cardinals holds a finite.cardinal, or is non.empty.
More than finitely many endsegments require infinitely many indices, >>>>>> i.e., all indices. No natnumbers are remaining in the contents.
infinitely many endsegments, then all contents has become indices.
As long as you intersect only finitely many segments.Two identical sequences have the same limit.The contents cannot disappear "in the limit". It has to be lost one by >>>> one if ∀k ∈ ℕ : E(k+1) = E(k) \ {k} is really true for all natnumbers.Same thing. Every finite number is "lost" in some segment.
All segments are infinite.
As long as all endsegments are infinite so is their intersection.
Yes. It is E(1) having all natnumbers as its content.And there is no empty segment.
Am 06.12.2024 um 20:46 schrieb Chris M. Thomasson:
On 12/5/2024 11:06 PM, Moebius wrote:
Am 06.12.2024 um 01:30 schrieb Chris M. Thomasson:Take any natural number and run it through Cantor Pairing to get a unique
On 12/5/2024 8:08 AM, WM wrote:
On 05.12.2024 13:26, Richard Damon wrote:
Which ones can not be "taken" or "given".
Those with less than infinitely many successors.
Mckenheim, bei Dir sind wirklich ein paar Schrauben locker.
Indeed! Numbers which do not exist can not be "taken" or "given"
Mckenheim is completely right here! On the other hand, "those"?!
Do you even know how to take any natural number, create a unique pair
and then get back to the original number from said pair?
I certainly don't know. Please tell me!
pair. From this pair alone we can also get back to the original number.
Where can I find those natural numbers? And what EXACTLY do you mean by
"run it through Cantor Pairing"? I and how does it _create_ pairs?
Strange things are going on.
On 12/7/2024 4:20 AM, Richard Damon wrote:
On 12/7/24 5:44 AM, WM wrote:
On 06.12.2024 14:30, joes wrote:
Am Thu, 05 Dec 2024 17:04:58 +0100 schrieb WM:
On 05.12.2024 13:56, joes wrote:Duh. There are no other numbers.
Am Thu, 05 Dec 2024 12:42:17 +0100 schrieb WM:It is impossible to choose a number outside of a tiny subset.
But every number you can take belongs to a vanishing subset of ℕ. >>>>>> What does that have to do with the ability to be "chosen"?
May be. My statements however concern completed infinity. The subset
is never complete.
Regards, WM
And thus the limit of the subsets is not complete either, per your
logic of a limit of constants is always that same constant. Thus, your
trying to define the complete infinity just fails.
WM seems to think that "completed" means there simply must be a largest natural number? I can only guess here.
On 12/07/2024 12:27 PM, Jim Burns wrote:
On 12/7/2024 9:37 AM, Ross Finlayson wrote:
On 12/06/2024 08:34 PM, Jim Burns wrote:
<RF>
</RF>
Yet, I think that I've always been
both forthcoming and forthright
in providing answers, and context, in
this loooong conversation [...]
...For.Certain.Values.Of forthcoming and forthright.
Sounds like your food safety with regards to clams
makes that it's not just seafood allergies
why I wouldn't trust the dish.
For example, "yin-yang ad-infinitum".
On 07.12.2024 20:59, Jim Burns wrote:
On 12/7/2024 6:09 AM, WM wrote:
On 06.12.2024 19:17, Jim Burns wrote:
On 12/6/2024 3:19 AM, WM wrote:
On 05.12.2024 23:20, Jim Burns wrote:
⎜ With {} NOT as an end.segment,
all endsegments hold content.
But no common.to.all finite.cardinals.
Show two endsegments which
do not hold common content.
I will, after you
show me a more.than.finitely.many two.
There are no more than finitely many
natural numbers which can be shown.
All which can be shown have common content.
⎜ there STILL are
⎜ more.than.finite.many end.segments,
Not actually infinitely many however.
More.than.finitely.many are enough to
break the rules we devise for finitely.many.
More than finitely many are finitely many,
You (WM) define it that way,
which turns your arguments into gibberish.
This is not gibberish but mathematics:
∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n).
Every counter argument has to violate this.
That is inacceptable.
More than finitely many are finitely many,
unless they are actually infinitely many.
Therefore they are not enough.
Call it 'potential'.
It is.
No empty intersection without an empty endsegment.
On 12/7/2024 1:38 PM, Moebius wrote:
Am 07.12.2024 um 22:20 schrieb Chris M. Thomasson:
Have you ever implemented a Cantor Pairing function that can go back
and forth wrt the original number to unique pair and back to the
original number? They are pretty fun to play around with.
Actually, I've implemented a complete library for (finite) "sets" in
C. :-)
Well, that's fine. Wrt this subject its all about a Cantor pairing. Take
any natural, (yes zero works as well) and be able to map it into a 100% unique pairing. Then say okay, we have this unique pair. Now, we are
able to take said unique pair and map it right back to the natural that created it to begin with.
Am Sat, 07 Dec 2024 22:50:27 +0100 schrieb WM:
∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n)That sentence only talks about a single n at a time, though,
All n are infinitely many.
not about the infinite intersection.
Am Sat, 07 Dec 2024 22:37:04 +0100 schrieb WM:
∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n).Incorrect. "More than finite" means infinite. Your formula only
Every counter argument has to violate this. That is inacceptable.
talks about finite intersections.
That is not an infinite intersection.Each finite.cardinality cannot be more.than.finitely.manySo it is. Each finite cardinal cannot turn a finite set into an infinite
set. But even for infinite sets we have ∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) =
E(n).
∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n).That happens, as you know, only in the limit.
No empty intersection without an empty endsegment.
On 12/7/2024 4:37 PM, WM wrote:
On 07.12.2024 20:59, Jim Burns wrote:
On 12/7/2024 6:09 AM, WM wrote:
On 06.12.2024 19:17, Jim Burns wrote:
On 12/6/2024 3:19 AM, WM wrote:
On 05.12.2024 23:20, Jim Burns wrote:
⎜ With {} NOT as an end.segment,
all endsegments hold content.
But no common.to.all finite.cardinals.
Show two endsegments which
do not hold common content.
I will, after you
show me a more.than.finitely.many two.
There are no more than finitely many
natural numbers which can be shown.
All which can be shown have common content.
Each end.segment is more.than.finite and
the intersection of the end.segments is empty.
This is not gibberish but mathematics:
∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n).
True.
Every counter argument has to violate this.
False.
For example, see above.
Each finite.cardinal is not in common with
more.than.finitely.many end.segments.
Each end.segment has, for each finite.cardinal,
a subset larger than that cardinal.
On 07.12.2024 20:59, Jim Burns wrote:Actually there are infinitely many naturals, and of every two segments
On 12/7/2024 6:09 AM, WM wrote:There are no more than finitely many natural numbers which can be shown.
On 06.12.2024 19:17, Jim Burns wrote:I will, after you show me a more.than.finitely.many two.
On 12/6/2024 3:19 AM, WM wrote:
On 05.12.2024 23:20, Jim Burns wrote:
Show two endsegments which do not hold common content.But no common.to.all finite.cardinals.⎜ With {} NOT as an end.segment,all endsegments hold content.
All which can be shown have common content.
Incorrect. "More than finite" means infinite. Your formula onlyThis is not gibberish but mathematics:You (WM) define it that way,More than finitely many are finitely many,More.than.finitely.many are enough to break the rules we devise for⎜ there STILL are more.than.finite.many end.segments,Not actually infinitely many however.
finitely.many.
which turns your arguments into gibberish.
∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n).
Every counter argument has to violate this. That is inacceptable.
That is not an infinite intersection.Each finite.cardinality cannot be more.than.finitely.manySo it is. Each finite cardinal cannot turn a finite set into an infinite
set. But even for infinite sets we have ∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) =
E(n).
*in⎜ If n is a finite.cardinal, then ⎜ n is.less.than n+1, and n+1 is
finite.
⎝ n cannot be more.than.finitely.many.
More than finitely many are *finitely many,
That happens, as you know, only in the limit.It is.unless they are actually infinitely many. Therefore they are notCall it 'potential'.
enough.
None of that changes that each finite.cardinal is followed by aand leaves the set finite. But even for infinitely many n:
finite.cardinal,
∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n).
No empty intersection without an empty endsegment.
On 07.12.2024 17:21, joes wrote:Only in the limit.
Am Thu, 05 Dec 2024 22:31:01 +0100 schrieb WM:Then we have the empty endsegment.
On 05.12.2024 21:11, joes wrote:Yes, and then we don't need any more "contents".
Am Thu, 05 Dec 2024 20:30:22 +0100 schrieb WM:If all natnumbers have been lost, then nothing remains. If there are
On 05.12.2024 18:12, Jim Burns wrote:There is no empty segment.
On 12/5/2024 4:00 AM, WM wrote:And it is the empty endsegment.
On 04.12.2024 21:36, Jim Burns wrote:⎛ That's the intersection.
No intersection of more.than.finitely.many end.segments of the >>>>>>>> finite.cardinals holds a finite.cardinal, or is non.empty.Small wonder.
More than finitely many endsegments require infinitely many
indices,
i.e., all indices. No natnumbers are remaining in the contents.
infinitely many endsegments, then all contents has become indices.
That sentence only talks about a single n at a time, though,∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n)As long as you intersect only finitely many segments.Two identical sequences have the same limit.The contents cannot disappear "in the limit". It has to be lost oneSame thing. Every finite number is "lost" in some segment.
by one if ∀k ∈ ℕ : E(k+1) = E(k) \ {k} is really true for all
natnumbers.
All segments are infinite.
As long as all endsegments are infinite so is their intersection.
All n are infinitely many.
I.e. at no natural.When you "don't need any more contents", there is the empty set.Yes. It is E(1) having all natnumbers as its content.And there is no empty segment.
On 08.12.2024 00:38, Jim Burns wrote:
Each end.segment is more.than.finite and
the intersection of the end.segments is empty.
∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n)
What can't you understand here?
On 12/8/2024 5:50 AM, WM wrote:non.empty end.segments.
∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n)
What can't you understand here?
{E(i):i} is the set.of.all non.empty end.segments.
⋂{E(i):i} is the intersection.of.all
∀n ∈ ℕ:
{E(i):i}∪{E(n+1)} = {E(i):i}
Each is "already" in.
∀n ∈ ℕ:
(⋂{E(i):i})∩E(n+1) = ⋂{E(i):i}
∀n ∈ ℕ:
⎛ E(1)∩E(2)∩...∩E(n) = E(n)
⎜ E(n)∩E(n+1) ≠ E(n)
⎜ (⋂{E(i):i})∩E(n+1) = ⋂{E(i):i}
⎝ E(n) ≠ ⋂{E(i):i}
⋂{E(i):i} ∉ {E(i):i}
The intersection.of.all non.empty end.segments
isn't any non.empty end.segment.
You (WM) are considering
infinite dark.finite.cardinals,
which do not exist.
On 08.12.2024 19:01, Jim Burns wrote:
You (WM) are consideringThen analysis is contradicted in set theory.
infinite dark.finite.cardinals,
which do not exist.
∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n).
The limit of the left-hand side is empty, the limit of the right-hand
side is full, i.e. not empty.
I do not tolerate that.
Regards, WM
On 12/9/24 4:04 AM, WM wrote:
On 08.12.2024 19:01, Jim Burns wrote:By your logic, 1 equals 0,
You (WM) are consideringThen analysis is contradicted in set theory.
infinite dark.finite.cardinals,
which do not exist.
∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n).
The limit of the left-hand side is empty, the limit of the right-hand
side is full, i.e. not empty.
I do not tolerate that.
WM wrote :
On 08.12.2024 19:01, Jim Burns wrote:
You (WM) are consideringThen analysis is contradicted in set theory.
infinite dark.finite.cardinals,
which do not exist.
∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n).
The limit of the left-hand side is empty, the limit of the right-hand
side is full, i.e. not empty.
I do not tolerate that.
So you just pretend that they have the same 'limit'?
On 08.12.2024 19:01, Jim Burns wrote:
You (WM) are considering
infinite dark.finite.cardinals,
which do not exist.
Then analysis is contradicted in set theory.
∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n).
The limit of the left-hand side is empty,
the limit of the right-hand side is full,
i.e. not empty.
I do not tolerate that.
On 08.12.2024 19:01, Jim Burns wrote:of all the finite.cardinals.
On 12/8/2024 5:50 AM, WM wrote:
∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n)
What can't you understand here?
{E(i):i} is the set.of.all
non.empty end.segments.
of all the finite.cardinals.⋂{E(i):i} is the intersection.of.allnon.empty end.segments.
∀n ∈ ℕ:
{E(i):i}∪{E(n+1)} = {E(i):i}
Each is "already" in.
Not the empty endsegment.
∀n ∈ ℕ: E(n) is non-empty.
But not every E(n+1).
∀n ∈ ℕ:
(⋂{E(i):i})∩E(n+1) = ⋂{E(i):i}
No. E(n+1) = { } is possible.
E(n+1) ≠ E(n)∀n ∈ ℕ:
⎛ E(1)∩E(2)∩...∩E(n) = E(n)
⎜ E(n)∩E(n+1) ≠ E(n)
E(n)∩E(n+1) = E(n+1)
⎜ (⋂{E(i):i})∩E(n+1) = ⋂{E(i):i}
No.
⋂{E(i):i} is the intersection of non-empty endsegments
and as such non-empty.
⎝ E(n) ≠ ⋂{E(i):i}
Not for all n.
⋂{E(i):i} ∉ {E(i):i}
The intersection.of.all non.empty end.segments
isn't any non.empty end.segment.
On 08.12.2024 19:01, Jim Burns wrote:What is the RHS limit? Not that you have written out a sequence.
You (WM) are considering infinite dark.finite.cardinals,Then analysis is contradicted in set theory.
which do not exist.
∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n).
The limit of the left-hand side is empty, the limit of the right-hand
side is full, i.e. not empty.
I do not tolerate that.
Le 08/12/2024 à 23:34, Crank Wolfgang Mückenheim from Hochschule
Augsburg aka WM a écrit :
On 08.12.2024 19:01, Jim Burns wrote:
On 12/8/2024 5:50 AM, WM wrote:non.empty end.segments.
∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n)
What can't you understand here?
{E(i):i} is the set.of.all non.empty end.segments.
⋂{E(i):i} is the intersection.of.all
∀n ∈ ℕ:
{E(i):i}∪{E(n+1)} = {E(i):i}
Each is "already" in.
Not the empty endsegment.
∀n ∈ ℕ: E(n) is non-empty. But not every E(n+1).
You could hardly write something worse and more wrong that that.
The very core property of N is that if n ∈ ℕ then n+1 ∈ ℕ.
On 12/9/2024 4:04 AM, WM wrote:
Your two sequences as you have written them
are equal, and have equal limits: the empty set.
I suspect that it is the distinction between
cardinality of limit #⋂{E(i):i} = #{} = 0 and
limit of cardinalities
On 08.12.2024 11:46, joes wrote:And not about the infinite intersection.
Am Sat, 07 Dec 2024 22:37:04 +0100 schrieb WM:
It talks about all finite intersections.∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n).Incorrect. "More than finite" means infinite. Your formula only talks
Every counter argument has to violate this. That is inacceptable.
about finite intersections.
Empty, like you said.It is the sequence of all endsegments and of all finite intersections.That is not an infinite intersection.Each finite.cardinality cannot be more.than.finitely.manySo it is. Each finite cardinal cannot turn a finite set into an
infinite set. But even for infinite sets we have ∀n ∈ ℕ:
E(1)∩E(2)∩...∩E(n) = E(n).
The limits of identical sequences are identical
And what is the limit?∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n).That happens, as you know, only in the limit.
No empty intersection without an empty endsegment.
On 08.12.2024 00:38, Jim Burns wrote:Nope, not in common with all, only for finite intersections.
On 12/7/2024 4:37 PM, WM wrote:All endsegments which can be shown (by their indices) have common
On 07.12.2024 20:59, Jim Burns wrote:
On 12/7/2024 6:09 AM, WM wrote:
On 06.12.2024 19:17, Jim Burns wrote:
On 12/6/2024 3:19 AM, WM wrote:
On 05.12.2024 23:20, Jim Burns wrote:
There are no more than finitely many natural numbers which can beI will, after you show me a more.than.finitely.many two.Show two endsegments which do not hold common content.But no common.to.all finite.cardinals.⎜ With {} NOT as an end.segment,all endsegments hold content.
shown.
All which can be shown have common content.
content.
That all those n only produce finite intersections.Each end.segment is more.than.finite and the intersection of the∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n)
end.segments is empty.
What can't I understand here?
There are definitely more than finitely many naturals and segmentsThis is not gibberish but mathematics:False. For example, see above.
∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n).
Every counter argument has to violate this.
Each finite.cardinal is not in common with more.than.finitely.manyOf course not. All non-empty endsegments belong to a finite set with an
end.segments.
upper bound.
What changes to what? Why exactly there?Each end.segment has, for each finite.cardinal,That is not true for the last dark endsegments. It changes at the dark
a subset larger than that cardinal.
finite cardinal ω/2.
On 09.12.2024 20:23, joes wrote:There is, and you referred to it. Now what is it?
Am Mon, 09 Dec 2024 10:04:23 +0100 schrieb WM:There is no limit in set theory, contrary to the LHS limit { }.
On 08.12.2024 19:01, Jim Burns wrote:What is the RHS limit?
You (WM) are considering infinite dark.finite.cardinals,Then analysis is contradicted in set theory.
which do not exist.
∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n).
The limit of the left-hand side is empty, the limit of the right-hand
side is full, i.e. not empty.
I do not tolerate that.
--∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n).
The limit of the left-hand side is empty, the limit of the right-hand
side is full, i.e. not empty.
On 09.12.2024 18:20, Jim Burns wrote:Very insightful.
On 12/9/2024 4:04 AM, WM wrote:
Your two sequences as you have written them are equal, and have equal
limits: the empty set.
I suspect that it is the distinction between cardinality of limit
#⋂{E(i):i} = #{} = 0 and limit of cardinalities
The cardinality of the limit is the cardinality of the limit set.ThisIt’s transparent how he does exactly that.
set is defined by the sequence of cardinalities from ∀k ∈ ℕ : E(k+1) = E(k) \ {k} and by nothing else.
On 09.12.2024 18:20, Jim Burns wrote:
Your two sequences as you have written them
are equal, and have equal limits: the empty set.
I suspect that it is the distinction between
cardinality of limit #⋂{E(i):i} = #{} = 0 and
limit of cardinalities ⋂{#E(i):i} = ⋂{ℵ₀:i} = ℵ₀
The cardinality of the limit is
the cardinality of the limit set.
This set is defined by
the sequence of cardinalities from
∀k ∈ ℕ : E(k+1) = E(k) \ {k}
and by nothing else.
Le 09/12/2024 à 20:40, Crank Wolfgang Mückenheim from Hochschule
Augsburg aka WM a écrit :
On 09.12.2024 20:18, Python wrote:
Le 08/12/2024 à 23:34, Crank Wolfgang Mückenheim from Hochschule
Augsburg aka WM a écrit :
On 08.12.2024 19:01, Jim Burns wrote:
On 12/8/2024 5:50 AM, WM wrote:non.empty end.segments.
∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n)
What can't you understand here?
{E(i):i} is the set.of.all non.empty end.segments.
⋂{E(i):i} is the intersection.of.all
∀n ∈ ℕ:
{E(i):i}∪{E(n+1)} = {E(i):i}
Each is "already" in.
Not the empty endsegment.
∀n ∈ ℕ: E(n) is non-empty. But not every E(n+1).
You could hardly write something worse and more wrong that that.
The very core property of N is that if n ∈ ℕ then n+1 ∈ ℕ.
That is correct for definable natural numbers and even for almost all
dark natural numbers.
The very core property of analysis is that equal sequences have equal
limits if they have limits at all.
E(1)∩E(2)∩...∩E(n) = E(n)
Lim E(1)∩E(2)∩...∩E(n) = {}
Lim E(n) = {}
The are equal.
Am Mon, 09 Dec 2024 10:04:23 +0100 schrieb WM:
On 08.12.2024 19:01, Jim Burns wrote:What is the RHS limit?
You (WM) are considering infinite dark.finite.cardinals,Then analysis is contradicted in set theory.
which do not exist.
∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n).
The limit of the left-hand side is empty, the limit of the right-hand
side is full, i.e. not empty.
I do not tolerate that.
Not that you have written out a sequence.
On 12/9/2024 2:45 PM, WM wrote:
On 09.12.2024 18:20, Jim Burns wrote:
Your two sequences as you have written them
are equal, and have equal limits: the empty set.
I suspect that it is the distinction between
cardinality of limit #⋂{E(i):i} = #{} = 0 and
limit of cardinalities ⋂{#E(i):i} = ⋂{ℵ₀:i} = ℵ₀
The cardinality of the limit is
the cardinality of the limit set.
The limit set is
the set of numbers in common with each end.segment
and isn't anything else.
WM explained :
On 09.12.2024 21:55, Jim Burns wrote:
On 12/9/2024 2:45 PM, WM wrote:
On 09.12.2024 18:20, Jim Burns wrote:
Your two sequences as you have written them
are equal, and have equal limits: the empty set.
I suspect that it is the distinction between
cardinality of limit #⋂{E(i):i} = #{} = 0 and
limit of cardinalities ⋂{#E(i):i} = ⋂{ℵ₀:i} = ℵ₀
The cardinality of the limit is
the cardinality of the limit set.
By the way, we need no cardinality. We need only the sequence of sets
with the empty set in the limit.
The limit set is
the set of numbers in common with each end.segment
and isn't anything else.
The limit set is the same for both sequences.
(E(1)∩E(2)∩...∩E(n)) and (E(n))
In order to stop tricksters we go without cardinality.
But size matters, or so I've heard.
WM presented the following explanation :
On 09.12.2024 22:45, FromTheRafters wrote:
WM explained :
On 09.12.2024 21:55, Jim Burns wrote:
On 12/9/2024 2:45 PM, WM wrote:
On 09.12.2024 18:20, Jim Burns wrote:
Your two sequences as you have written them
are equal, and have equal limits: the empty set.
I suspect that it is the distinction between
cardinality of limit #⋂{E(i):i} = #{} = 0 and
limit of cardinalities ⋂{#E(i):i} = ⋂{ℵ₀:i} = ℵ₀
The cardinality of the limit is
the cardinality of the limit set.
By the way, we need no cardinality. We need only the sequence of
sets with the empty set in the limit.
The limit set is
the set of numbers in common with each end.segment
and isn't anything else.
The limit set is the same for both sequences.
(E(1)∩E(2)∩...∩E(n)) and (E(n))
In order to stop tricksters we go without cardinality.
But size matters, or so I've heard.
In sequences of sets only sets matter.
The why are you averse to the emptyset?
On 09.12.2024 13:03, Richard Damon wrote:
On 12/9/24 4:04 AM, WM wrote:
On 08.12.2024 19:01, Jim Burns wrote:By your logic, 1 equals 0,
You (WM) are consideringThen analysis is contradicted in set theory.
infinite dark.finite.cardinals,
which do not exist.
∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n).
The limit of the left-hand side is empty, the limit of the right-hand
side is full, i.e. not empty.
I do not tolerate that.
No, that are two different sequences.
Regards, WM
On 12/9/24 8:04 AM, WM wrote:
On 09.12.2024 13:03, Richard Damon wrote:
On 12/9/24 4:04 AM, WM wrote:
On 08.12.2024 19:01, Jim Burns wrote:By your logic, 1 equals 0,
You (WM) are consideringThen analysis is contradicted in set theory.
infinite dark.finite.cardinals,
which do not exist.
∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n).
The limit of the left-hand side is empty, the limit of the
right-hand side is full, i.e. not empty.
I do not tolerate that.
No, that are two different sequences.
But since both 0^x and x^0 as x approaches 0 approach 0^0, your logic
says that 0^0 is both 0 and 1.
Just because you have a sequence, doesn't mean you can talk about the
end infinite state at the "end" of the sequence.
you have two sequences
that seem to go to the same infinte set at the end,
On 12/9/2024 2:13 PM, WM wrote:
On 09.12.2024 23:05, FromTheRafters wrote:
WM presented the following explanation :
In sequences of sets only sets matter.
Then why are you averse to the empty set?
Not at all! It proves the existence of dark numbers.
Yawn...
On 10.12.2024 01:45, Richard Damon wrote:
On 12/9/24 8:04 AM, WM wrote:
On 09.12.2024 13:03, Richard Damon wrote:
On 12/9/24 4:04 AM, WM wrote:
On 08.12.2024 19:01, Jim Burns wrote:By your logic, 1 equals 0,
You (WM) are consideringThen analysis is contradicted in set theory.
infinite dark.finite.cardinals,
which do not exist.
∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n).
The limit of the left-hand side is empty, the limit of the right-
hand side is full, i.e. not empty.
I do not tolerate that.
No, that are two different sequences.
But since both 0^x and x^0 as x approaches 0 approach 0^0, your logic
says that 0^0 is both 0 and 1.
You should check your "logic". When two different sequences have
different limits, this does not mean that the limits are identical. By
the way 0^0 = 1 is simply a definition.
Just because you have a sequence, doesn't mean you can talk about the
end infinite state at the "end" of the sequence.
The end infinite state is a set.
you have two sequences that seem to go to the same infinte set at the
end,
Two sequences that are identical term by term cannot have different
limits. 0^x and x^0 are different term by term.
REgards, WM
On 12/10/24 3:50 AM, WM wrote:
Two sequences that are identical term by term cannot have different
limits. 0^x and x^0 are different term by term.
Which isn't the part I am talking of, it is that just because each step
of a sequence has a value, doesn't mean the thing that is at that limit,
has the same value.
Well, the empty set contains each and every dark number, that's for sure!
On 12/10/2024 1:06 AM, Moebius wrote:
Am 09.12.2024 um 23:21 schrieb Chris M. Thomasson:
On 12/9/2024 2:13 PM, WM wrote:
On 09.12.2024 23:05, FromTheRafters wrote:
WM presented the following explanation :
In sequences of sets only sets matter.
Or rather _their contents_.
Then why are you averse to the empty set?
Not at all! It proves the existence of dark numbers.
Yawn...
Well, the empty set contains each and every dark number, that's for sure!
{ } = infinite dark?
Again, it is pointless to try here to convince this disgusting charlatan
of anything.
[...] This guy is a criminal. Not joking.
On 12/10/2024 1:06 AM, Moebius wrote:
Am 09.12.2024 um 23:21 schrieb Chris M. Thomasson:
On 12/9/2024 2:13 PM, WM wrote:
On 09.12.2024 23:05, FromTheRafters wrote:
WM presented the following explanation :
In sequences of sets only sets matter.
Or rather _their contents_.
Then why are you averse to the empty set?
Not at all! It proves the existence of dark numbers.
Yawn...
Well, the empty set contains each and every dark number, that's for sure!
{ } = infinite dark?
On 12/10/2024 12:20 PM, Moebius wrote:
Am 10.12.2024 um 20:31 schrieb Chris M. Thomasson:
On 12/10/2024 1:06 AM, Moebius wrote:
Am 09.12.2024 um 23:21 schrieb Chris M. Thomasson:
On 12/9/2024 2:13 PM, WM wrote:
On 09.12.2024 23:05, FromTheRafters wrote:
WM presented the following explanation :
In sequences of sets only sets matter.
Or rather _their contents_.
Then why are you averse to the empty set?
Not at all! It proves the existence of dark numbers.
Yawn...
Well, the empty set contains each and every dark number, that's for
sure!
{ } = infinite dark?
Nope. Actually,
for all x: If x is a dark number, then x is in {}.
See?!
Shit. So, any natural number that WM cannot think of is dark? I guess.
On 09.12.2024 20:23, joes wrote:Dafuq? How do you derive that? Why not for the RHS?
Am Mon, 09 Dec 2024 10:04:23 +0100 schrieb WM:There is no limit in set theory, contrary to the LHS limit { }.
On 08.12.2024 19:01, Jim Burns wrote:What is the RHS limit?
You (WM) are considering infinite dark.finite.cardinals,Then analysis is contradicted in set theory.
which do not exist.
∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n).
The limit of the left-hand side is empty, the limit of the
right-hand side is full, i.e. not empty.
I do not tolerate that.
--Not that you have written out a sequence.I wrote a general term. The sequences are (E(1)∩E(2)∩...∩E(n)) and (E(n)).
On 10.12.2024 13:17, Richard Damon wrote:
On 12/10/24 3:50 AM, WM wrote:
Two sequences that are identical term by term cannot have different
limits. 0^x and x^0 are different term by term.
Which isn't the part I am talking of, it is that just because each
step of a sequence has a value, doesn't mean the thing that is at that
limit, has the same value.
Of course not. But if each step of two sequences has the same value,
then the limits are the same too. This is the case for
(E(1)∩E(2)∩...∩E(n)) and (E(n)).
Regards, WM
On 12/9/2024 2:45 PM, WM wrote:
On 12/9/2024 4:04 AM, WM wrote:
On 12/8/2024 5:50 AM, WM wrote:
On 08.12.2024 00:38, Jim Burns wrote:
Each end.segment has, for each finite.cardinal,
a subset larger than that cardinal.
That is not true for the last dark endsegments.
It changes at the dark finite cardinal ω/2.
∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n).
The limit of the left-hand side is empty,
the limit of the right-hand side is full,
i.e. not empty.
I do not tolerate that.
The cardinality of the limit is
the cardinality of the limit set.
This set is defined
by the sequence of cardinalities from
∀k ∈ ℕ : E(k+1) = E(k) \ {k}
and by nothing else.
By the way, we need no cardinality.
We need only the sequence of sets
with the empty set in the limit.
The limit set is the same
for both sequences.
(E(1)∩E(2)∩...∩E(n)) and (E(n))
In order to stop tricksters
we go without cardinality.
It changes at the dark finite cardinal ω/2.
∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n).
The limit of the left-hand side is empty,
the limit of the right-hand side is full,
i.e. not empty.
By the way, we need no cardinality.
We need only the sequence of sets
with the empty set in the limit.
The limit set is the same
for both sequences.
(E(1)∩E(2)∩...∩E(n)) and (E(n))
In order to stop tricksters
we go without cardinality.
Am Mon, 09 Dec 2024 22:28:41 +0100 schrieb WM:
>> ∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n).
>> The limit of the left-hand side is empty, the limit of the
>> right-hand side is full, i.e. not empty.
>> I do not tolerate that.
> What is the RHS limit?
There is no limit in set theory, contrary to the LHS limit { }.Why not for the RHS?
On 12/10/24 12:30 PM, WM wrote:
On 10.12.2024 13:17, Richard Damon wrote:
On 12/10/24 3:50 AM, WM wrote:
Two sequences that are identical term by term cannot have different
limits. 0^x and x^0 are different term by term.
Which isn't the part I am talking of, it is that just because each
step of a sequence has a value, doesn't mean the thing that is at
that limit, has the same value.
Of course not. But if each step of two sequences has the same value,
then the limits are the same too. This is the case for
(E(1)∩E(2)∩...∩E(n)) and (E(n)).
But the limit of the sequence isn't necessary what is at the "end" of
the sequence.
On 12/9/2024 4:41 PM, WM wrote:
The limit set is the same
for both sequences.
(E(1)∩E(2)∩...∩E(n)) and (E(n))
In order to stop tricksters
we go without cardinality.
We need only the sequence of sets with the empty set in the limit.
The limit set is the same
for both sequences.
(E(1)∩E(2)∩...∩E(n)) and (E(n))
In order to stop tricksters
we go without cardinality.
What you (WM) mean by "going without cardinality"
is
pretending that
a sequence of sets
and
the sequence of the cardinalities of those sets
are the same thing.
Which, for end.segments of final.cardinals,
they are not the same.
#⋂{E(i):i} = #{} = 0
⋂{#E(i):i} = ⋂{ℵ₀:i} = ℵ₀
Le 09/12/2024 à 22:50, WM a écrit :
The very core property of analysis is that equal sequences have
equal limits if they have limits at all.
E(1)∩E(2)∩...∩E(n) = E(n)
Lim E(1)∩E(2)∩...∩E(n) = {}
Lim E(n) = {}
They are equal.
Not in a set theory where every endsegment is infinite.
Nonsense.
I once asked a whole classroom what they think of your claims, providing
your posts/site as well as Ben's paper : http://bsb.me.uk/dd-wealth.pdf.
Le 11/12/2024 à 16:51, Crank Wolfgang Mückenheim from Hochschuleelementary school?
I once asked a whole classroom what they think of your claims,
Ben Bacarisse's example : To keep things simple, let's number the
notes 1; 2; 3; : : : and we'll give McDuck only $2 a day. One has to
be returned a day.
He accuses me to confuse lim card and card lim and to use the wrong
limit.
He is right.
On 11.12.2024 05:06, Jim Burns wrote:
Which, for end.segments of final.cardinals,
they are not the same.
#⋂{E(i):i} = #{} = 0
⋂{#E(i):i} = ⋂{ℵ₀:i} = ℵ₀
There is no use for # but only for
⋂{E(i):i} = {}.
The problem is that
every definable endsegment is infinite.
But if all natural numbers are applied
in a Cantor bijection,
then they all must leave the endsegments
by the only possible process:
E(k+1) = E(k) \ {k}
⋂{E(i):i} = {}.
Therefore,
one.element.emptier ℕ\{0}
is not.smaller.than ℕ
That is the reason that
the end.segments of ℕ
stay the same larger.than.any.finite size
as they lose each element of ℕ.
On 11.12.2024 20:27, Jim Burns wrote:
⋂{E(i):i} = {}.
Of course. But
all intersections with finite contents
are invisible.
Therefore,
one.element.emptier ℕ\{0}
is not.smaller.than ℕ
It is a smaller set.
Cardinalities are not useful.
... you are the fool.
On 12/11/2024 2:57 PM, WM wrote:
On 11.12.2024 20:27, Jim Burns wrote:
⋂{E(i):i} = {}.
Of course. But
all intersections with finite contents
are invisible.
We know about what's invisible by
assembling finite sequences holding only
claims which are true.or.not.first.false.
We know that
each claim in the claim.sequence is true
by _looking at the claims_
independently of _looking at the invisible_
_It doesn't matter_
whether any finite.cardinals are invisible.
Each finite cardinal is finite, and
that is enough to start
a finite sequence of claims holding only
claims which are true.or.not.first.false
-- claims about each finite.cardinal, visible or not.
Some claims seem too dull to need verifying.
"Is a finite.cardinal finite?"
Better to ask "Is the Pope Catholic?"
But such obviously.true claims start us off.
Other claims, the more interesting claims,
can be verified as not.first.false
_by looking at the claims_
NOT by looking at finite.cardinals
Look at q in ⟨p p⇒q q⟩
There is no way in which q can be first.false.
It doesn't matter what q means, or what p means.
We can see q is not.first.false in that sequence.
Repeat the pattern ⟨p p⇒q q⟩ and a few others
for a whole finite sequence of claims,
and
that whole finite sequences of claims
holds no first false claim,
and thus holds no false claim.
Which we know by _looking at the claims_
Therefore,
one.element.emptier ℕ\{0}
is not.smaller.than ℕ
It is a smaller set.
For each k in ℕ
there is unique k+1 in ℕ\{0}
Cardinalities are not useful.
And yet, by ignoring them,
you (WM) end up wrong about
⎛ For each k in ℕ
⎝ there is unique k+1 in ℕ\{0}
On 11.12.2024 21:58, Jim Burns wrote:
On 12/11/2024 2:57 PM, WM wrote:
On 11.12.2024 20:27, Jim Burns wrote:
Each "leaves" by
not.being.in.common.with.all.end.segments.
∀k ∈ ℕ:
k ∉ E(k)\{k} = E(k+1) ⊇ ⋂{E(i):i} ∌ k
⋂{E(i):i} = {}.
Therefore,
one.element.emptier ℕ\{0}
is not.smaller.than ℕ
It is a smaller set.
For each k in ℕ
there is unique k+1 in ℕ\{0}
Cardinalities are not useful.
And yet, by ignoring them,
you (WM) end up wrong about
⎛ For each k in ℕ
⎝ there is unique k+1 in ℕ\{0}
Is the complete removal of natural numbers
from the sequence of intersections
bound by the law
∀k ∈ ℕ :
∩{E(1), E(2), ..., E(k+1)} =
∩{E(1), E(2), ..., E(k)} \ {k}
or not?
On 11.12.2024 03:04, Richard Damon wrote:
On 12/10/24 12:30 PM, WM wrote:
On 10.12.2024 13:17, Richard Damon wrote:
On 12/10/24 3:50 AM, WM wrote:
Two sequences that are identical term by term cannot have different
limits. 0^x and x^0 are different term by term.
Which isn't the part I am talking of, it is that just because each
step of a sequence has a value, doesn't mean the thing that is at
that limit, has the same value.
Of course not. But if each step of two sequences has the same value,
then the limits are the same too. This is the case for
(E(1)∩E(2)∩...∩E(n)) and (E(n)).
But the limit of the sequence isn't necessary what is at the "end" of
the sequence.
The end of the sequence is defined by ∀k ∈ ℕ : E(k+1) = E(k) \ {k}.
Regards, WM
On 12/11/24 9:32 AM, WM wrote:
On 11.12.2024 03:04, Richard Damon wrote:
On 12/10/24 12:30 PM, WM wrote:
On 10.12.2024 13:17, Richard Damon wrote:
On 12/10/24 3:50 AM, WM wrote:
Two sequences that are identical term by term cannot have
different limits. 0^x and x^0 are different term by term.
Which isn't the part I am talking of, it is that just because each
step of a sequence has a value, doesn't mean the thing that is at
that limit, has the same value.
Of course not. But if each step of two sequences has the same value,
then the limits are the same too. This is the case for
(E(1)∩E(2)∩...∩E(n)) and (E(n)).
But the limit of the sequence isn't necessary what is at the "end" of
the sequence.
The end of the sequence is defined by ∀k ∈ ℕ : E(k+1) = E(k) \ {k}.
None of which are an infinite sets, so trying to take a "limit" of
combining them is just improper.
On 12/11/2024 4:53 PM, WM wrote:
Is the complete removal of natural numbers from the sequence of
intersections
bound by the law
∀k ∈ ℕ :
∩{E(1), E(2), ..., E(k+1)} =
∩{E(1), E(2), ..., E(k)} \ {k}
or not?
Each "leaves" by
not.being.in.common.with.all.end.segments.
∀k ∈ ℕ:
k ∉ E(k)\{k} = E(k+1) ⊇ ⋂{E(i):i} ∌ k
⋂{E(i):i} = {}.
On 12.12.2024 01:32, Richard Damon wrote:
On 12/11/24 9:32 AM, WM wrote:
On 11.12.2024 03:04, Richard Damon wrote:
On 12/10/24 12:30 PM, WM wrote:
On 10.12.2024 13:17, Richard Damon wrote:
On 12/10/24 3:50 AM, WM wrote:
Two sequences that are identical term by term cannot have
different limits. 0^x and x^0 are different term by term.
Which isn't the part I am talking of, it is that just because each >>>>>> step of a sequence has a value, doesn't mean the thing that is at
that limit, has the same value.
Of course not. But if each step of two sequences has the same
value, then the limits are the same too. This is the case for
(E(1)∩E(2)∩...∩E(n)) and (E(n)).
But the limit of the sequence isn't necessary what is at the "end"
of the sequence.
The end of the sequence is defined by ∀k ∈ ℕ : E(k+1) = E(k) \ {k}.
None of which are an infinite sets, so trying to take a "limit" of
combining them is just improper.
Most endsegments are infinite. But if Cantor can apply all natural
numbers as indices for his sequences, then all must leave the sequence
of endsegments. Then the sequence (E(k)) must end up empty. And there
must be a continuous staircase from E(k) to the empty set.
Regards, WM
On 12.12.2024 01:32, Richard Damon wrote:It makes no sense not being able to „apply” numbers. Clearly Cantor does. The sequence IS continuous. It’s just that you misconceive of the
On 12/11/24 9:32 AM, WM wrote:
On 11.12.2024 03:04, Richard Damon wrote:
On 12/10/24 12:30 PM, WM wrote:
On 10.12.2024 13:17, Richard Damon wrote:
On 12/10/24 3:50 AM, WM wrote:
Two sequences that are identical term by term cannot have
different limits. 0^x and x^0 are different term by term.
Which isn't the part I am talking of, it is that just because each >>>>>> step of a sequence has a value, doesn't mean the thing that is at
that limit, has the same value.
Of course not. But if each step of two sequences has the same value, >>>>> then the limits are the same too. This is the case for
(E(1)∩E(2)∩...∩E(n)) and (E(n)).
But the limit of the sequence isn't necessary what is at the "end" of
the sequence.
The end of the sequence is defined by ∀k ∈ ℕ : E(k+1) = E(k) \ {k}. The sequence is endless, has no end, is infinite.
None of which are an infinite sets, so trying to take a "limit" of
combining them is just improper.
Most endsegments are infinite. But if Cantor can apply all natural
numbers as indices for his sequences, then all must leave the sequence
of endsegments. Then the sequence (E(k)) must end up empty. And there
must be a continuous staircase from E(k) to the empty set.
WM wrote :
Most endsegments are infinite.
Many endsegments are infinite too.
On 12/12/24 4:12 AM, WM wrote:
Note, "inifinite" isn't a Natural Number,
or a Real Number, so NO
segement, specified by values, can have an "infinte endsegment".
On 12.12.2024 12:30, FromTheRafters wrote:They are all infinite (there are no „dark” numbers).
WM wrote :
Most endsegments are infinite. But if Cantor can apply all naturalMost endsegments are infinite.Many endsegments are infinite too.
numbers as indices for his bijections, then all must leave the sequence
of endsegments. Then the sequence (E(k)) must end up empty. And there
must be a continuous staircase from E(k) to the empty set.
On 12.12.2024 15:23, joes wrote:For you: no, an infinite set cannot be exhausted in finite steps.
Am Thu, 12 Dec 2024 10:12:26 +0100 schrieb WM:
If a bijection with ℕ is possible, the sequence can be exhausted so thatThe end of the sequence is defined by ∀k ∈ ℕ : E(k+1) = E(k) \ {k}. >> The sequence is endless, has no end, is infinite.
no natural numbers remains in an endsegment.
As you haven’t disproven.He claims it. That means no numbers remain unpaired in endsegments.It makes no sense not being able to „apply” numbers. Clearly CantorNone of which are an infinite sets, so trying to take a "limit" ofMost endsegments are infinite. But if Cantor can apply all natural
combining them is just improper.
numbers as indices for his sequences, then all must leave the sequence
of endsegments. Then the sequence (E(k)) must end up empty. And there
must be a continuous staircase from E(k) to the empty set.
does.
The limit is only reachable… at infinity.The sequence IS continuous. It’s just that you misconceive of the limitCantor does. If the limit is not reachable, then complete bijections
as reachable.
cannot be established.
"If we think the numbers p/q in such an order [...] then every numberTrue: no number is in all endsegments.
p/q comes at an absolutely fixed position of a simple infinite sequence"
[E. Zermelo: "Georg Cantor – Gesammelte Abhandlungen mathematischen und philosophischen Inhalts", Springer, Berlin (1932) p. 126]
"The infinite sequence thus defined has the peculiar property to contain
the positive rational numbers completely, and each of them only once at
a determined place." [G. Cantor, letter to R. Lipschitz (19 Nov 1883)]
If you accept these claims, then no number must remain in an endsegment.
On 11.12.2024 23:28, Jim Burns wrote:
On 12/11/2024 4:53 PM, WM wrote:
Is the complete removal of natural numbers
from the sequence of intersections
bound by the law
∀k ∈ ℕ :
∩{E(1), E(2), ..., E(k+1)} =
∩{E(1), E(2), ..., E(k)} \ {k}
or not?
Each "leaves" by
not.being.in.common.with.all.end.segments.
That means no natural number remains
in all endsegments.
But every endsegment has only
one number less than its predecessor.
This closes the bridge between
infinite intersections of endsegments and
the empty intersection of endsegments.
∀k ∈ ℕ:
k ∉ E(k)\{k} = E(k+1) ⊇ ⋂{E(i):i} ∌ k
⋂{E(i):i} = {}.
There must be
a continuous sequence of steps of height 1
from many elements to none.
There must be
a continuous sequence of steps of height 1
from many elements to none.
Can you confirm this?
On 12/12/2024 4:07 AM, WM wrote:
The existence of a bridge implies that,
somewhere we can't see,
a size which cannot change by 1
changes by 1 to
a size which can change by 1
There must be
a continuous sequence of steps of height 1
from many elements to none.
What you describe is many, not infinity.
Can you confirm this?
For many, yes,
For infinity, never.
Having a continuous sequence of steps of height 1
from many elements to none
is what makes it finite.
On 12.12.2024 13:26, Richard Damon wrote:
On 12/12/24 4:12 AM, WM wrote:
Note, "inifinite" isn't a Natural Number,
But a state,often dscribed byan infinite whole number.
or a Real Number, so NO segement, specified by values, can have an
"infinte endsegment".
Every infinite segment is an infinite segment and therefore has an
infinite segment as subset. But if Cantor can apply all natural numbers
as indices for his bijections, then all must leave the sequence of endsegments. Then the sequence (E(k)) must end up empty. And there must
be a continuous staircase from E(k) to the empty set.
Regards, WM
On 12.12.2024 17:15, Jim Burns wrote:
The existence of a bridge implies that,
somewhere we can't see,
a size which cannot change by 1
changes by 1 to
a size which can change by 1
Every set can change by one element.
Every set can change by one element.
No size is required and no size is possible
if this is forbidden.
Am Thu, 12 Dec 2024 15:23:04 +0100 schrieb WM:
Most endsegments are infinite. But if Cantor can apply all naturalThey are all infinite
numbers as indices for his bijections, then all must leave the sequence
of endsegments. Then the sequence (E(k)) must end up empty. And there
must be a continuous staircase from E(k) to the empty set.
Am Thu, 12 Dec 2024 15:33:20 +0100 schrieb WM:
On 12.12.2024 15:23, joes wrote:For you: no, an infinite set cannot be exhausted in finite steps.
Am Thu, 12 Dec 2024 10:12:26 +0100 schrieb WM:If a bijection with ℕ is possible, the sequence can be exhausted so that >> no natural numbers remains in an endsegment.
The sequence is endless, has no end, is infinite.The end of the sequence is defined by ∀k ∈ ℕ : E(k+1) = E(k) \ {k}.
"If we think the numbers p/q in such an order [...] then every number
p/q comes at an absolutely fixed position of a simple infinite sequence"
[E. Zermelo: "Georg Cantor – Gesammelte Abhandlungen mathematischen und
philosophischen Inhalts", Springer, Berlin (1932) p. 126]
"The infinite sequence thus defined has the peculiar property to contain
the positive rational numbers completely, and each of them only once at
a determined place." [G. Cantor, letter to R. Lipschitz (19 Nov 1883)]
If you accept these claims, then no number must remain in an endsegment.True: no number is in all endsegments.
False: there is a segment with no numbers.
WM was thinking very hard :
On 12.12.2024 12:30, FromTheRafters wrote:
WM wrote :
But if Cantor can apply all natural numbers as indices for his
bijections, then all must leave the sequence of endsegments. Then the
sequence (E(k)) must end up empty.
Impossible, since k and k+1 etcetera are natural numbers.
And there must be a continuous staircase from E(k) to the empty set.
A limit set or element does not have to be completely within a set or interval.
On 12/12/2024 2:11 PM, WM wrote:
Again, Cantor Pairing works with any natural number. Not just many of
them... ALL of them.
After serious thinking WM wrote :
On 12.12.2024 13:26, Richard Damon wrote:
On 12/12/24 4:12 AM, WM wrote:
Note, "inifinite" isn't a Natural Number,
But a state,often dscribed byan infinite whole number.
Infinite whole number? Sheeesh!!
On 12/12/24 9:25 AM, WM wrote:
if Cantor can apply all naturalBut a segment that is infinite in length is, by definiton, missing at
numbers as indices for his bijections, then all must leave the
sequence of endsegments. Then the sequence (E(k)) must end up empty.
And there must be a continuous staircase from E(k) to the empty set.
least on end.
So, which bijection from Cantor are you talking about? Of are you
working on a straw man that Cantor never talked about?
"erhalten wir nicht bloß eine einzige unendliche ganze Zahl, sondern
eine unendliche Folge von solchen, die voneinander wohl unterschieden
sind und in gesetzmäßigen zahlentheoretischen Beziehungen zueinander
sowohl wie zu den endlichen ganzen Zahlen stehen."
On 12/12/2024 5:11 PM, WM wrote:
On 12.12.2024 17:15, Jim Burns wrote:
The existence of a bridge implies that,
somewhere we can't see,
a size which cannot change by 1
changes by 1 to
a size which can change by 1
Every set can change by one element.
No size is required and no size is possible
if this is forbidden.
Ignoring cardinality changes nothing
about f(k) = k∪{k} and ⟦0,w⦆ and ⟦1,w⦆
On 12/12/2024 6:23 AM, WM wrote:
WM wrote :
Most endsegments are infinite.
Most endsegments <bla>
On 13.12.2024 03:29, Richard Damon wrote:It changes the domain from D to N. What operation is „attachment”?
On 12/12/24 9:44 AM, WM wrote:
On 12.12.2024 13:26, Richard Damon wrote:
On 12/12/24 4:53 AM, WM wrote:
On 12.12.2024 01:38, Richard Damon wrote:
On 12/11/24 9:04 AM, WM wrote:It is precisely this.
In mathematics, a set A is Dedekind-infinite (named after the
German mathematician Richard Dedekind) if some proper subset B of >>>>>>> A is equinumerous to A. [Wikipedia].
So? That isn't what Cantor was talking about in his pairings
No, Cantors pairing is between two SETS, not a set and its subset.
Yes, we can call the subset a set, since it is, but then when we look
at it for the pairing, we need to be looking at its emancipated
version, not the version tied into the original set.
Both is the same. In emancipated version it is not as obvious as in
the subset version.
Nope, when the subset is considered as its own independent set, the
operation you want to do isn't part of its operations.
The subset is considered as its own independent set D = {10n | n ∈ ℕ}
and then it is attached to the set ℕ = {1, 2, 3, ...}. R´That does not change the subset.
On 12/13/24 4:57 AM, WM wrote:
"erhalten wir nicht bloß eine einzige unendliche ganze Zahl, sondern
eine unendliche Folge von solchen, die voneinander wohl unterschieden
sind und in gesetzmäßigen zahlentheoretischen Beziehungen zueinander
sowohl wie zu den endlichen ganzen Zahlen stehen."
Which, if I understand the context he says this in, are the transfinite numbers.
The "infinite integers" are not claimes to be part of the Natural
Numbers, but of the extended set of numbers created by the rules of transfinite mathematics.
Am 12.12.2024 um 23:03 schrieb Chris M. Thomasson:
On 12/12/2024 6:23 AM, WM wrote:
WM wrote :
Most endsegments are infinite.
_All_ endsegments are infinite.
Am Fri, 13 Dec 2024 09:42:36 +0100 schrieb WM:You can also say pairing. The elements of D are paired with the elements
The subset is considered as its own independent set D = {10n | n ∈ ℕ}It changes the domain from D to N. What operation is „attachment”?
and then it is attached to the set ℕ = {1, 2, 3, ...}. That does not
change the subset.
On 13.12.2024 13:12, Moebius wrote:lol. Infinite segments don’t exist?
Am 12.12.2024 um 23:03 schrieb Chris M. Thomasson:Maybe. But then there are no endsegments at all, and Cantors theory is nonsense from the scratch.
On 12/12/2024 6:23 AM, WM wrote:_All_ endsegments are infinite.
WM wrote :
Most endsegments are infinite.
On 13.12.2024 15:41, joes wrote:Aha, and what is paired with the rest of N? This is not the identity.
Am Fri, 13 Dec 2024 09:42:36 +0100 schrieb WM:
You can also say pairing. The elements of D are paired with the elementsThe subset is considered as its own independent set D = {10n | n ∈It changes the domain from D to N. What operation is „attachment”?
ℕ}
and then it is attached to the set ℕ = {1, 2, 3, ...}. That does not
change the subset.
10n of ℕ. After this small detour everything proceeds as usual.
On 13.12.2024 06:06, Jim Burns wrote:
On 12/12/2024 5:11 PM, WM wrote:
On 12.12.2024 17:15, Jim Burns wrote:
The existence of a bridge implies that,
somewhere we can't see,
a size which cannot change by 1
changes by 1 to
a size which can change by 1
Every set can change by one element.
No size is required and no size is possible
if this is forbidden.
All sets have one.emptier and
one.fuller counterparts.
Some sets cannot match
their one.emptier and one.fuller counterparts.
We call them finite sets.
Other sets, which aren't finite, can match
their one.emptier and one.fuller counterparts.
We call them infinite sets.
Ignoring cardinality changes nothing
about f(k) = k∪{k} and ⟦0,w⦆ and ⟦1,w⦆
Ignoring that Cantor's claim requires to
empty the endsegments from all natural numbers
in order to use them as indices in mappings
(ℕ to the set of endsegments, ℕ to ℚ,
ℕ to the lines of the Cantor list
for "proving" the uncountability of ℝ)
shows inconsistency.
Am Fri, 13 Dec 2024 18:12:28 +0100 schrieb WM:
On 13.12.2024 13:12, Moebius wrote:lol. Infinite segments don’t exist?
Am 12.12.2024 um 23:03 schrieb Chris M. Thomasson:Maybe. But then there are no endsegments at all, and Cantors theory is
On 12/12/2024 6:23 AM, WM wrote:_All_ endsegments are infinite.
WM wrote :
Most endsegments are infinite.
nonsense from the scratch.
Am Fri, 13 Dec 2024 18:20:59 +0100 schrieb WM:
On 13.12.2024 15:41, joes wrote:Aha, and what is paired with the rest of N? This is not the identity.
> Am Fri, 13 Dec 2024 09:42:36 +0100 schrieb WM:
>> The subset is considered as its own independent set D = {10n | n ∈
>> ℕ}
>> and then it is attached to the set ℕ = {1, 2, 3, ...}. That does not >> >> change the subset.
> It changes the domain from D to N. What operation is „attachment”? >> You can also say pairing. The elements of D are paired with the elements
10n of ℕ. After this small detour everything proceeds as usual.
I was under the impression the bijection step came later.
On 12/13/2024 6:25 AM, WM wrote:
Each finite.cardinal
⎛ is first in an end.segment.
⎝ is used as a index of an end.segment.
Each finite.cardinal
⎛ is absent from an end segment.
⎝ is emptied from the end.segments.
(ℕ to the set of endsegments, ℕ to ℚ,
ℕ to the lines of the Cantor list
for "proving" the uncountability of ℝ)
shows inconsistency.
Each set has
its emptier.by.one and fuller.by.one counterparts.
On 12/13/2024 2:55 AM, WM wrote:
But if <bla>
when I say "all" I am not implying some sort of largest natural number.
On 12/10/2024 1:12 PM, Moebius wrote:
Am 10.12.2024 um 22:09 schrieb Chris M. Thomasson:
Shit. So, any natural number that WM cannot think of is dark? I guess.
Who knows? It's complicated. :-P
:-)
Yikes!
On 13.12.2024 13:11, Richard Damon wrote:
On 12/13/24 4:57 AM, WM wrote:
"erhalten wir nicht bloß eine einzige unendliche ganze Zahl, sondern
eine unendliche Folge von solchen, die voneinander wohl unterschieden
sind und in gesetzmäßigen zahlentheoretischen Beziehungen zueinander
sowohl wie zu den endlichen ganzen Zahlen stehen."
Which, if I understand the context he says this in, are the
transfinite numbers.
The "infinite integers" are not claimes to be part of the Natural
Numbers, but of the extended set nof numbers created by the rules of
transfinite mathematics.
In German he does not say integer but whole number. Of course they are
not natural numbers, but numbers without decimal point.
Regards, WM
On 13.12.2024 20:00, Jim Burns wrote:
On 12/13/2024 6:25 AM, WM wrote:
Ignoring that Cantor's claim requires to
empty the endsegments from all natural numbers
in order to use them as indices in mappings
Each finite.cardinal
⎛ is first in an end.segment.
⎝ is used as a index of an end.segment.
Each finite.cardinal
⎛ is absent from an end segment.
⎝ is emptied from the end.segments.
Require at will, sir.
If endegments were defined as
E(n) = {n+1, n+2, ...}:
E(0) = {1, 2, 3, ...}
E(1) = {2, 3, 4, ...}
E(2) = {3, 4, 5, ...}
...
E(ω-1} = { }.
Then this change from content to index
would even be more obvious.
Each set has
its emptier.by.one and fuller.by.one counterparts.
If all natnumbers can be used for mappings as indices
then every natnumber has to leave the sequence of endsegments.
Do you agree?
Therefore their intersection is empty.
Do you agree?
Emptying by one only implies
finite endsegment intersetcions.
Do you agree?
If not
describe the process according to your opinion.
On 13.12.2024 19:29, joes wrote:I mean, you apply your function before the bijection, but then also
Am Fri, 13 Dec 2024 18:20:59 +0100 schrieb WM:No, it came earlier and it was no bijecton.
On 13.12.2024 15:41, joes wrote:Aha, and what is paired with the rest of N? This is not the identity.
> Am Fri, 13 Dec 2024 09:42:36 +0100 schrieb WM:
>> The subset is considered as its own independent set D = {10n | n
>> ∈ ℕ}
>> and then it is attached to the set ℕ = {1, 2, 3, ...}. That does
>> not change the subset.
> It changes the domain from D to N. What operation is „attachment”? >>> You can also say pairing. The elements of D are paired with the
elements 10n of ℕ. After this small detour everything proceeds as
usual.
I was under the impression the bijection step came later.
On 13.12.2024 03:23, Richard Damon wrote:Nah. Just imagine all of the inf.many steps as a whole - y’know, ACTUAL infinity.
On 12/12/24 9:25 AM, WM wrote:
That means that the premise "if Cantor can apply all natural numbers as indices for his bijections" is false.if Cantor can apply all natural numbers as indices for his bijections,But a segment that is infinite in length is, by definiton, missing at
then all must leave the sequence of endsegments. Then the sequence
(E(k)) must end up empty.
And there must be a continuous staircase from E(k) to the empty set.
least on end.
Then the list were finite. It isn’t, though.So, which bijection from Cantor are you talking about? Of are you
working on a straw man that Cantor never talked about?
There are many. The mapping from natumbers to the rationals, for
instance, needs all natural numbers. That means all must leave the endsegments. Another example is Cantor's list "proving" uncountable
sets. If not every natural number has left the endsegment and is applied
as an index of a line of the list, the list is useless.
But if every natural number has left the endsegments, then theYes.
intersection of all endsegments is empty.
Then the infinite sequence ofNo. It is literally „without an end”, and yet can be „completed”, if only you were able to conceive of infinity.
endegments has a last term (and many finite predecessors, because of ∀k
∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}).
On 12.12.2024 13:26, Richard Damon wrote:Yes, well.
On 12/12/24 4:12 AM, WM wrote:
Note, "inifinite" isn't a Natural Number,But a state,often dscribed byan infinite whole number.
or a Real Number, so NO segement, specified by values, can have anEvery infinite segment is an infinite segment
"infinte endsegment".
and therefore has an infinite segment as subset.Getting close here…
But if Cantor can apply all natural numbersYes, in the limit. There is no natural k such that E(k) is empty.
as indices for his bijections, then all must leave the sequence of endsegments. Then the sequence (E(k)) must end up empty.
And there must be a continuous staircase from E(k) to the empty set.This is where you are mistaken. How do you get from naturals to
On 12/13/2024 2:31 PM, WM wrote:
On 13.12.2024 20:00, Jim Burns wrote:
On 12/13/2024 6:25 AM, WM wrote:
Ignoring that Cantor's claim requires to
empty the endsegments from all natural numbers
in order to use them as indices in mappings
Each finite.cardinal
⎛ is first in an end.segment.
⎝ is used as a index of an end.segment.
Each finite.cardinal
⎛ is absent from an end segment.
⎝ is emptied from the end.segments.
Require at will, sir.
If endegments were defined as
E(n) = {n+1, n+2, ...}:
E(0) = {1, 2, 3, ...}
E(1) = {2, 3, 4, ...}
E(2) = {3, 4, 5, ...}
...
E(ω-1} = { }.
Then this change from content to index
would even be more obvious.
The index of Eᑉ(2) is content of Eᑉ(1), etc.
The number doesn't change.
Which after.segment changes.
----
One problem which
Eᑉ(ω-1) = {}
has is that
'finite' is NOT defined the way in which
you (WM) think 'finite' should be,
which means
ω is NOT defined the way in which
you (WM) think ω should be.
Which means
⎛ if k is finite
⎝ then k+1 is finite
Therefore,
k ∈ ⟦0,ω⦆ ⇒ k+1 ∈ ⟦0,ω⦆
If all natnumbers can be used for mappings as indices
then every natnumber has to leave the sequence of endsegments.
Do you agree?
Yes.
Therefore their intersection is empty.
Do you agree?
Yes.
Emptying by one only implies
finite endsegment intersetcions.
Do you agree?
No.
If not describe the process according to your opinion.
Each end.segment ⟦k,ω⦆ of ⟦0,ω⦆ contains,
for each finite.cardinal j
a subset ⟦k,k+j⟧ holding more.than.j.many.
That contradicts |⟦k,ω⦆| being finite.
On 10.12.2024 23:37, joes wrote:Like the quantifier says. But what is the limit?
You (WM) are considering infinite dark.finite.cardinals,Then analysis is contradicted in set theory.
which do not exist.
∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n).
Am Mon, 09 Dec 2024 22:28:41 +0100 schrieb WM:All sets are claimed to be infinite.
>> The limit of the left-hand side is empty, the limit of the
>> right-hand side is full, i.e. not empty.
>> I do not tolerate that.
> What is the RHS limit?
Yes, they can. That’s what the limit means - taken to infinity.All sets are claimed to be infinite. A sequence of infinitely manyThere is no limit in set theory, contrary to the LHS limit { }.Why not for the RHS?
numbers cannot lose all numbers "in the limit"
On 12.12.2024 18:29, Jeff Barnett wrote:YOU try to map N onto itself.
On 12/12/2024 6:59 AM, joes wrote:Dedekind maps the elements of a subset to the elements of its superset.
Am Tue, 10 Dec 2024 18:01:04 +0100 schrieb WM:But that's sort of what mappings are for! Aren't they?
On 10.12.2024 13:19, Richard Damon wrote:What Richard meant: do not confuse the set being mapped with the one
The pairing is between TWO sets, not the members of a set withThe pairing is between the elements. Otherwise you could pair R and Q
itself.
by simply claiming it.
"The infinite sequence thus defined has the peculiar property to
contain the positive rational numbers completely, and each of them
only once at a determined place." [Cantor] Note the numbers, not the
set.
being mapped onto.
Same do I.
On 12.12.2024 14:59, joes wrote:You change the domain from D to N.
What Richard meant: do not confuse the set being mapped with the oneI don't. D = {10n | n ∈ ℕ} is the set being mapped. The set D being mapped does not change when it is attached to the set ℕ being mapped in form of black hats.
being mapped onto.
On 08.12.2024 11:43, joes wrote:No, it says nothing about the intersection of all E(n),
Am Sat, 07 Dec 2024 22:50:27 +0100 schrieb WM:
It talks about all n and therefore about all E(n).∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n)That sentence only talks about a single n at a time, though,
All n are infinitely many.
not about the infinite intersection.
Am Fri, 13 Dec 2024 11:55:16 +0100 schrieb WM:
Just imagine all of the inf.many steps as a whole - y’know, ACTUAL infinity.
If not every natural number has left the endsegment and is appliedThen the list were finite. It isn’t, though.
as an index of a line of the list, the list is useless.
But if every natural number has left the endsegments, then theYes.
intersection of all endsegments is empty.
Then the infinite sequence ofNo. It is literally „without an end”,
endegments has a last term (and many finite predecessors, because of ∀k
∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}).
and yet can be „completed”, if
only you were able to conceive of infinity.
Am Thu, 12 Dec 2024 15:25:59 +0100 schrieb WM:
But if Cantor can apply all natural numbersYes, in the limit. There is no natural k such that E(k) is empty.
as indices for his bijections, then all must leave the sequence of
endsegments. Then the sequence (E(k)) must end up empty.
And there must be a continuous staircase from E(k) to the empty set.This is where you are mistaken.
How do you get from naturals to
finite dark numbers anyway?
Am Sun, 08 Dec 2024 11:55:34 +0100 schrieb WM:
On 08.12.2024 11:43, joes wrote:No, it says nothing about the intersection of all E(n),
Am Sat, 07 Dec 2024 22:50:27 +0100 schrieb WM:It talks about all n and therefore about all E(n).
∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n)That sentence only talks about a single n at a time, though,
All n are infinitely many.
not about the infinite intersection.
only about finite intersections
On 13.12.2024 03:29, Richard Damon wrote:It changes the set from D to N and thus the function.
On 12/12/24 9:44 AM, WM wrote:The subset is considered as its own independent set D = {10n | n ∈ ℕ}
On 12.12.2024 13:26, Richard Damon wrote:Nope, when the subset is considered as its own independent set, the
On 12/12/24 4:53 AM, WM wrote:Both is the same. In emancipated version it is not as obvious as in
On 12.12.2024 01:38, Richard Damon wrote:No, Cantors pairing is between two SETS, not a set and its subset.
On 12/11/24 9:04 AM, WM wrote:It is precisely this.
In mathematics, a set A is Dedekind-infinite (named after theSo? That isn't what Cantor was talking about in his pairings
German mathematician Richard Dedekind) if some proper subset B of >>>>>>> A is equinumerous to A. [Wikipedia].
Yes, we can call the subset a set, since it is, but then when we look
at it for the pairing, we need to be looking at its emancipated
version, not the version tied into the original set.
the subset version.
operation you want to do isn't part of its operations.
and then it is attached to the set ℕ = {1, 2, 3, ...}. R´That does not change the subset.
On 13.12.2024 03:29, Richard Damon wrote:Why do you want to map N\D to N?
On 12/12/24 4:57 PM, WM wrote:
No such element can be named. But 9/10 of all ℕ cannot get mappedD = {10n | n ∈ ℕ} is the set being mapped. The set D being mapped does >>> not change when it is attached to the set ℕ being mapped in form ofAnd so, which element of which set didn't get mapped to a member of the
black hats.
other by the defined mapping?
because the limit of the constant sequence 1/9, 1/9, 1/9, ... is 1/9.
This proves the existence of numbers which cannot be named.
Am Fri, 13 Dec 2024 09:54:12 +0100 schrieb WM:
On 13.12.2024 03:29, Richard Damon wrote:Why do you want to map N\D to N?
On 12/12/24 4:57 PM, WM wrote:No such element can be named. But 9/10 of all ℕ cannot get mapped
D = {10n | n ∈ ℕ} is the set being mapped. The set D being mapped does >>>> not change when it is attached to the set ℕ being mapped in form ofAnd so, which element of which set didn't get mapped to a member of the
black hats.
other by the defined mapping?
because the limit of the constant sequence 1/9, 1/9, 1/9, ... is 1/9.
This proves the existence of numbers which cannot be named.
Am Fri, 13 Dec 2024 09:42:36 +0100 schrieb WM:
The subset is considered as its own independent set D = {10n | n ∈ ℕ}It changes the set from D to N and thus the function.
and then it is attached to the set ℕ = {1, 2, 3, ...}. That does not
change the subset.
On 14.12.2024 16:22, joes wrote:Oh, so your prepended function IS the identity D->D after all.
Am Fri, 13 Dec 2024 09:42:36 +0100 schrieb WM:
It changes the set from D to a subset of ℕ with same number of elementsThe subset is considered as its own independent set D = {10n | n ∈ ℕ} >>> and then it is attached to the set ℕ = {1, 2, 3, ...}. That does notIt changes the set from D to N and thus the function.
change the subset.
and proves that it cannot be changed to ℕ.
On 14.12.2024 05:54, Jim Burns wrote:
On 12/13/2024 2:31 PM, WM wrote:
On 13.12.2024 20:00, Jim Burns wrote:
On 12/13/2024 6:25 AM, WM wrote:
If endegments were defined as
E(n) = {n+1, n+2, ...}:
E(0) = {1, 2, 3, ...}
E(1) = {2, 3, 4, ...}
E(2) = {3, 4, 5, ...}
...
E(ω-1} = { }.
Then this change from content to index
would even be more obvious.
One problem which
Eᑉ(ω-1) = {}
has is that
'finite' is NOT defined the way in which
you (WM) think 'finite' should be,
which means
ω is NOT defined the way in which
you (WM) think ω should be.
Don't say what not is.
Which means
⎛ if k is finite
⎝ then k+1 is finite
This contradicts the fact that
nothing remains but
every element can go only as a single.
E(ω-1} = { }.
Don't say what not is.
On 14.12.2024 16:20, joes wrote:
Am Fri, 13 Dec 2024 09:54:12 +0100 schrieb WM:
On 13.12.2024 03:29, Richard Damon wrote:Why do you want to map N\D to N?
On 12/12/24 4:57 PM, WM wrote:No such element can be named. But 9/10 of all ℕ cannot get mapped
D = {10n | n ∈ ℕ} is the set being mapped. The set D being mapped doesAnd so, which element of which set didn't get mapped to a member of the >>>> other by the defined mapping?
not change when it is attached to the set ℕ being mapped in form of >>>>> black hats.
because the limit of the constant sequence 1/9, 1/9, 1/9, ... is 1/9.
This proves the existence of numbers which cannot be named.
I don't. I show that it is impossible to map D to ℕ.
Regards, WM
On 12/14/2024 5:26 AM, WM wrote:
Don't say what not is.
For sets A and B with one.to.one A.to.B
B is not.smaller.than A
Don't say what not is.
If ω-1 exists
then
⎛ ω-1 is last.before.ω
⎜ ω is first bound of the finites
⎜ ω-1 is not any before.ω bound of the finites
⎜ ω-1 is not infinite (see [2])
⎜ ω-1 is finite
⎜ ω-1 is smaller.than (ω-1)+1, also finite (see [3])
⎜ (ω-1)+1 is smaller.than (ω-1)+2, also finite
⎜ ω bounds w-1, (ω-1)+1, (ω-1)+2
⎜ ω-1 < (ω-1)+1 < (ω-1)+2 ≤ ω
⎝ ω-1 is not last.before.ω
On 14.12.2024 05:54, Jim Burns wrote:
[...]
Explain your vision of the problem:
If
ℕ is a set, i.e.
if it is complete such that
all numbers can be used
for indexing sequences or in other mappings,
then
it can also be exhausted such that
no element remains.
Then
the set of what remains unused, i.e.,
of intersections of endsegments
(1) E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
loses all content.
Then,
by the law
(2) ∀k ∈ ℕ :
∩{E(1),E(2),...,E(k+1)} =
∩{E(1),E(2),...,E(k)}\{k}
the content must become finite.
Explain your vision of the problem:
On 14.12.2024 15:08, Richard Damon wrote:Oh, you mean including the pair (n, 10n) in the bijection. Note that
On 12/14/24 3:38 AM, WM wrote:To put a hat on n is to attach a hat to n.
On 14.12.2024 01:03, Richard Damon wrote:No, they are PAIR with elements of N.
On 12/13/24 12:00 PM, WM wrote:They are elements of D and become attached to elements of ℕ.
On 13.12.2024 13:11, Richard Damon wrote:No, the black hats are attached to the element of D, not N.
Note, the pairing is not between some elements of N that are alsoYes all elements of D, as black hats attached to the elements 10n of >>>>> ℕ, have to get attached to all elements of ℕ. There the simple shift >>>>> from 10n to n (division by 10) is applied.
in D, with other elements in N, but the elements of D and the
elements on N.
There is no operatation to "Attach" sets.
??? The bijection is not finite.Then deal with all infinitely many intervals [1, n].No, we are not forbiding "detailed" analysisThose who try to forbid the detailed analysis are dishonest swindlersThat pairs the elements of D with the elements of ℕ. Alas, it can be >>>>> proved that for every interval [1, n] the deficit of hats amounts to >>>>> at least 90 %. And beyond all n, there are no further hats.But we aren't dealing with intervals of [1, n] but of the full set.
and tricksters and not worth to participate in scientific discussion.
Only in the limit.The intervals [1, n] cover the full set.Why can't he? The problem is in the space of the full set, not theThe problem is that you can't GET to "beyond all n" in the pairing,If this is impossible, then also Cantor cannot use all n.
as there are always more n to get to.
finite sub sets.
There is the logic of the infinite.There is no other logic.Nope, it proves it is incompatible with finite logic.Yes, there are only 1/10th as many Black Hats as White Hats, butThis example proves that aleph_0 is nonsense.
since that number is Aleph_0/10, which just happens to also equal
Aleph_0, there is no "deficit" in the set of Natual Numbers.
On 12/14/2024 5:26 AM, WM wrote:
the set of what remains unused, i.e.,
of intersections of endsegments
(1) E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
loses all content.
Then,
by the law
(2) ∀k ∈ ℕ :
∩{E(1),E(2),...,E(k+1)} =
∩{E(1),E(2),...,E(k)}\{k}
the content must become finite.
Explain your vision of the problem:
A finite member ⟦0,ψ⦆ of the (well.ordered) ordinals
is smaller.than its successor ⟦0,ψ⦆∪{ψ}
If ⟦0,ψ⦆ is smaller than its successor ⟦0,ψ⦆∪{ψ}
then ⟦0,ψ+1⦆ = ⟦0,ψ⦆∪{ψ} is smaller.than
its successor ⟦0,ψ+1⦆∪{ψ+1}
which means
If ψ is finite, then ψ+1 is finite.
If ψ+1 is finite, then ψ+2 is finite.
ω is the first upper bound of finite ordinals.
If ψ < ω, then ψ < ψ+1 < ψ+2 ≤ ω
If ω-1 exists
then
ω-1 is last.before.ω
α < β < ω ⇒ α ≠ ω-1
If ω-1 exists
then
ω-1 < (ω-1)+1 < (ω-1)+2 ≤ ω
ω-1 ≠ ω-1
Therefore,
ω-1 doesn't exist
On 14.12.2024 23:04, Jim Burns wrote:
On 12/14/2024 5:26 AM, WM wrote:
the set of what remains unused, i.e.,
of intersections of endsegments
(1) E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
loses all content.
Then,
by the law
(2) ∀k ∈ ℕ :
∩{E(1),E(2),...,E(k+1)} =
∩{E(1),E(2),...,E(k)}\{k}
the content must become finite.
Explain your vision of the problem:
A finite member ⟦0,ψ⦆ of the (well.ordered) ordinals
is smaller.than its successor ⟦0,ψ⦆∪{ψ}
If ⟦0,ψ⦆ is smaller than its successor ⟦0,ψ⦆∪{ψ}
then ⟦0,ψ+1⦆ = ⟦0,ψ⦆∪{ψ} is smaller.than
its successor ⟦0,ψ+1⦆∪{ψ+1}
which means
If ψ is finite, then ψ+1 is finite.
If ψ+1 is finite, then ψ+2 is finite.
Yes, that is the potentially infinite collection of definable numbers.
But it explains nothing.
Not as a definable number. That is common knowledge. But you should not
ω is the first upper bound of finite ordinals.
If ψ < ω, then ψ < ψ+1 < ψ+2 ≤ ω
If ω-1 exists
then
ω-1 is last.before.ω
α < β < ω ⇒ α ≠ ω-1
If ω-1 exists
then
ω-1 < (ω-1)+1 < (ω-1)+2 ≤ ω
ω-1 ≠ ω-1
Therefore,
ω-1 doesn't exist
only say what not exists.
(1) E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ... loses all content.
By the law
(2) ∀k ∈ ℕ :
∩{E(1),E(2),...,E(k+1)} = ∩{E(1),E(2),...,E(k)}\{k}
the sequence gets empty one by one.
Regards, WM
On 15.12.2024 12:03, Mikko wrote:Shame.
On 2024-12-14 09:50:52 +0000, WM said:
On 14.12.2024 09:52, Mikko wrote:
On 2024-12-12 22:06:58 +0000, WM said:
We cannot name dark numbers as individuals.So you say that there is a natural number that does not have a nextis Dedekind-infinte:This "bijection" appears possible but it is not.
the successor function is a bijection between the set of all natural
numbers and non-zero natural numbers.
natural number. What number is that?
All numbers which can beAnd thus all n e N do.
used a individuals belong to a potentially infinite collection ℕ_def.
There is no firm end. When n belongs to ℕ_def, then also n+1 and 2n and n^n^n belong to ℕ_def.
The only common property is that all the numbersIf all successors belong to N_def, it can’t be finite and the
belong to a finite set and have an infinite set of dark successors.
This is the only way to explain that the intersection of endegmentsThe explanation is that the sequence is infinitely long.
E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
loses all content in a sequences which allow the loss of only one number
per step.
Am Sat, 14 Dec 2024 17:00:43 +0100 schrieb WM:
??? The bijection is not finite.Then deal with all infinitely many intervals [1, n].No, we are not forbiding "detailed" analysisThose who try to forbid the detailed analysis are dishonest swindlersThat pairs the elements of D with the elements of ℕ. Alas, it can be >>>>>> proved that for every interval [1, n] the deficit of hats amounts to >>>>>> at least 90 %. And beyond all n, there are no further hats.But we aren't dealing with intervals of [1, n] but of the full set.
and tricksters and not worth to participate in scientific discussion.
Only in the limit.The intervals [1, n] cover the full set.Why can't he? The problem is in the space of the full set, not theThe problem is that you can't GET to "beyond all n" in the pairing,If this is impossible, then also Cantor cannot use all n.
as there are always more n to get to.
finite sub sets.
There is the logic of the infinite.There is no other logic.Nope, it proves it is incompatible with finite logic.Yes, there are only 1/10th as many Black Hats as White Hats, butThis example proves that aleph_0 is nonsense.
since that number is Aleph_0/10, which just happens to also equal
Aleph_0, there is no "deficit" in the set of Natual Numbers.
Am Sun, 15 Dec 2024 12:33:15 +0100 schrieb WM:
On 15.12.2024 12:03, Mikko wrote:
All numbers which can beAnd thus all n e N do.
used a individuals belong to a potentially infinite collection ℕ_def.
There is no firm end. When n belongs to ℕ_def, then also n+1 and 2n and
n^n^n belong to ℕ_def.
The only common property is that all the numbersIf all successors belong to N_def, it can’t be finite and the
belong to a finite set and have an infinite set of dark successors.
successors can’t be dark.
This is the only way to explain that the intersection of endegmentsThe explanation is that the sequence is infinitely long.
E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
loses all content in a sequences which allow the loss of only one number
per step.
On 12/15/24 7:00 AM, WM wrote:
On 14.12.2024 23:04, Jim Burns wrote:
If ψ is finite, then ψ+1 is finite.
If ψ+1 is finite, then ψ+2 is finite.
Yes, that is the potentially infinite collection of definable numbers.
But it explains nothing.
That is the collection of numbers known as the Natural Numbers, so I
guess you are admitting that your "Definable Numbers" include *ALL* of
the Natural Numbers
On 14.12.2024 23:04, Jim Burns wrote:
On 12/14/2024 5:26 AM, WM wrote:
the set of what remains unused, i.e.,
of intersections of endsegments
(1) E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
loses all content.
Then,
by the law
(2) ∀k ∈ ℕ :
∩{E(1),E(2),...,E(k+1)} =
∩{E(1),E(2),...,E(k)}\{k}
the content must become finite.
Explain your vision of the problem:
A finite member ⟦0,ψ⦆ of the (well.ordered) ordinals
is smaller.than its successor ⟦0,ψ⦆∪{ψ}
If ⟦0,ψ⦆ is smaller than its successor ⟦0,ψ⦆∪{ψ}
then ⟦0,ψ+1⦆ = ⟦0,ψ⦆∪{ψ} is smaller.than
its successor ⟦0,ψ+1⦆∪{ψ+1}
which means
If ψ is finite, then ψ+1 is finite.
If ψ+1 is finite, then ψ+2 is finite.
Yes, that is
the potentially infinite collection of
definable numbers.
But it explains nothing.
Explain your vision of the problem:
ω is the first upper bound of finite ordinals.
If ψ < ω, then ψ < ψ+1 < ψ+2 ≤ ω
If ω-1 exists
then
ω-1 is last.before.ω
α < β < ω ⇒ α ≠ ω-1
If ω-1 exists
then
ω-1 < (ω-1)+1 < (ω-1)+2 ≤ ω
ω-1 ≠ ω-1
Therefore,
ω-1 doesn't exist
Not as a definable number.
That is common knowledge.
But you should not only say what not exists.
(1) E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
loses all content.
By the law
(2) ∀k ∈ ℕ :
∩{E(1),E(2),...,E(k+1)} =
∩{E(1),E(2),...,E(k)}\{k}
the sequence gets empty one by one.
On 12/15/2024 7:00 AM, WM wrote:
Unless you have changed whatᵂᴹ you (WM) mean,
to completeᵂᴹ a potentiallyᵂᴹ infinite set means
to insert an epilogue (presumably darkᵂᴹ) so that
set+epilogue is actuallyᵂᴹ infinite.
(1) E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
loses all content.
the set of common finite.ordinals is empty.
By the law
(2) ∀k ∈ ℕ :
∩{E(1),E(2),...,E(k+1)} =
∩{E(1),E(2),...,E(k)}\{k}
the sequence gets empty one by one.
The limit set {}
⎛ holds all common finite.ordinals.
⎝ isn't in the sequence.
On 15.12.2024 12:15, joes wrote:Those are all finite.
Am Sat, 14 Dec 2024 17:00:43 +0100 schrieb WM:
Therefore we use all [1, n].??? The bijection is not finite.Then deal with all infinitely many intervals [1, n].No, we are not forbiding "detailed" analysisswindlers and tricksters and not worth to participate in scientificThat pairs the elements of D with the elements of ℕ. Alas, it can >>>>>>> be proved that for every interval [1, n] the deficit of hatsBut we aren't dealing with intervals of [1, n] but of the full set. >>>>> Those who try to forbid the detailed analysis are dishonest
amounts to at least 90 %. And beyond all n, there are no further >>>>>>> hats.
discussion.
Wonrg. There is no natural n that „covers N”.With and without limit.Only in the limit.The intervals [1, n] cover the full set.Why can't he? The problem is in the space of the full set, not theThe problem is that you can't GET to "beyond all n" in the pairing, >>>>>> as there are always more n to get to.If this is impossible, then also Cantor cannot use all n.
finite sub sets.
Even that about infinities.All logic is finite.There is the logic of the infinite.There is no other logic.Nope, it proves it is incompatible with finite logic.Yes, there are only 1/10th as many Black Hats as White Hats, butThis example proves that aleph_0 is nonsense.
since that number is Aleph_0/10, which just happens to also equal
Aleph_0, there is no "deficit" in the set of Natual Numbers.
On 15.12.2024 19:53, Jim Burns wrote:Duh. All naturals are finite. You need to actually remove all inf.many
On 12/15/2024 7:00 AM, WM wrote:
Unless you have changed whatᵂᴹ you (WM) mean,This epilogue is required to empty ℕ by |ℕ \ {1, 2, 3, ...}| = 0. All definable numbers fail: ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
to completeᵂᴹ a potentiallyᵂᴹ infinite set means to insert an epilogue
(presumably darkᵂᴹ) so that set+epilogue is actuallyᵂᴹ infinite.
What’s your problem then?Fine.(1) E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...the set of common finite.ordinals is empty.
loses all content.
By the law (2) ∀k ∈ ℕ :
∩{E(1),E(2),...,E(k+1)} = ∩{E(1),E(2),...,E(k)}\{k}
the sequence gets empty one by one.
The limit set {}
⎛ holds all common finite.ordinals.
⎝ isn't in the sequence.
If all natural numbers are individually available for indexing then they
are available for individually leaving the intersection.
On 15.12.2024 11:56, Mikko wrote:Which sequence do you get a limit of 0 from?
On 2024-12-14 08:53:19 +0000, WM said:
Please refer to the simplest example I gave you on 2024-11-27:
The possibility of a bijection between the sets ℕ = {1, 2, 3, ...} and >>> D = {10n | n ∈ ℕ} is contradicted because for every interval (0, n]
the relative covering is not more than 1/10, and there are no further
numbers 10n beyond all natural numbers n.
It is already proven that there is such bijection. What is proven
cannot be contradicted unless you can prove that 1 = 2.
What is proven under false (self-contradictory) premises can be shown to
be false. Here we have a limit of 1/10 from analysis and a limit of 0
from set theory. That shows that if set theory is right, we have 1/10 =
0 ==> 1 = 0 ==> 2 = 1.
That sequence does not appear in the bijection.But the disproof of the bijection does. There is no reason to forbidThe sequence 1/10, 1/10, 1/10, ... has limit 1/10.Irrelevant as the proof of the exitence of the bijection does not
mention that sequence.
that sequence.
On 15.12.2024 13:39, joes wrote:Omfg. N is infinite. All n are finite. There is no n such that N =
Am Sun, 15 Dec 2024 12:33:15 +0100 schrieb WM:
On 15.12.2024 12:03, Mikko wrote:
Never an n can be named which is responsible for ℕ\ℕ = { }. But ℕ\ℕ = {} can happen.All numbers which can be used a individuals belong to a potentiallyAnd thus all n e N do.
infinite collection ℕ_def. There is no firm end. When n belongs to
ℕ_def, then also n+1 and 2n and n^n^n belong to ℕ_def.
Yes, it does. They are equal. You think N is finite and made up „dark” numbers to patch it up.ℕ_def is potentially infinite. But it does not contain the numbers which complete the set ℕ.The only common property is that all the numbers belong to a finiteIf all successors belong to N_def, it can’t be finite and the
set and have an infinite set of dark successors.
successors can’t be dark.
Not in any finite number of steps.And that means what? The set ℕ cannot be emptied? The set cannot beThis is the only way to explain that the intersection of endegmentsThe explanation is that the sequence is infinitely long.
E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
loses all content in a sequences which allow the loss of only one
number per step.
emptied one by one?
Not all elements can be used as indices?Not for any k.
Either this is fact or ∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1),
E(2), ..., E(k)} \ {k} empties the set.
On 15.12.2024 13:52, Richard Damon wrote:
On 12/15/24 7:00 AM, WM wrote:
On 14.12.2024 23:04, Jim Burns wrote:
If ψ is finite, then ψ+1 is finite.
If ψ+1 is finite, then ψ+2 is finite.
Yes, that is the potentially infinite collection of definable
numbers. But it explains nothing.
That is the collection of numbers known as the Natural Numbers, so I
guess you are admitting that your "Definable Numbers" include *ALL* of
the Natural Numbers
Dark numbers are required to empty ℕ by |ℕ \ {1, 2, 3, ...}| = 0. All definable numbers fail: ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
Regards, WM
Duh. All naturals are finite. You need to actually remove all inf.many
of them.
Am Sun, 15 Dec 2024 16:25:55 +0100 schrieb WM:
On 15.12.2024 12:15, joes wrote:Those are all finite.
Am Sat, 14 Dec 2024 17:00:43 +0100 schrieb WM:Therefore we use all [1, n].
??? The bijection is not finite.Then deal with all infinitely many intervals [1, n].No, we are not forbiding "detailed" analysisswindlers and tricksters and not worth to participate in scientific >>>>>> discussion.That pairs the elements of D with the elements of ℕ. Alas, it can >>>>>>>> be proved that for every interval [1, n] the deficit of hatsBut we aren't dealing with intervals of [1, n] but of the full set. >>>>>> Those who try to forbid the detailed analysis are dishonest
amounts to at least 90 %. And beyond all n, there are no further >>>>>>>> hats.
Wonrg. There is no natural n that „covers N”.With and without limit.Only in the limit.The intervals [1, n] cover the full set.Why can't he? The problem is in the space of the full set, not theThe problem is that you can't GET to "beyond all n" in the pairing, >>>>>>> as there are always more n to get to.If this is impossible, then also Cantor cannot use all n.
finite sub sets.
Am Sun, 15 Dec 2024 12:25:26 +0100 schrieb WM:
On 15.12.2024 11:56, Mikko wrote:Which sequence do you get a limit of 0 from?
On 2024-12-14 08:53:19 +0000, WM said:
Please refer to the simplest example I gave you on 2024-11-27:
The possibility of a bijection between the sets ℕ = {1, 2, 3, ...} and >>>> D = {10n | n ∈ ℕ} is contradicted because for every interval (0, n] >>>> the relative covering is not more than 1/10, and there are no further
numbers 10n beyond all natural numbers n.
It is already proven that there is such bijection. What is proven
cannot be contradicted unless you can prove that 1 = 2.
What is proven under false (self-contradictory) premises can be shown to
be false. Here we have a limit of 1/10 from analysis and a limit of 0
from set theory. That shows that if set theory is right, we have 1/10 =
0 ==> 1 = 0 ==> 2 = 1.
That sequence does not appear in the bijection.But the disproof of the bijection does. There is no reason to forbidThe sequence 1/10, 1/10, 1/10, ... has limit 1/10.Irrelevant as the proof of the exitence of the bijection does not
mention that sequence.
that sequence.
Am Sun, 15 Dec 2024 16:23:19 +0100 schrieb WM:
On 15.12.2024 13:39, joes wrote:Omfg. N is infinite. All n are finite. There is no n such that N =
Am Sun, 15 Dec 2024 12:33:15 +0100 schrieb WM:Never an n can be named which is responsible for ℕ\ℕ = { }. But ℕ\ℕ =
On 15.12.2024 12:03, Mikko wrote:
All numbers which can be used a individuals belong to a potentiallyAnd thus all n e N do.
infinite collection ℕ_def. There is no firm end. When n belongs to
ℕ_def, then also n+1 and 2n and n^n^n belong to ℕ_def.
{} can happen.
{1, …, n}.
Yes, it does.ℕ_def is potentially infinite. But it does not contain the numbers which >> complete the set ℕ.The only common property is that all the numbers belong to a finiteIf all successors belong to N_def, it can’t be finite and the
set and have an infinite set of dark successors.
successors can’t be dark.
They are equal. You think N is finite and made up „dark”
numbers to patch it up.
Not in any finite number of steps.And that means what? The set ℕ cannot be emptied? The set cannot beThis is the only way to explain that the intersection of endegmentsThe explanation is that the sequence is infinitely long.
E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
loses all content in a sequences which allow the loss of only one
number per step.
emptied one by one?
Not all elements can be used as indices?Not for any k.
Either this is fact or ∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1),
E(2), ..., E(k)} \ {k} empties the set.
On 12/15/24 1:57 PM, WM wrote:
Dark numbers are required to empty ℕ by |ℕ \ {1, 2, 3, ...}| = 0. All
definable numbers fail: ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
And why do you need to do that?
On 15.12.2024 21:32, joes wrote:What sequence are you talking about? You can’t compare relative and
Am Sun, 15 Dec 2024 12:25:26 +0100 schrieb WM:Sorry, the limit of not indexed numbers is 9/10 according to analysis
On 15.12.2024 11:56, Mikko wrote:Which sequence do you get a limit of 0 from?
On 2024-12-14 08:53:19 +0000, WM said:
Please refer to the simplest example I gave you on 2024-11-27:
The possibility of a bijection between the sets ℕ = {1, 2, 3, ...} >>>>> and D = {10n | n ∈ ℕ} is contradicted because for every interval (0, >>>>> n] the relative covering is not more than 1/10, and there are no
further numbers 10n beyond all natural numbers n.
It is already proven that there is such bijection. What is proven
cannot be contradicted unless you can prove that 1 = 2.
What is proven under false (self-contradictory) premises can be shown
to be false. Here we have a limit of 1/10 from analysis and a limit of
0 from set theory. That shows that if set theory is right, we have
1/10 = 0 ==> 1 = 0 ==> 2 = 1.
and 0 according to set theory, resulting in 9/10 = 0.
But how does it relate to it?Therefore people were unaware of its failure.That sequence does not appear in the bijection.But the disproof of the bijection does. There is no reason to forbidThe sequence 1/10, 1/10, 1/10, ... has limit 1/10.Irrelevant as the proof of the exitence of the bijection does not
mention that sequence.
that sequence.
On 15.12.2024 21:21, joes wrote:Contrary to the bijection.
Am Sun, 15 Dec 2024 16:25:55 +0100 schrieb WM:All n are finite.
On 15.12.2024 12:15, joes wrote:Those are all finite.
Am Sat, 14 Dec 2024 17:00:43 +0100 schrieb WM:Therefore we use all [1, n].
??? The bijection is not finite.Then deal with all infinitely many intervals [1, n].No, we are not forbiding "detailed" analysisThose who try to forbid the detailed analysis are dishonestThat pairs the elements of D with the elements of ℕ. Alas, it >>>>>>>>> can be proved that for every interval [1, n] the deficit of hats >>>>>>>>> amounts to at least 90 %. And beyond all n, there are no further >>>>>>>>> hats.But we aren't dealing with intervals of [1, n] but of the full >>>>>>>> set.
swindlers and tricksters and not worth to participate in
scientific discussion.
It does not. It applies to every single finite „interval” (whyever those matter), but not to the whole N.All intervals do it because there is no n outside of all intervals [1,Wonrg. There is no natural n that „covers N”.With and without limit.Only in the limit.The intervals [1, n] cover the full set.Why can't he? The problem is in the space of the full set, not the >>>>>> finite sub sets.The problem is that you can't GET to "beyond all n" in theIf this is impossible, then also Cantor cannot use all n.
pairing,
as there are always more n to get to.
n]. My proof applies all intervals.
On 15.12.2024 22:13, Richard Damon wrote:It just shows the infinity of n. No natural is as large as N.
On 12/15/24 1:57 PM, WM wrote:
Dark numbers are required to empty ℕ by |ℕ \ {1, 2, 3, ...}| = 0. All >>> definable numbers fail: ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
And why do you need to do that?In order to show the existence of dark numbers.
On 15.12.2024 21:20, joes wrote:Duh again. No natural is infinite.
Duh. All naturals are finite. You need to actually remove all inf.manyThat is not possible with definable naturals:
of them.
And numbers which succeed ∀k ∈ ℕThere is no natural larger than all others.
On 15.12.2024 21:20, joes wrote:
Duh. All naturals are finite. You need to actually remove all inf.many
of them.
That is not possible with definable naturals:
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
And numbers which succeed
∀k ∈ ℕ: ∩{E(1),E(2),...,E(k+1)} = ∩{E(1),E(2),...,E(k)}\{k}
produce finite endsegments and therefore are invisible.
Regards, WM
On 15.12.2024 21:32, joes wrote:
Am Sun, 15 Dec 2024 12:25:26 +0100 schrieb WM:
On 15.12.2024 11:56, Mikko wrote:Which sequence do you get a limit of 0 from?
On 2024-12-14 08:53:19 +0000, WM said:
Please refer to the simplest example I gave you on 2024-11-27:
The possibility of a bijection between the sets ℕ = {1, 2, 3, ...} and >>>>> D = {10n | n ∈ ℕ} is contradicted because for every interval (0, n] >>>>> the relative covering is not more than 1/10, and there are no further >>>>> numbers 10n beyond all natural numbers n.
It is already proven that there is such bijection. What is proven
cannot be contradicted unless you can prove that 1 = 2.
What is proven under false (self-contradictory) premises can be shown to >>> be false. Here we have a limit of 1/10 from analysis and a limit of 0
from set theory. That shows that if set theory is right, we have 1/10 =
0 ==> 1 = 0 ==> 2 = 1.
Sorry, the limit of not indexed numbers is 9/10 according to analysis
and 0 according to set theory, resulting in 9/10 = 0.
That sequence does not appear in the bijection.But the disproof of the bijection does. There is no reason to forbidThe sequence 1/10, 1/10, 1/10, ... has limit 1/10.Irrelevant as the proof of the exitence of the bijection does not
mention that sequence.
that sequence.
Therefore people were unaware of its failure.
Regards, WM
On 13.12.2024 03:29, Richard Damon wrote:I bet you that 2, …, 9 can be mapped.
On 12/12/24 4:57 PM, WM wrote:
No such element can be named. But 9/10 of all ℕ cannot get mappedD = {10n | n ∈ ℕ} is the set being mapped. The set D being mapped does >>> not change when it is attached to the set ℕ being mapped in form ofAnd so, which element of which set didn't get mapped to a member of the
black hats.
other by the defined mapping?
because the limit of the constant sequence 1/9, 1/9, 1/9, ... is 1/9.
This proves the existence of numbers which cannot be named.
Am Mon, 16 Dec 2024 09:30:18 +0100 schrieb WM:
All intervals do it because there is no n outside of all intervals [1,It does not. It applies to every single finite „interval”
n]. My proof applies all intervals.
On 16.12.2024 12:55, joes wrote:You do not cover N, only finite parts.
Am Mon, 16 Dec 2024 09:30:18 +0100 schrieb WM:
What element is not covered by all intervals that I use?All intervals do it because there is no n outside of all intervals [1,It does not. It applies to every single finite „interval”,
n]. My proof applies all intervals.
but not to the whole N.
On 15.12.2024 19:53, Jim Burns wrote:
On 12/15/2024 7:00 AM, WM wrote:
Unless you have changed whatᵂᴹ you (WM) mean,
an actuallyᵂᴹ infinite set is smaller.than
a fuller.by.one set, but
it contains a potentiallyᵂᴹ infinite subset, meaning
a subset not.smaller.than a fuller.by.one set.
Unless you have changed whatᵂᴹ you (WM) mean,
to completeᵂᴹ a potentiallyᵂᴹ infinite set means
to insert an epilogue (presumably darkᵂᴹ) so that
set+epilogue is actuallyᵂᴹ infinite.
This epilogue is required to empty ℕ
by |ℕ \ {1, 2, 3, ...}| = 0.
All definable numbers fail:
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
(1) E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
loses all content.
the set of common finite.ordinals is empty.
Fine.
By the law
(2) ∀k ∈ ℕ :
∩{E(1),E(2),...,E(k+1)} =
∩{E(1),E(2),...,E(k)}\{k}
the sequence gets empty one by one.
The limit set {}
⎛ holds all common finite.ordinals.
⎝ isn't in the sequence.
If all natural numbers are
individually available for indexing
then they are available for
individually leaving the intersection.
Am Mon, 16 Dec 2024 14:59:27 +0100 schrieb WM:
On 16.12.2024 12:55, joes wrote:You do not cover N, only finite parts.
Am Mon, 16 Dec 2024 09:30:18 +0100 schrieb WM:What element is not covered by all intervals that I use?
All intervals do it because there is no n outside of all intervals [1, >>>> n]. My proof applies all intervals.It does not. It applies to every single finite „interval”,
but not to the whole N.
On 16.12.2024 16:40, joes wrote:Inf.many numbers for every n. N is infinite.
Am Mon, 16 Dec 2024 14:59:27 +0100 schrieb WM:What do I miss to cover?
On 16.12.2024 12:55, joes wrote:You do not cover N, only finite parts.
Am Mon, 16 Dec 2024 09:30:18 +0100 schrieb WM:What element is not covered by all intervals that I use?
All intervals do it because there is no n outside of all intervalsIt does not. It applies to every single finite „interval”,
[1, n]. My proof applies all intervals.
but not to the whole N.
On 15.12.2024 21:21, joes wrote:
Am Sun, 15 Dec 2024 16:25:55 +0100 schrieb WM:
On 15.12.2024 12:15, joes wrote:Those are all finite.
Am Sat, 14 Dec 2024 17:00:43 +0100 schrieb WM:Therefore we use all [1, n].
??? The bijection is not finite.Then deal with all infinitely many intervals [1, n].No, we are not forbiding "detailed" analysisswindlers and tricksters and not worth to participate in scientific >>>>>>> discussion.That pairs the elements of D with the elements of ℕ. Alas, it can >>>>>>>>> be proved that for every interval [1, n] the deficit of hats >>>>>>>>> amounts to at least 90 %. And beyond all n, there are no further >>>>>>>>> hats.But we aren't dealing with intervals of [1, n] but of the full set. >>>>>>> Those who try to forbid the detailed analysis are dishonest
All n are finite.
Wonrg. There is no natural n that „covers N”.With and without limit.Only in the limit.The intervals [1, n] cover the full set.Why can't he? The problem is in the space of the full set, not the >>>>>> finite sub sets.The problem is that you can't GET to "beyond all n" in the pairing, >>>>>>>> as there are always more n to get to.If this is impossible, then also Cantor cannot use all n.
All intervals do it because there is no n outside of all intervals [1,
n]. My proof applies all intervals.
Regards, WM
On 15.12.2024 21:32, joes wrote:
Am Sun, 15 Dec 2024 12:25:26 +0100 schrieb WM:
On 15.12.2024 11:56, Mikko wrote:Which sequence do you get a limit of 0 from?
On 2024-12-14 08:53:19 +0000, WM said:
Please refer to the simplest example I gave you on 2024-11-27:
The possibility of a bijection between the sets ℕ = {1, 2, 3, ...} and >>>>> D = {10n | n ∈ ℕ} is contradicted because for every interval (0, n] >>>>> the relative covering is not more than 1/10, and there are no further >>>>> numbers 10n beyond all natural numbers n.
It is already proven that there is such bijection. What is proven
cannot be contradicted unless you can prove that 1 = 2.
What is proven under false (self-contradictory) premises can be shown to >>> be false. Here we have a limit of 1/10 from analysis and a limit of 0
from set theory. That shows that if set theory is right, we have 1/10 =
0 ==> 1 = 0 ==> 2 = 1.
Sorry, the limit of not indexed numbers is 9/10 according to analysis
and 0 according to set theory, resulting in 9/10 = 0.
That sequence does not appear in the bijection.But the disproof of the bijection does. There is no reason to forbidThe sequence 1/10, 1/10, 1/10, ... has limit 1/10.Irrelevant as the proof of the exitence of the bijection does not
mention that sequence.
that sequence.
Therefore people were unaware of its failure.
Regards, WM
Am Mon, 16 Dec 2024 17:49:20 +0100 schrieb WM:
On 16.12.2024 16:40, joes wrote:Inf.many numbers for every n.
Am Mon, 16 Dec 2024 14:59:27 +0100 schrieb WM:What do I miss to cover?
On 16.12.2024 12:55, joes wrote:You do not cover N, only finite parts.
Am Mon, 16 Dec 2024 09:30:18 +0100 schrieb WM:What element is not covered by all intervals that I use?
All intervals do it because there is no n outside of all intervals >>>>>> [1, n]. My proof applies all intervals.It does not. It applies to every single finite „interval”,
but not to the whole N.
N is infinite.
On 12/16/24 3:41 AM, WM wrote:
Sorry, the limit of not indexed numbers is 9/10 according to analysis
and 0 according to set theory, resulting in 9/10 = 0.
Which shows that one of them is likely wrong.
On 12/16/24 3:30 AM, WM wrote:
On 15.12.2024 21:21, joes wrote:
Therefore we use all [1, n].Those are all finite.
All n are finite.
But N isn't, so the sets [1, n] aren't what the bijection is defined on.
All intervals do it because there is no n outside of all intervals [1,
n]. My proof applies all intervals.
And all the intervals are finite, and thus not the INFINITE set N, which
is where the bijection occurs.
On 12/15/2024 2:16 PM, WM wrote:
(1) E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
loses all content.
the set of common finite.ordinals is empty.
Fine.
Each finite.cardinal
leaves after that.many steps,
with further steps to follow,
more steps than any other finite.cardinal
By the law
(2) ∀k ∈ ℕ :
∩{E(1),E(2),...,E(k+1)} =
∩{E(1),E(2),...,E(k)}\{k}
the sequence gets empty one by one.
If all natural numbers are
individually available for indexing
then they are available for
individually leaving the intersection.
Yes.
Each finite.cardinal
indexes its own end.segment,
and leaves at the next end.segment.
WM wrote :
On 17.12.2024 00:52, Richard Damon wrote:
On 12/16/24 3:30 AM, WM wrote:
On 15.12.2024 21:21, joes wrote:
Therefore we use all [1, n].Those are all finite.
All n are finite.
But N isn't, so the sets [1, n] aren't what the bijection is defined on.
Every element is the last element of a FISON [1, n].
Why does your FISON look like a real interval?
On 16.12.2024 11:23, Mikko wrote:It is not even in a sequence. How does that matter?
On 2024-12-15 11:33:15 +0000, WM said:
The set, i.e. all numbers together, has no successor.We cannot name dark numbers as individuals.We needn't. The axioms of natural numbers ensure that every natural
number has a successor,
You misunderstood: Mikko was naming the axioms of the naturals.If that is not possible then there are no natural numbers.That is not possible for an actually infinite set. It is only possible
for numbers coming into being.
On 15.12.2024 21:29, joes wrote:You don’t. You only ever apply a finite number (except in the limit).
Am Sun, 15 Dec 2024 16:23:19 +0100 schrieb WM:I apply all of them. Is there more in ℕ?
On 15.12.2024 13:39, joes wrote:Omfg. N is infinite. All n are finite. There is no n such that N =
Am Sun, 15 Dec 2024 12:33:15 +0100 schrieb WM:Never an n can be named which is responsible for ℕ\ℕ = { }. But ℕ\ℕ =
On 15.12.2024 12:03, Mikko wrote:
All numbers which can be used a individuals belong to a potentiallyAnd thus all n e N do.
infinite collection ℕ_def. There is no firm end. When n belongs to >>>>> ℕ_def, then also n+1 and 2n and n^n^n belong to ℕ_def.
{} can happen.
{1, …, n}.
Well yes, the set {1, 2, …, n} != N.∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵoYes, it does.ℕ_def is potentially infinite. But it does not contain the numbersThe only common property is that all the numbers belong to a finiteIf all successors belong to N_def, it can’t be finite and the
set and have an infinite set of dark successors.
successors can’t be dark.
which complete the set ℕ.
Big deal. N is not a FISON. BTW, the union of them does exhaust it.They are equal. You think N is finite and made up „dark” numbers toI know that ℕ is infinite and that all FISONs {1, 2, 3, ..., n} fail to exhaust it.
patch it up.
Yes, in the limit.But in an infinite number of steps.Not in any finite number of steps.And that means what? The set ℕ cannot be emptied? The set cannot beThis is the only way to explain that the intersection of endegmentsThe explanation is that the sequence is infinitely long.
E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
loses all content in a sequences which allow the loss of only one
number per step.
emptied one by one?
Sure there is.Other steps are not possible, in particular there is no limit.Not all elements can be used as indices?Not for any k.
Either this is fact or ∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1),
E(2), ..., E(k)} \ {k} empties the set.
CantorCantor applies the infinite set.
applies omly finite k and claims that none remains unused (as the
content of all endsegments).
On 16.12.2024 18:25, joes wrote:
Am Mon, 16 Dec 2024 17:49:20 +0100 schrieb WM:
On 16.12.2024 16:40, joes wrote:Inf.many numbers for every n.
Am Mon, 16 Dec 2024 14:59:27 +0100 schrieb WM:What do I miss to cover?
On 16.12.2024 12:55, joes wrote:You do not cover N, only finite parts.
Am Mon, 16 Dec 2024 09:30:18 +0100 schrieb WM:What element is not covered by all intervals that I use?
All intervals do it because there is no n outside of all intervals >>>>>>> [1, n]. My proof applies all intervals.It does not. It applies to every single finite „interval”,
but not to the whole N.
But Cantor using every n does not miss to cover anything?
N is infinite.
Every element is the last element of a FISON [1, n]. ℕ is the set of all FISONs. I use all FISONs. ∀n ∈ ℕ: f([1, n]) =< 1/10.
Ever heard of the effect of the universal quantifier?
Regards, WM
Regards, WM
On 17.12.2024 00:52, Richard Damon wrote:No, it is the union.
On 12/16/24 3:30 AM, WM wrote:
On 15.12.2024 21:21, joes wrote:
Every element is the last element of a FISON [1, n]. ℕ is the set of all FISONs.But N isn't, so the sets [1, n] aren't what the bijection is definedAll n are finite.Therefore we use all [1, n].Those are all finite.
on.
I use all FISONs.No, you don’t. N is not a FISON.
What else do you think it uses?According to Cantor the "bijection" uses all n and nothing more.All intervals do it because there is no n outside of all intervals [1,And all the intervals are finite, and thus not the INFINITE set N,
n]. My proof applies all intervals.
which is where the bijection occurs.
On 17.12.2024 11:07, FromTheRafters wrote:But why real intervals?
WM wrote :Originally I had used unit intervals [n, n+1] of real intervals (0,
On 17.12.2024 00:52, Richard Damon wrote:Why does your FISON look like a real interval?
On 12/16/24 3:30 AM, WM wrote:Every element is the last element of a FISON [1, n].
On 15.12.2024 21:21, joes wrote:
But N isn't, so the sets [1, n] aren't what the bijection is definedAll n are finite.Therefore we use all [1, n].Those are all finite.
on.
10n]. The argument remains the same.
On 17.12.2024 00:57, Richard Damon wrote:As long as it is finite.
On 12/16/24 3:59 AM, WM wrote:An unbounded number can be subtracted individually.
On 15.12.2024 22:14, Richard Damon wrote:Sure an infinite number of members can be subtracted individually,
On 12/15/24 2:44 PM, WM wrote:Many members can be subtracted individually but infinitely many
On 15.12.2024 13:54, Richard Damon wrote:Which just shows that the full set in infinte, and any member in it
because they are dark.
You can't "name" your dark numbers,
|ℕ \ {1, 2, 3, ...}| = 0 cannot be accomplished by visible numbers >>>>> because ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
is finite, and not the last member.
members cannot be subtracted individually. They are belonging to the
set. They are dark.
However, if all areWhatever do you mean by that? There is no last to inf.many. „All” are
subtracted individually, then a last one is subtracted. That cannot
happen.
On 16.12.2024 18:08, Jim Burns wrote:
On 12/15/2024 2:16 PM, WM wrote:
On 15.12.2024 19:53, Jim Burns wrote:
On 12/15/2024 7:00 AM, WM wrote:
(1) E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
loses all content.
the set of common finite.ordinals is empty.
Fine.
Each finite.cardinal
leaves after that.many steps,
with further steps to follow,
more steps than any other finite.cardinal
which is following upom it.
Explain your vision of the problem:
Your logic that if it holds for all FISONs, it holds for N,
Am Mon, 16 Dec 2024 09:37:13 +0100 schrieb WM:
You don’t. You only ever apply a finite number (except in the limit).
CantorCantor applies the infinite set.
applies only finite k and claims that none remains unused (as the
content of all endsegments).
According to Cantor the "bijection" uses all n and nothing more.
Right, but no FISON uses contains ALL n.
Am Tue, 17 Dec 2024 10:13:48 +0100 schrieb WM:
Every element is the last element of a FISON [1, n]. ℕ is the set of all >> FISONs.No, it is the union.
I use all FISONs.No, you don’t. N is not a FISON.
What else do you think it uses?According to Cantor the "bijection" uses all n and nothing more.All intervals do it because there is no n outside of all intervals [1, >>>> n]. My proof applies all intervals.And all the intervals are finite, and thus not the INFINITE set N,
which is where the bijection occurs.
Am Tue, 17 Dec 2024 11:30:46 +0100 schrieb WM:
An unbounded number can be subtracted individually.As long as it is finite.
However, if all areWhatever do you mean by that? There is no last to inf.many. „All” are
subtracted individually, then a last one is subtracted. That cannot
happen.
not finite.
On 12/17/2024 4:00 AM, WM wrote:
Each finite.cardinal i after 0
is countable.to from.0
∃⟦0,i⦆ smaller than Bobbed ⟦0,i⦆∪{Bob}
On 17.12.2024 19:52, Jim Burns wrote:
Each finite.cardinal i after 0
is countable.to from.0
∃⟦0,i⦆ smaller than Bobbed ⟦0,i⦆∪{Bob}
Stop inventing new nonsense words.
We have the sequence of
intersections of endsegments
f(k) = ∩{E(1), E(2), ..., E(k)}
with E(1) = ℕ
and the definition of that function
∀k ∈ ℕ :
∩{E(1), E(2), ..., E(k+1)} =
∩{E(1), E(2), ..., E(k)} \ {k}
and the fact that
∩{E(1), E(2), ...} is empty.
More is not required to prove
the existence of finite endsegments.
On 17.12.2024 13:34, Richard Damon wrote:He „applies“ the set of all N, as opposed to every single n.
Your logic that if it holds for all FISONs, it holds for N,Please explain what Cantor does to apply more than what I apply, namely
all n ∈ ℕ.
On 17.12.2024 13:34, Richard Damon wrote:
According to Cantor the "bijection" uses all n and nothing more.
Right, but no FISON uses contains ALL n.
But all FISONs contain/are all n.
Regards, WM
On 12/17/2024 5:12 PM, WM wrote:
We have the sequence of
intersections of endsegments
f(k) = ∩{E(1), E(2), ..., E(k)}
with E(1) = ℕ
and the definition of that function
∀k ∈ ℕ :
∩{E(1), E(2), ..., E(k+1)} =
∩{E(1), E(2), ..., E(k)} \ {k}
and the fact that
∩{E(1), E(2), ...} is empty.
More is not required to prove
the existence of finite endsegments.
Here is a counter.example to your claimed requirement:
There are no finite.cardinals common to
all the infinite end.segments of finite.cardinals.
The infinite end.segments of finite.cardinals
do not include any finite end.segments and
they have an empty intersection.
----
More generally,
for limit.set Lim.⟨A₀,A₁,A₂,…⟩ of
On 17.12.2024 13:34, Richard Damon wrote:
Your logic that if it holds for all FISONs, it holds for N,
Please explain what Cantor does to apply more than what I apply, namely
all n ∈ ℕ.
Regards, WM
On 17.12.2024 18:07, joes wrote:
Am Tue, 17 Dec 2024 10:13:48 +0100 schrieb WM:
Every element is the last element of a FISON [1, n]. ℕ is the set of all >>> FISONs.No, it is the union.
It is also the set because every FISON can represent one natnumber.
I use all FISONs.No, you don’t. N is not a FISON.
It is all FISONs.
What else do you think it uses?According to Cantor the "bijection" uses all n and nothing more.All intervals do it because there is no n outside of all intervals [1, >>>>> n]. My proof applies all intervals.And all the intervals are finite, and thus not the INFINITE set N,
which is where the bijection occurs.
You claimed that he uses more than I do, namely all natural numbers.
Regards, WM
On 12/17/24 4:57 PM, WM wrote:
You claimed that he uses more than I do, namely all natural numbers.
Right, you never use ALL the natural numbers, only a finite subset of them.
Am Tue, 17 Dec 2024 22:49:51 +0100 schrieb WM:
On 17.12.2024 13:34, Richard Damon wrote:He „applies“ the set of all N, as opposed to every single n.
Your logic that if it holds for all FISONs, it holds for N,Please explain what Cantor does to apply more than what I apply, namely
all n ∈ ℕ.
On 18.12.2024 02:16, Jim Burns wrote:
On 12/17/2024 5:12 PM, WM wrote:
We have the sequence of
intersections of endsegments
f(k) = ∩{E(1), E(2), ..., E(k)}
with E(1) = ℕ
and the definition of that function
∀k ∈ ℕ :
∩{E(1), E(2), ..., E(k+1)} =
∩{E(1), E(2), ..., E(k)} \ {k}
and the fact that
∩{E(1), E(2), ...} is empty.
More is not required to prove
the existence of finite endsegments.
Here is a counter.example to your claimed requirement:
There are no finite.cardinals common to
all the infinite end.segments of finite.cardinals.
There are no definable cardinals
common to all endsegments of definable cardinals.
There are no definable cardinals
common to all endsegments of definable cardinals.
Finite cardinals are in all non-empty endsegments.
The infinite end.segments of finite.cardinals
do not include any finite end.segments and
they have an empty intersection.
Explicitly wrong.
As long as only infinite endsegments are concerned
their intersection is infinite.
----
More generally,
for limit.set Lim.⟨A₀,A₁,A₂,…⟩ of
There are no limits involved.
On 12/18/2024 7:40 AM, WM wrote:
The infinite end.segments of finite.cardinals
do not include any finite end.segments and
they have an empty intersection.
Explicitly wrong.
As long as only infinite endsegments are concerned their intersection
is infinite.
There are no limits involved.
Then it would be great if you (WM)
stopped calling things 'limit sets'
On 18.12.2024 10:35, joes wrote:Why is it wrong?
Am Tue, 17 Dec 2024 22:49:51 +0100 schrieb WM:Please quote the text from which you have obtained that wrong idea.
On 17.12.2024 13:34, Richard Damon wrote:He „applies“ the set of all N, as opposed to every single n.
Your logic that if it holds for all FISONs, it holds for N,Please explain what Cantor does to apply more than what I apply,
namely all n ∈ ℕ.
On 18.12.2024 19:22, Jim Burns wrote:
On 12/18/2024 7:40 AM, WM wrote:
On 18.12.2024 02:16, Jim Burns wrote:
The infinite end.segments of finite.cardinals
do not include any finite end.segments and
they have an empty intersection.
Explicitly wrong.
As long as
only infinite endsegments are concerned
their intersection is infinite.
There are no limits involved.
Then it would be great if you (WM)
stopped calling things 'limit sets'
The sets described here are not limit sets.
f(k) = ∩{E(1), E(2), ..., E(k)}
with E(1) = ℕ
∀k ∈ ℕ :
∩{E(1), E(2), ..., E(k+1)} =
∩{E(1), E(2), ..., E(k)} \ {k},
∩{E(1), E(2), ...} is empty.
Other sets are not in the argument.
On 18.12.2024 13:29, Richard Damon wrote:
On 12/17/24 4:57 PM, WM wrote:
You claimed that he uses more than I do, namely all natural numbers.
Right, you never use ALL the natural numbers, only a finite subset of
them.
Please give the quote from which you obtain a difference between
"The infinite sequence thus defined has the peculiar property to contain
the positive rational numbers completely, and each of them only once at
a determined place." [G. Cantor, letter to R. Lipschitz (19 Nov 1883)]
and my "the infinite sequence f(n) = [1, n] contains all natural numbers
n completely, and each of them only once at a determined place."
Regards, WM
Am Wed, 18 Dec 2024 20:07:35 +0100 schrieb WM:
On 18.12.2024 10:35, joes wrote:Why is it wrong?
> Am Tue, 17 Dec 2024 22:49:51 +0100 schrieb WM:
>> On 17.12.2024 13:34, Richard Damon wrote:
>>
>>> Your logic that if it holds for all FISONs, it holds for N,
>> Please explain what Cantor does to apply more than what I apply,
>> namely all n ∈ ℕ.
> He „applies“ the set of all N, as opposed to every single n.
Please quote the text from which you have obtained that wrong idea.
Am Wed, 18 Dec 2024 20:06:19 +0100 schrieb WM:
On 18.12.2024 13:29, Richard Damon wrote:You deny the limit.
On 12/17/24 4:57 PM, WM wrote:Please give the quote from which you obtain a difference between "The
You claimed that he uses more than I do, namely all natural numbers.Right, you never use ALL the natural numbers, only a finite subset of
them.
infinite sequence thus defined has the peculiar property to contain the
positive rational numbers completely, and each of them only once at a
determined place." [G. Cantor, letter to R. Lipschitz (19 Nov 1883)] and
my "the infinite sequence f(n) = [1, n] contains all natural numbers n
completely, and each of them only once at a determined place."
On 12/18/24 2:06 PM, WM wrote:
On 18.12.2024 13:29, Richard Damon wrote:How is your f(n) an "infinite sequence, since n is a finite number in
On 12/17/24 4:57 PM, WM wrote:
You claimed that he uses more than I do, namely all natural numbers.
Right, you never use ALL the natural numbers, only a finite subset of
them.
Please give the quote from which you obtain a difference between
"The infinite sequence thus defined has the peculiar property to
contain the positive rational numbers completely, and each of them
only once at a determined place." [G. Cantor, letter to R. Lipschitz
(19 Nov 1883)]
and my "the infinite sequence f(n) = [1, n] contains all natural
numbers n completely, and each of them only once at a determined place."
each instance.
NONE of your f(n) contains *ALL* natural numbers, since no "n" is the
highest natural number,
Your problem is you just don't understand what "infinity" is
On 12/18/2024 2:14 PM, WM wrote:
On 18.12.2024 19:22, Jim Burns wrote:
On 12/18/2024 7:40 AM, WM wrote:
On 18.12.2024 02:16, Jim Burns wrote:
The infinite end.segments of finite.cardinals
do not include any finite end.segments and
they have an empty intersection.
Explicitly wrong.
As long as
only infinite endsegments are concerned
their intersection is infinite.
There are no limits involved.
Then it would be great if you (WM)
stopped calling things 'limit sets'
The sets described here are not limit sets.
Q. What is a limit set?
f(k) = ∩{E(1), E(2), ..., E(k)}
with E(1) = ℕ
∀k ∈ ℕ :
∩{E(1), E(2), ..., E(k+1)} =
∩{E(1), E(2), ..., E(k)} \ {k},
∩{E(1), E(2), ...} is empty.
Other sets are not in the argument.
∀k ∈ ℕ :
k ∉ E(k+1) ⊇ ⋂{E(1),E(2),...} ∌ k
⋂{E(1),E(2),...} is empty.
On 18.12.2024 21:21, Jim Burns wrote:
On 12/18/2024 2:14 PM, WM wrote:
On 18.12.2024 19:22, Jim Burns wrote:
On 12/18/2024 7:40 AM, WM wrote:
On 18.12.2024 02:16, Jim Burns wrote:
The infinite end.segments of finite.cardinals
do not include any finite end.segments and
they have an empty intersection.
Explicitly wrong.
As long as
only infinite endsegments are concerned
their intersection is infinite.
f(k) = ∩{E(1), E(2), ..., E(k)}
with E(1) = ℕ
∀k ∈ ℕ :
∩{E(1), E(2), ..., E(k+1)} =
∩{E(1), E(2), ..., E(k)} \ {k},
∩{E(1), E(2), ...} is empty.
Other sets are not in the argument.
∀k ∈ ℕ :
k ∉ E(k+1) ⊇ ⋂{E(1),E(2),...} ∌ k
⋂{E(1),E(2),...} is empty.
Other numbers are not in the argument.
But E(1) is full.
The only way to get rid of content is
to proceed by
∀k ∈ ℕ:
∩{E(1), E(2), ..., E(k+1)} =
∩{E(1), E(2), ..., E(k)} \ {k},
i.e. to lose one number per term of
the function
f(k) = ∩{E(1), E(2), ..., E(k)}.
The empty term
The empty term
has lost all natnumbers one by one.
On 12/19/2024 9:50 AM, WM wrote:
The empty term
has lost all natnumbers one by one.
One.by.one and (darkᵂᴹ or visibleᵂᴹ) endlessly.
For each set smaller.than fuller.by.one sets,
fuller.by.ONE sets are smaller.than
fuller.by.TWO sets.
No set is largest among
the sets smaller.than fuller.by.one sets.
On 12/19/2024 9:50 AM, WM wrote
⋂{E(1),E(2),...} is empty.
Other numbers are not in the argument.
But E(1) is full.
The only way to get rid of content is
to proceed by
∀k ∈ ℕ:
∩{E(1), E(2), ..., E(k+1)} =
∩{E(1), E(2), ..., E(k)} \ {k},
i.e. to lose one number per term of
the function
f(k) = ∩{E(1), E(2), ..., E(k)}.
f(k) = E(k) = ⟦k,ℵ₀⦆
Yes,
after infinitely.many one.number.losses,
infinitely.many numbers are lost.
However,
each number is not.after infinitely.many numbers,
each loss is not.after infinitely.many losses.
The empty term
There is no darkᵂᴹ or visibleᵂᴹ empty term.
⎛ Assume otherwise.
The empty term
has lost all natnumbers one by one.
One.by.one and (darkᵂᴹ or visibleᵂᴹ) endlessly.
On 19.12.2024 04:29, Richard Damon wrote:
On 12/18/24 2:06 PM, WM wrote:
On 18.12.2024 13:29, Richard Damon wrote:How is your f(n) an "infinite sequence, since n is a finite number in
On 12/17/24 4:57 PM, WM wrote:
You claimed that he uses more than I do, namely all natural numbers.
Right, you never use ALL the natural numbers, only a finite subset
of them.
Please give the quote from which you obtain a difference between
"The infinite sequence thus defined has the peculiar property to
contain the positive rational numbers completely, and each of them
only once at a determined place." [G. Cantor, letter to R. Lipschitz
(19 Nov 1883)]
and my "the infinite sequence f(n) = [1, n] contains all natural
numbers n completely, and each of them only once at a determined place." >>>
each instance.
How is Cantor's sequence infinite since every positive rational number
is finite?
NONE of your f(n) contains *ALL* natural numbers, since no "n" is the
highest natural number,
None of Cantor's terms q_n contains all rational numbers, sice no n is
the highest natural number.
Your problem is you just don't understand what "infinity" is
Your problem is that you believe to understand it.
Regards, WM
He builds an infinite sequence that pairs a natural number to every
rational number .
On 12/20/2024 2:13 AM, WM wrote:
On 20.12.2024 03:52, Richard Damon wrote:
He builds an infinite sequence that pairs a natural number to every
rational number .
And I build an infinite sequence of intervals [1, n] that contains
every natural number.
Cantor Pairing works will any of them.
On 19.12.2024 19:29, Jim Burns wrote:
On 12/19/2024 9:50 AM, WM wrote
On 18.12.2024 21:21, Jim Burns wrote:
⋂{E(1),E(2),...} is empty.
Other numbers are not in the argument.
But E(1) is full.
The only way to get rid of content is
to proceed by
∀k ∈ ℕ:
∩{E(1), E(2), ..., E(k+1)} =
∩{E(1), E(2), ..., E(k)} \ {k},
i.e. to lose one number per term of
the function
f(k) = ∩{E(1), E(2), ..., E(k)}.
f(k) = E(k) = ⟦k,ℵ₀⦆
Yes,
after infinitely.many one.number.losses,
infinitely.many numbers are lost.
That means all numbers are lost by loss of
one number per term.
That implies finite endsegments.
That means all numbers are lost by loss of
one number per term.
That implies finite endsegments.
However,
each number is not.after infinitely.many numbers,
each loss is not.after infinitely.many losses.
As long as
any natural number remains,
the sequence of lost numbers is finite
(ended by the remaining number).
On 12/20/2024 6:38 AM, WM wrote:
On 20.12.2024 12:55, Chris M. Thomasson wrote:
On 12/20/2024 2:13 AM, WM wrote:
On 20.12.2024 03:52, Richard Damon wrote:
He builds an infinite sequence that pairs a natural number to every
rational number .
And I build an infinite sequence of intervals [1, n] that contains
every natural number.
Cantor Pairing works will any of them.
Therefore no interval [1, n] and therefore no n is missing in my proof
that the pairing
10n --> n, n ∈ ℕ
does not work. For every interval at least 9/10 are missing.
Huh? Have you ever implemented Cantor Pairing and tried it out? It works
with any unsigned integer. So, it works with all of them. Why do you
seem to have trouble with it?
[Mückenheim's problem] is he isn't using the [complete] set of Natural Numbers, only
a FISON of 1 to n [where n is some natural number].
Strange to me. Here is Cantor Pairing in MIDI notes, lol:
https://youtu.be/XkwgJt5bxKI
On 12/19/2024 4:37 PM, WM wrote:
That means all numbers are lost by loss of
one number per term.
That implies finite endsegments.
Q. What does 'finite' mean?
That means all numbers are lost by loss of
one number per term.
That implies finite endsegments.
No.
Yes, each number is lost by loss of
one number per term.
However,
each end.segment is not finite.
As long as
any natural number remains,
the sequence of lost numbers is finite
(ended by the remaining number).
Claims P⇒Q and ¬Q⇒¬P are both.true or both.false.
P⇒Q above, ¬Q⇒¬P below.
As long as
the sequence of lost numbers is infinite,
no natural number remains.
I agree.
His problem is he isn't using the actual set of Natural Numbers, only a
FISON of 1 to n,
On 20.12.2024 19:48, Jim Burns wrote:
On 12/19/2024 4:37 PM, WM wrote:
That means all numbers are lost by loss of
one number per term.
That implies finite endsegments.
Q. What does 'finite' mean?
Here is a new and better definition of endsegments
E(n) = {n+1, n+2, n+3, ...} with E(0) = ℕ.
∀n ∈ ℕ : E(n+1) = E(n) \ {n+1}
means that the sequence of endsegments
can decrease only by one natnumber per step.
Therefore the sequence of endsegments
cannot become empty
(i.e., not all natnumbers can be applied as indices)
unless the empty endsegment is reached,
unless the empty endsegment is reached,
and
before finite endsegments,
endsegments containing only 1, 2, 3, or n ∈ ℕ numbers,
have been passed.
These however, if existing at all, cannot be seen.
They are dark.
That means all numbers are lost by loss of
one number per term.
That implies finite endsegments.
No.
Yes, each number is lost by loss of
one number per term.
However,
each end.segment is not finite.
Then the last endsegment is empty.
On 12/21/2024 6:34 AM, WM wrote:
On 20.12.2024 19:48, Jim Burns wrote:
On 12/19/2024 4:37 PM, WM wrote:
That means all numbers are lost by loss of
one number per term.
That implies finite endsegments.
Q. What does 'finite' mean?
Consider end.segments of the finite cardinals.
Q. What does 'finite' mean,
'finite', whether darkᵂᴹ or visibleᵂᴹ?
Here is a new and better definition of endsegments
E(n) = {n+1, n+2, n+3, ...} with E(0) = ℕ.
∀n ∈ ℕ : E(n+1) = E(n) \ {n+1}
means that the sequence of endsegments can decrease only by one
natnumber per step.
E(n+1) is larger.than each of
the sets for which there are smaller.by.one sets.
E(n+1) isn't any of
the sets for which there are smaller.by.one sets.
E(n+1) isn't smaller.by.one than E(n).
E(n+1) is emptier.by.one than E(n)
Therefore the sequence of endsegments
cannot become empty
Yes, because
the sequence of end.segments
can become emptier.one.by.one, but
it cannot become smaller.one.by.one.
(i.e., not all natnumbers can be applied as indices)
Each finite.cardinal can be applied,
which makes the sequence emptier.by.one
but does not make the sequence smaller.by.one.
On 21.12.2024 20:32, Jim Burns wrote:
E(n+1) is larger.than each of
the sets for which there are smaller.by.one sets.
E(n+1) isn't any of
the sets for which there are smaller.by.one sets.
No.
E(n+2) is smaller than E(n+1)
by one element, namely n+2.
What does 'finite' mean?
E(n+2) is the set of all finite.cardinals > n+2
E(n+1) = E(n+2)∪{n+2}
E(n+2)∪{n+2} isn't larger.than E(n+2)
#E(n+2) isn't any of the finite.cardinals in ℕ
E(n+2) isn't any of the sets
smaller.than fuller.by.one sets.
E(n+2) is emptier.by.one than E(n+1), but
E(n+2) isn't smaller.by.one than E(n+1).
On 21.12.2024 04:37, Richard Damon wrote:
His problem is he isn't using the actual set of Natural Numbers, only
a FISON of 1 to n,
Wrong. I am using all FISONs. No natural number remains unused.
Regards, WM
On 12/21/24 4:58 PM, WM wrote:
Finite endsegments have a natural number of elements.
SO, none of your E(n) are finite endsegments, since they all have an
INFINITE number of elements, being the INFINITE set of
{ n+1, n+2, n+3, ... } by your definition.
On 12/21/24 12:46 PM, WM wrote:
On 21.12.2024 04:37, Richard Damon wrote:
His problem is he isn't using the actual set of Natural Numbers, only
a FISON of 1 to n,
And you don't use ALL natural numbers, as you logic doesn't allow you to
do the needed infiite operations to use ALL, only ANY,
No FISION uses ALL the Natural Numbers,
On 22.12.2024 13:28, Richard Damon wrote:
On 12/21/24 4:58 PM, WM wrote:
Finite endsegments have a natural number of elements.
SO, none of your E(n) are finite endsegments, since they all have an
INFINITE number of elements, being the INFINITE set of
{ n+1, n+2, n+3, ... } by your definition.
The intersection of all endsegments is empty. It cannot get empty other
than by one element per endsegment.
Regards, WM
On 12/22/24 8:11 AM, WM wrote:
Find a natural number that is not in all intervals [1, n] which I use:
∀n ∈ ℕ [1, n].
But you can't use *ALL* intervals, becaue you need to use them
individually
On 22.12.2024 11:16, Jim Burns wrote:
What does 'finite' mean?
A set is finiteᵂᴹ if it contains
a visibleᵂᴹ naturalᵂᴹ number of elements.
All naturalᵂᴹ numbers are finiteᵂᴹ but
the realm of darkᵂᴹ numbers
appears like infinityᵂᴹ.
The function Eᵂᴹ(n) decreases
from infinityᵂᴹ to zero
by single steps of height 1
like the function NUFᵂᴹ(x) increases
by single steps of height 1.
Set theorists must accept
magic steps of infiniteᵂᴹ size or
refuse to describe these transitions at all.
On 12/22/2024 6:32 AM, WM wrote:
The function E(n) decreases
from infinity to zero
by single steps of height 1
like the function NUF(x) increases
by single steps of height 1. Set theorists must accept
magic steps of infiniteᵂᴹ size or
refuse to describe these transitions at all.
Alternatively,
set theorists could continue to talk about
what set theorists talk about, instead of
what you (WM) talk about.
On 22.12.2024 20:10, Richard Damon wrote:
On 12/22/24 8:11 AM, WM wrote:
Find a natural number that is not in all intervals [1, n] which I use:
∀n ∈ ℕ [1, n].
But you can't use *ALL* intervals, becaue you need to use them
individually
No, I do as Cantor did.
Regards, WM
WM brought next idea :
The function E(n) decreases from infinity to zero because in set
theory ℕ \ {1, 2, 3, ...} = { } is an accepted formula.
In what way is it decreasing?
On 12/22/24 5:11 PM, WM wrote:
On 22.12.2024 20:10, Richard Damon wrote:
On 12/22/24 8:11 AM, WM wrote:
Find a natural number that is not in all intervals [1, n] which I use: >>>> ∀n ∈ ℕ [1, n].
But you can't use *ALL* intervals, becaue you need to use them
individually
No, I do as Cantor did.
No, you do what you THINK Cantor did,
On 22.12.2024 22:22, Jim Burns wrote:
On 12/22/2024 6:32 AM, WM wrote:
A set is finiteᵂᴹ if it contains
a visibleᵂᴹ naturalᵂᴹ number of elements.
All naturalᵂᴹ numbers are finiteᵂᴹ but
the realm of darkᵂᴹ numbers
appears like infinityᵂᴹ.
Set A is finiteⁿᵒᵗᐧᵂᴹ if A∪{x} is larger, for A∌x
Set Y is infiniteⁿᵒᵗᐧᵂᴹ if Y∪{x} isn't larger, for Y∌x
ℕⁿᵒᵗᐧᵂᴹ is the set of finiteⁿᵒᵗᐧᵂᴹ.cardinals.
which means
ℕⁿᵒᵗᐧᵂᴹ is the set of cardinals #A of
sets A smaller.than A∪{x} for A∌x
For each finiteⁿᵒᵗᐧᵂᴹ set A,
its cardinality #A is in ℕⁿᵒᵗᐧᵂᴹ
For each set Y without #Y in ℕⁿᵒᵗᐧᵂᴹ
Y is infiniteⁿᵒᵗᐧᵂᴹ, and
Y∪{x} isn't larger than Y, for Y∌x
The function E(n) decreases
from infinity to zero
by single steps of height 1
like the function NUF(x) increases
by single steps of height 1.
Set theorists must accept
magic steps of infiniteᵂᴹ size or
refuse to describe these transitions at all.
Alternatively,
set theorists could continue to talk about
what set theorists talk about, instead of
what you (WM) talk about.
They do not wish to recognize that
their theory is self-contradictory.
But perhaps students would be interested.
I tell them the following story.
The function Eᵂᴹ(n) decreases
from infinityᵂᴹ to zero
because in set theory
ℕᵂᴹ \ {1, 2, 3, ...} = { }
is an accepted formula.
The set ℕᵂᴹ can get empty
by subtracting its elements.
Either this is possible one by one,
then finite endsegments do exist,
or it is only possible be removing
(after the first elements one by one)
the remaining elements collectively.
This shows the existence of
numbers which can be handled collectively only.
How should we call them?
Another approach is to call [prove] the empty set
the limit of the sequence E(n).
But note that a limit is a set [JB:]
⎡ in.which is each element in each set of
⎢ almost.all of infinitelyⁿᵒᵗᐧᵂᴹ.many sets
⎢ and
⎢ not.in.which is each element not.in each set of
⎣ almost.all of infinitelyⁿᵒᵗᐧᵂᴹ.many sets
between which [it] and all terms of the sequence
nothing fits.
Therefore the limit empty set causes
sets with few elements only.
The empty limit set is caused by
each finiteⁿᵒᵗᐧᵂᴹ.cardinal being not.in
almost all of the end.segments
(in only finitely.many end.segment.exceptions)
On 23.12.2024 01:05, FromTheRafters wrote:
WM brought next idea :
The function E(n) decreases from infinity to zero because in set
theory ℕ \ {1, 2, 3, ...} = { } is an accepted formula.
In what way is it decreasing?
One by one. Every endsegment transforms a natural number from content
inside to index outside. That is guaranteed by mathematics:
∀k ∈ ℕ : ∩{E(0), E(1), E(2), ..., E(k+1)} = ∩{E(0), E(1), E(2), ...,
E(k)} \ {k+1}. Note the universal quantifier.
Regards, WM
On 23.12.2024 02:01, Richard Damon wrote:
On 12/22/24 5:11 PM, WM wrote:
On 22.12.2024 20:10, Richard Damon wrote:
On 12/22/24 8:11 AM, WM wrote:
Find a natural number that is not in all intervals [1, n] which I use: >>>>> ∀n ∈ ℕ [1, n].
But you can't use *ALL* intervals, becaue you need to use them
individually
No, I do as Cantor did.
No, you do what you THINK Cantor did,
Show an n that I do not use with all intervals [1, n].
Regards, WM
On 23.12.2024 12:02, Jim Burns wrote:
On 12/22/2024 5:24 PM, WM wrote:
Another approach is to call [prove] the empty set
the limit of the sequence E(n).
But note that a limit is a set [JB:]
⎡ in.which is each element in each set of
⎢ almost.all of infinitelyⁿᵒᵗᐧᵂᴹ.many sets
⎢ and
⎢ not.in.which is each element not.in each set of
⎣ almost.all of infinitelyⁿᵒᵗᐧᵂᴹ.many sets
between which [it] and all terms of the sequence
nothing fits.
Therefore the limit empty set causes
sets with few elements only.
The empty limit set is caused by
each finiteⁿᵒᵗᐧᵂᴹ.cardinal being not.in
almost all of the end.segments
(in only finitely.many end.segment.exceptions)
And why is this so?
Because
every endsegment removes exactly one number
from the content.
There is a bijection between
endsegments and removed numbers.
On 12/23/2024 6:32 AM, Richard Damon wrote:
On 12/23/24 4:31 AM, WM wrote:
On 23.12.2024 02:01, Richard Damon wrote:
On 12/22/24 5:11 PM, WM wrote:
On 22.12.2024 20:10, Richard Damon wrote:
On 12/22/24 8:11 AM, WM wrote:
Find a natural number that is not in all intervals [1, n] which I >>>>>>> use: ∀n ∈ ℕ [1, n].
But you can't use *ALL* intervals, becaue you need to use them
individually
No, I do as Cantor did.
No, you do what you THINK Cantor did,
Show an n that I do not use with all intervals [1, n].
The LAST one, which you say must exist to use your logic.
LOL! n+1?
Am I close here? Or way off?
Am 24.12.2024 um 00:49 schrieb Chris M. Thomasson:
On 12/23/2024 6:32 AM, Richard Damon wrote:
On 12/23/24 4:31 AM, WM wrote:
On 23.12.2024 02:01, Richard Damon wrote:
On 12/22/24 5:11 PM, WM wrote:
On 22.12.2024 20:10, Richard Damon wrote:
On 12/22/24 8:11 AM, WM wrote:
Find a natural number that is not in all intervals [1, n] which >>>>>>>> I use: ∀n ∈ ℕ [1, n].
But you can't use *ALL* intervals, becaue you need to use them
individually
No, I do as Cantor did.
No, you do what you THINK Cantor did,
Show an n that I do not use with all intervals [1, n].
The LAST one, which you say must exist to use your logic.
LOL! n+1?
Am I close here? Or way off?
Yeah, you are close to the edge.
See: https://www.youtube.com/watch?v=51oPKLSuyQY
On 12/23/2024 3:56 PM, Moebius wrote:
Am 24.12.2024 um 00:49 schrieb Chris M. Thomasson:
On 12/23/2024 6:32 AM, Richard Damon wrote:
On 12/23/24 4:31 AM, WM wrote:
On 23.12.2024 02:01, Richard Damon wrote:
On 12/22/24 5:11 PM, WM wrote:
On 22.12.2024 20:10, Richard Damon wrote:
On 12/22/24 8:11 AM, WM wrote:
Find a natural number that is not in all intervals [1, n] which >>>>>>>>> I use: ∀n ∈ ℕ [1, n].
But you can't use *ALL* intervals, becaue you need to use them >>>>>>>> individually
No, I do as Cantor did.
No, you do what you THINK Cantor did,
Show an n that I do not use with all intervals [1, n].
The LAST one, which you say must exist to use your logic.
LOL! n+1?
Am I close here? Or way off?
Yeah, you are close to the edge.
See: https://www.youtube.com/watch?v=51oPKLSuyQY
Does
{ 0, 1, 2, 3, ...}
equal:
{ 0 + 0, 0 + 1, 0 + 2, 0 + 3, ... }
Does WM mean ...
if WM thinks that the unsigned integers are NOT unbounded, well he
is wrong?
On 12/23/2024 4:08 PM, Moebius wrote:
Actually, I'm not interested in anything this asshole full of shit
means or doesn't mean, sorry.
Fair enough. Trying to engage with him (WM) is useless?
While, conversing with you is, actually, productive?
On 12/23/2024 4:12 PM, Moebius wrote:
Am 24.12.2024 um 01:09 schrieb Chris M. Thomasson:Ohhh, shit! The "almost" aspect scares me. His, so called, Students? YIKES!!!!! ;^o
if WM thinks that the unsigned integers are NOT unbounded, well he is
wrong?
WM is wrong about almost everything he's talking about (if
mathematics, especially set theory, is concerned).
On 12/23/24 3:52 AM, WM wrote:
On 23.12.2024 01:05, FromTheRafters wrote:
WM brought next idea :
The function E(n) decreases from infinity to zero because in set
theory ℕ \ {1, 2, 3, ...} = { } is an accepted formula.
In what way is it decreasing?
One by one. Every endsegment transforms a natural number from content
inside to index outside. That is guaranteed by mathematics:
∀k ∈ ℕ : ∩{E(0), E(1), E(2), ..., E(k+1)} = ∩{E(0), E(1), E(2), ...,
E(k)} \ {k+1}. Note the universal quantifier.
Except that you don't understand that infinitiy minus 1 is still
infinity, and thus did not "decrease".
On 12/23/24 4:31 AM, WM wrote:
No, I do as Cantor did.
No, you do what you THINK Cantor did,
Show an n that I do not use with all intervals [1, n].
The LAST one, which you say must exist to use your logic.
On 12/23/2024 6:18 AM, WM wrote:
The empty limit set is caused by
each finiteⁿᵒᵗᐧᵂᴹ.cardinal being not.in
almost all of the end.segments
(in only finitely.many end.segment.exceptions)
And why is this so?
Because
every endsegment removes exactly one number from the content.
There is a bijection between
endsegments and removed numbers.
There is a bijection between
more.than any.finiteⁿᵒᵗᐧᵂᴹ.cardinal.many
end.segments and
more.than any.finiteⁿᵒᵗᐧᵂᴹ.cardinal.many
removed numbers?
Have you told your students yet?
On 23.12.2024 19:20, Jim Burns wrote:
On 12/23/2024 6:18 AM, WM wrote:
The empty limit set is caused by
each finiteⁿᵒᵗᐧᵂᴹ.cardinal being not.in
almost all of the end.segments
(in only finitely.many end.segment.exceptions)
And why is this so?
Because
every endsegment removes exactly one number from the content.
There is a bijection between
endsegments and removed numbers.
There is a bijection between
more.than any.finiteⁿᵒᵗᐧᵂᴹ.cardinal.many
end.segments and
more.than any.finiteⁿᵒᵗᐧᵂᴹ.cardinal.many
removed numbers?
Nonsense.
Have you told your students yet?
Of course not. There is a bijection {n} <--> E(n). No cardinal number ℵ₀ is involved.
Regards, WM
On 23.12.2024 15:32, Richard Damon wrote:
On 12/23/24 4:31 AM, WM wrote:
No, I do as Cantor did.
No, you do what you THINK Cantor did,
Show an n that I do not use with all intervals [1, n].
The LAST one, which you say must exist to use your logic.
I do what Cantor did. There is no last one. You cannot show an n that I
do not use. There is none. Therefore all your arguing breaks down.
Regards, WM
On 23.12.2024 19:20, Jim Burns wrote:
[...]
There is a bijection {n} <--> E(n).
No cardinal number ℵ₀ is involved.
Cantor went insane, ....
Cantor Pairing works with any unsigned integer.
On 12/24/2024 02:45 AM, WM wrote:
On 23.12.2024 15:32, Richard Damon wrote:
On 12/23/24 4:31 AM, WM wrote:
Show an n that <bla bla bla> (WM)
The LAST one, which you say must exist to use your logic. (RD)
[...] There is no last one. (WM)
On 12/24/2024 4:07 PM, Ross Finlayson wrote:
On 12/24/2024 12:27 PM, Chris M. Thomasson wrote:
Cantor's Pairing works with any unsigned integer.
No, it works with two copies of all the integers, [...]
It works with any unsigned integer.
On 12/24/24 5:45 AM, WM wrote:
On 23.12.2024 15:32, Richard Damon wrote:No, you do NOT do what Cantor did,
On 12/23/24 4:31 AM, WM wrote:
No, I do as Cantor did.
No, you do what you THINK Cantor did,
Show an n that I do not use with all intervals [1, n].
The LAST one, which you say must exist to use your logic.
I do what Cantor did. There is no last one. You cannot show an n that
I do not use. There is none. Therefore all your arguing breaks down.
On 12/24/24 5:53 AM, WM wrote:
Of course not. There is a bijection {n} <--> E(n). No cardinal numberthe set of E(n), are Aleph_0.
ℵ₀ is involved.
Except that both sets, this size of the Natural Numbers and the size of
On 24.12.2024 15:06, Richard Damon wrote:
On 12/24/24 5:45 AM, WM wrote:
On 23.12.2024 15:32, Richard Damon wrote:No, you do NOT do what Cantor did,
On 12/23/24 4:31 AM, WM wrote:
No, I do as Cantor did.
No, you do what you THINK Cantor did,
Show an n that I do not use with all intervals [1, n].
The LAST one, which you say must exist to use your logic.
I do what Cantor did. There is no last one. You cannot show an n that
I do not use. There is none. Therefore all your arguing breaks down.
What n do I not use?
Regards, WM
ℕ is the set of finite.cardinals,
all the finite.cardinals,
the visibleᵂᴹ and the darkᵂᴹ.
I gather that we don't know the name of
any darkᵂᴹ finite.cardinal δ
Still, we know (to start with)
one fact about δ
#⟦0,δ⦆ < #(⟦0,δ⦆∪{δ})
because, being a finite.cardinal, δ ∈ ℕ
or, more.verbosely, we can write
δ ∈ ⦃i: #⟦0,i⦆<#(⟦0,i⦆∪{i}) ⦄
----
For finite sets,
emptier.by.one sets are smaller.by.one, and
fuller.by.one sets are larger.by one.
That can serve as what.we.mean.by 'finite'
On 12/26/24 6:36 AM, WM wrote:
On 24.12.2024 15:06, Richard Damon wrote:The LAST one, that completes the set.
On 12/24/24 5:45 AM, WM wrote:
On 23.12.2024 15:32, Richard Damon wrote:No, you do NOT do what Cantor did,
On 12/23/24 4:31 AM, WM wrote:
No, I do as Cantor did.
No, you do what you THINK Cantor did,
Show an n that I do not use with all intervals [1, n].
The LAST one, which you say must exist to use your logic.
I do what Cantor did. There is no last one. You cannot show an n
that I do not use. There is none. Therefore all your arguing breaks
down.
What n do I not use?
The fact that after any finite number of removals, there are still
elements does not mean that when you remove *ALL* the elements there
will still be some left.#
On 22.12.2024 11:16, Jim Burns wrote:
What does 'finite' mean?
A set is finite if it contains
a visible natural number of elements.
All natural numbers are finite but
the realm of dark numbers
appears like infinity.
E(n+2) is
the set of all finite.cardinals > n+2
E(n+1) = E(n+2)∪{n+2}
E(n+2)∪{n+2} isn't larger.than E(n+2)
Wrong.
#E(n+2) isn't any of the finite.cardinals in ℕ
It is an infinite number but
even infinite numbers differ like |ℕ| =/= |ℕ| + 1.
On 18.12.2024 21:15, joes wrote:And we know everything not written by Cantor is false!
Am Wed, 18 Dec 2024 20:07:35 +0100 schrieb WM:It is wrong because there is no such text from Cantor.
On 18.12.2024 10:35, joes wrote:Why is it wrong?
> Am Tue, 17 Dec 2024 22:49:51 +0100 schrieb WM:
>> On 17.12.2024 13:34, Richard Damon wrote:
>>
>>> Your logic that if it holds for all FISONs, it holds for N,
>> Please explain what Cantor does to apply more than what I apply,
>> namely all n ∈ ℕ.
> He „applies“ the set of all N, as opposed to every single n.
Please quote the text from which you have obtained that wrong idea.
On 12/22/2024 6:32 AM, WM wrote:
I (JB) think that it may be that
'almost.all' '(∀)' refers concisely to
the differences in definition at
the center of our discussion.
For each finite.cardinal,
almost.all finite.cardinals are larger.
∀j ∈ ℕⁿᵒᵗᐧᵂᴹ: (∀)k ∈ ℕⁿᵒᵗᐧᵂᴹ: j < k
I think that you (WM) would deny that.
You would say, instead,
ᵂᴹ⎛ for each definable finite.cardinal
ᵂᴹ⎜ almost.all finite cardinals are larger.
For sequence ⟨Sₙ⟩ₙ᳹₌₀ of sets
Sₗᵢₘ is a limit.set of ⟨Sₙ⟩ₙ᳹₌₀
if
each x ∈ Sₗᵢₘ is ∈ almost.all Sₖ ∈ ⟨Sₙ⟩ₙ᳹₌₀ and each y ∉ Sₗᵢₘ is ∉ almost.all Sₖ ∈ ⟨Sₙ⟩ₙ᳹₌₀ ⟨Sₙ⟩ₙ᳹₌₀ ⟶ Sₗᵢₘ ⇐
⎛ x ∈ Sₗᵢₘ ⇐ (∀)Sₖ ∈ ⟨Sₙ⟩ₙ᳹₌₀: x ∈ Sₖ
⎝ y ∉ Sₗᵢₘ ⇐ (∀)Sₖ ∈ ⟨Sₙ⟩ₙ᳹₌₀: y ∉ Sₖ
---
The notation a͚ or S͚ for aₗᵢₘ or Sₗᵢₘ
is tempting, but
it gives the unfortunate impression that
a͚ and S͚ are the infinitieth entries of
their respective infinite.sequences.
They aren't infinitieth entries.
They are defined differently.
E(n+2) is
the set of all finite.cardinals > n+2
E(n+1) = E(n+2)∪{n+2}
E(n+2)∪{n+2} isn't larger.than E(n+2)
Wrong.
Almost all finite.cardinals are larger than
finite.cardinal n+1
{n+2} isn't large enough to change that.
Almost all finite.cardinals are larger than
finite cardinal n+2.
#E(n+2) isn't any of the finite.cardinals in ℕ
It is an infinite number but
even infinite numbers differ like |ℕ| =/= |ℕ| + 1.
Infiniteᵂᴹ numbers which differ like |ℕ| ≠ |ℕ| + 1.
are finiteⁿᵒᵗᐧᵂᴹ numbers.
The concept of 'limit' is a cornerstone of
calculus and analysis and topology.
That cornerstone rests upon 'almost.all'.
Each finite.cardinal has infinitely.more
finite.cardinals after it than before it.
If one re.defines things away from that,
it is only an odd coincidence if, after that,
any part of calculus or analysis or topology
continues to make sense.
Am Tue, 24 Dec 2024 11:42:56 +0100 schrieb WM:
The sets E(n) decrease. If the sequence (E(n)) could not get empty oneThe „sequence” doesn’t „get” empty. No element is empty.
by one then Cantor could not set up an infinite sequence using all
indices n of that sequence.
On 26.12.2024 20:59, joes wrote:
Am Tue, 24 Dec 2024 11:42:56 +0100 schrieb WM:
The sets E(n) decrease. If the sequence (E(n)) could not get empty oneThe „sequence” doesn’t „get” empty. No element is empty.
by one then Cantor could not set up an infinite sequence using all
indices n of that sequence.
The sequence does not get empty in the visible domain. That proves that
not all indices can be applied and therefore there is no bijection with ℕ.
Regards, WM
On 26.12.2024 19:41, Jim Burns wrote:
[...]
A limit is a set S͚ such that nothing fits
between it and all sets of the sequence.
The notation a͚ or S͚ for aₗᵢₘ or Sₗᵢₘ
is tempting, but
it gives the unfortunate impression that
a͚ and S͚ are the infinitieth entries of
their respective infinite.sequences.
They aren't infinitieth entries.
They are defined differently.
The last natural number is finite,
But like all dark numbers
it has no FISON
#E(n+2) isn't any of the finite.cardinals in ℕ
It is an infinite number but
even infinite numbers differ like |ℕ| =/= |ℕ| + 1.
Infiniteᵂᴹ numbers which differ like |ℕ| ≠ |ℕ| + 1.
are finiteⁿᵒᵗᐧᵂᴹ numbers.
No.
They are invariable numbers like ω and ω+1.
Am 26.12.2024 um 05:02 schrieb Chris M. Thomasson:
On 12/24/2024 4:07 PM, Ross Finlayson wrote:
On 12/24/2024 12:27 PM, Chris M. Thomasson wrote:
Cantor's Pairing works with any unsigned integer.
No, it works with two copies of all the integers, [...]
It works with any unsigned integer.
It works especially with the (elements in the) two sets {0, 2, 4, ...}
and {1, 3, 5, ...}:
n <-> n+1 .
P := {(n, n+1) : n e {0, 2, 4, ...}} .
Then P is {(0, 1}, (2, 3}, (4, 5}, ...}.
.
.
.
On 12/17/24 4:51 PM, WM wrote:
all FISONs contain/are all n.
On 12/17/24 4:51 PM, WM wrote:
all FISONs contain/are all n.
On 12/17/24 4:51 PM, WM wrote:
all FISONs contain/are all n.
Mumbo-jumbo.
Hint: For each and every n e IN there is a FISON such that n is in it.
BUT there's no FISON such that each and every n e IN is in it.
Mückenheim, you are insane (mad, gaga, loco loco).
.
.
.
On 12/17/24 4:51 PM, WM wrote:
all FISONs contain/are all n.
Hint: For each and every n e IN there is a FISON such that n is in it.
BUT there's no FISON such that each and every n e IN is in it.
On 12/17/24 4:51 PM, WM wrote:
all FISONs contain/are all n.
Hint: For each and every n e IN there is a FISON such that n is in it.
On 12/27/2024 5:14 AM, WM wrote:
They are invariable numbers like ω and ω+1.
ω is
the set of (well.ordered) ordinals k such that
#⟦0,k⦆ ≠ #(⟦0,k⦆∪⦃k⦄)
(such that k is finite)
A separate fact is that
⟦0,ω⦆ ≠ ⟦0,ω⦆∪⦃ω⦄
On 12/27/2024 01:00 PM, Jim Burns wrote:
[...]
The, "almost all", or, "almost everywhere",
does _not_ equate to "all" or "everywhere",
and these days in
sub-fields of mathematics like to do with
topology and the ultrafilter,
it's a usual conceit to
in at least one sense,
not being "actually" correct.
On 28.12.2024 06:23, Moebius wrote:
On 12/17/24 4:51 PM, WM wrote:
all FISONs contain/are all n.
Hint: For each and every n e IN there is a FISON such that n is in it.
No. If so, then the union of all FISONs would contain all n. But fact is
that the union of all FISONs is a FISON (all FISONs have infinitely many successors) and leaves almost all numbers outside
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
Regards, WM
On 28.12.2024 06:23, Moebius wrote:That is the case.
No. If so, then the union of all FISONs would contain all n.Hint: For each and every n e IN there is a FISON such that n is in it.On 12/17/24 4:51 PM, WM wrote:
all FISONs contain/are all n.
But fact isNo. There are infinitely many, so the union must be infinite too.
that the union of all FISONs is a FISON (all FISONs have infinitely many successors) and leaves almost all numbers outside ∀n ∈ ℕ_def: |ℕ \ {1,
2, 3, ..., n}| = ℵo.
doing an INFIITE union (which a union of all FISONs would be) can
result in something different in type then the union of a finite number
of FISONs.
On 12/27/2024 5:24 PM, Ross Finlayson wrote:
On 12/27/2024 01:00 PM, Jim Burns wrote:
[...]
The, "almost all", or, "almost everywhere",
does _not_ equate to "all" or "everywhere",
Correct.
⎛ In mathematics, the term "almost all" means
⎜ "all but a negligible quantity".
⎜ More precisely, if X is a set,
⎜ "almost all elements of X" means
⎜ "all elements of X but those in
⎜ a negligible subset of X".
⎜ The meaning of "negligible" depends on
⎜ the mathematical context; for instance,
Am Sat, 28 Dec 2024 15:03:44 +0100 schrieb WM:
ω is the limit of the sequence 1, 2, 3, ... .It is not a proper limit because the sequence doesn’t converge.
Undefined. Also you have the infinity of „dark numbers” consecutiveA separate fact is that ⟦0,ω⦆ ≠ ⟦0,ω⦆∪⦃ω⦄[0, ω-1] = [0,ω⦆ = ℕ =/= [0, ω]
to the naturals.
On 28.12.2024 15:12, Jim Burns wrote:A relative size is useless because it is still exactly as infinite,
On 12/27/2024 5:24 PM, Ross Finlayson wrote:
On 12/27/2024 01:00 PM, Jim Burns wrote:
The, "almost all", or, "almost everywhere",
does _not_ equate to "all" or "everywhere",
Correct.
⎛ In mathematics, the term "almost all" means ⎜ "all but a negligible
quantity".
⎜ More precisely, if X is a set,
⎜ "almost all elements of X" means ⎜ "all elements of X but those in ⎜ >> a negligible subset of X".
⎜ The meaning of "negligible" depends on ⎜ the mathematical context;
A good example is the set of FISONs. Every FISON contains only a
negligible quantity of natural numbers. A generous estimation is: Every
FISON contains less than 1 % of all natural numbers. There is no FISON
that contains more than 1 %. Therefore the union of all FISONs contains
less than 1 % of all natural numbers. Outside of the union of FISONs are almost all natural numbers.
On 27.12.2024 22:00, Jim Burns wrote:
On 12/27/2024 5:14 AM, WM wrote:
They are invariable numbers like ω and ω+1.
ω is
the set of (well.ordered) ordinals k such that
#⟦0,k⦆ ≠ #(⟦0,k⦆∪⦃k⦄)
(such that k is finite)
That is one interpretation.
My interpretation is Cantor's original one:
ω is the limit of the sequence 1, 2, 3, ... .
A separate fact is that
⟦0,ω⦆ ≠ ⟦0,ω⦆∪⦃ω⦄
[0, ω-1] = [0,ω⦆ = ℕ =/= [0, ω]
On 28.12.2024 15:12, Jim Burns wrote:
On 12/27/2024 5:24 PM, Ross Finlayson wrote:
The, "almost all", or, "almost everywhere",
does _not_ equate to "all" or "everywhere",
Correct.
⎛ In mathematics, the term "almost all" means
⎜ "all but a negligible quantity".
⎜ More precisely, if X is a set,
⎜ "almost all elements of X" means
⎜ "all elements of X but those in
⎜ a negligible subset of X".
⎜ The meaning of "negligible" depends on
⎜ the mathematical context; for instance,
A good example is the set of FISONs.
Every FISON contains only
a negligible quantity of natural numbers.
A generous estimation is:
Every FISON contains
less than 1 % of all natural numbers.
There is no FISON that contains more than 1 %.
Therefore the union of all FISONs contains
less than 1 % of all natural numbers.
Outside of the union of FISONs are
almost all natural numbers.
On 28.12.2024 15:15, Richard Damon wrote:
doing an INFIITE union (which a union of all FISONs would be) can
result in something different in type then the union of a finite
number of FISONs.
Every FISON covers less than half of all natural numbers. That implies
that all FISONs and the union of all FISONs cover less than half of all natural numbers.
Regards, WM
On 28.12.2024 06:23, Moebius wrote:Which, in standard mathematics, is all inf. many of them
All FISONs contain all *definable* n.On 12/17/24 4:51 PM, WM wrote:
all FISONs contain/are all n.
Almost none, actually, N being infinite and all.Hint: For each and every n e IN there is a FISON such that n is in it.Every FISON contains less than half of all natural numbers.
BUT there's no FISON such that each and every n e IN is in it.
ThereforeWorng.
the union of all of them contains less than half of all natural numbers.
On 28.12.2024 15:12, Jim Burns wrote:
On 12/27/2024 5:24 PM, Ross Finlayson wrote:
On 12/27/2024 01:00 PM, Jim Burns wrote:
[...]
The, "almost all", or, "almost everywhere",
does _not_ equate to "all" or "everywhere",
Correct.
⎛ In mathematics, the term "almost all" means
⎜ "all but a negligible quantity".
⎜ More precisely, if X is a set,
⎜ "almost all elements of X" means
⎜ "all elements of X but those in
⎜ a negligible subset of X".
⎜ The meaning of "negligible" depends on
⎜ the mathematical context; for instance,
A good example is the set of FISONs. Every FISON contains only a
negligible quantity of natural numbers. A generous estimation is: Every
FISON contains less than 1 % of all natural numbers. There is no FISON
that contains more than 1 %. Therefore the union of all FISONs contains
less than 1 % of all natural numbers. Outside of the union of FISONs are almost all natural numbers.
Regards, WM
Regards, WM
Am 28.12.2024 um 06:23 schrieb Moebius:
On 12/17/24 4:51 PM, WM wrote:
all FISONs contain/are all n.
Mumbo-jumbo.
Hint: For each and every n e IN there is a FISON such that n is in it.
BUT there's no FISON such that each and every n e IN is in it.
Mückenheim: "all singletons contain/are all n".
Hint: For each and every n e IN there is a singleton such that n is in
it (namely {n}).
BUT there's no singleton such that each and every n e IN is in it.
<facepalm>
Say, 1875520200019832...
Each digit is random. This [represents] a number and its within the infinite set
of unsigned integers.
Am 29.12.2024 um 01:48 schrieb Chris M. Thomasson:
Say, 1875520200019832...
Each digit is random. This [represents] a number and its within the
infinite set of unsigned integers.
Nope. Such a "number" is not an integer.
Hint: Each and every integer has to be "represented" by a finite (!)
string of digits.
.
.
.
On 12/28/2024 11:17 AM, Richard Damon wrote:
[...]
Consider
a random uniform distribution of natural integers,
same probability of each integer.
Now, you might aver
"that can't exist, because it would be
non-standard or not-a-real-function".
Then it's like
"no, it's distribution is non-standard,
not-a-real-function,
with real-analytical-character".
On 12/28/24 11:50 AM, WM wrote:
On 12/27/2024 01:00 PM, Jim Burns wrote:
"almost all" does _not_ equate to "all".
A good example [are the endsegments]. Every [endsegments] contains ["all but finitely many"] [aka "almost all"] natural numbers.
On 12/28/24 11:50 AM, WM wrote:
On 12/27/2024 01:00 PM, Jim Burns wrote:
"almost all" does _not_ equate to "all".
A good example [are the endsegments]. Every [endsegment] contains ["all but finitely many"] [aka "almost all"] natural numbers.
Am 29.12.2024 um 01:48 schrieb Chris M. Thomasson:
Say, 1875520200019832...
Each digit is random. This [represents] a number and its within the
infinite set of unsigned integers.
Nope. Such a "number" is not an integer.
Hint: Each and every "unsigned" integer has to be "represented" by a finite (!)
string of digits.
.
.
.
Say, 1875520200019832...
Each digit is random. This [represents] a number and its within the infinite set
of unsigned integers.
Say, 1875520200019832...
Each digit is random. This [represents] a number and it's within the infinite set
of unsigned integers.
Am 29.12.2024 um 01:48 schrieb Chris M. Thomasson:
Say, 1875520200019832...
Each digit is random. This [represents] a number and it's within the
infinite set of unsigned integers.
Nope. Such a "number" is not an integer.
Hint: Each and every "unsigned" integer has to be "represented" by a finite (!)
string of digits.
.
.
.
On 12/28/2024 5:58 PM, Moebius wrote:
Am 29.12.2024 um 01:48 schrieb Chris M. Thomasson:
Say, 1875520200019832...
Each digit is random. This [represents] a number and it's within the
infinite set of unsigned integers.
Nope. Such a "number" is not an integer.
It's an unsigned integer at every step?
Am 29.12.2024 um 03:07 schrieb Chris M. Thomasson:
On 12/28/2024 5:58 PM, Moebius wrote:
Am 29.12.2024 um 01:48 schrieb Chris M. Thomasson:
Say, 1875520200019832...
Each digit is random. This [represents] a number and it's within the
infinite set of unsigned integers.
Nope. Such a "number" is not an integer.
It's an unsigned integer at every step?
Sure.
For each and every n e IN the string
d_n d_(n-1) ... d_0
where d_i e {"0", "1", ..., "9"} for all i e {0, ..., n} represents an unsigned integer.
Am 29.12.2024 um 03:07 schrieb Chris M. Thomasson:
On 12/28/2024 5:58 PM, Moebius wrote:
Am 29.12.2024 um 01:48 schrieb Chris M. Thomasson:
Say, 1875520200019832...
Each digit is random. This [represents] a number and it's within the
infinite set of unsigned integers.
Nope. Such a "number" is not an integer.
It's an unsigned integer at every step?
Sure.
For each and every n e IN the string
d_n d_(n-1) ... d_0
where d_i e {"0", "1", ..., "9"} for all i e {0, ..., n} represents an unsigned integer.
On 12/28/24 11:50 AM, WM wrote:
Every Natural Number is less that almost all other natural numbers, so
its %-tile of progress is effectively 0, but together they make up the
whole infinite set.
Am Sat, 28 Dec 2024 17:50:49 +0100 schrieb WM:
A good example is the set of FISONs. Every FISON contains only aA relative size is useless because it is still exactly as infinite,
negligible quantity of natural numbers. A generous estimation is: Every
FISON contains less than 1 % of all natural numbers. There is no FISON
that contains more than 1 %. Therefore the union of all FISONs contains
less than 1 % of all natural numbers. Outside of the union of FISONs are
almost all natural numbers.
i.e. countable.
On 12/28/2024 11:50 AM, WM wrote:
A good example is the set of FISONs.
Every FISON contains only
a negligible quantity of natural numbers.
Yes.
A generous estimation is:
Every FISON contains
less than 1 % of all natural numbers.
There is no FISON that contains more than 1 %.
Yes.
Therefore the union of all FISONs contains less than 1 % of all
natural numbers.
No.
The union of FISONS contains
100% of all members of any FISON.
Still,
each FISON contains
less than 1% of the union of FISONs.
Outside of the union of FISONs are almost all natural numbers.
No.
Hint: For each and every n e IN there is a singleton such that n is in
it (namely {n}).
BUT there's no singleton such that each and every n e IN is in it.
On 12/28/2024 9:03 AM, WM wrote:
[0, ω-1] = [0,ω⦆ = ℕ =/= [0, ω]
Yes,
⟦0,ω⦆ = ℕ ≠ ⟦0,ω⟧
However,
⎛ Assume k < ω
⎜ #⟦0,k⦆ ≠ #(⟦0,k⦆∪⦃k⦄)
⎜ #⟦0,k+1⦆ ≠ #(⟦0,k+1⦆∪⦃k+1⦄)
⎝ k+1 < ω
Am Sat, 28 Dec 2024 14:58:50 +0100 schrieb WM:My theorem: If all definable numbers (FISONs) stay below a certain
ThereforeWorng.
the union of all of them contains less than half of all natural numbers.
On 28.12.2024 18:48, joes wrote:
Am Sat, 28 Dec 2024 17:50:49 +0100 schrieb WM:
A good example is the set of FISONs. Every FISON contains only aA relative size is useless because it is still exactly as infinite,
negligible quantity of natural numbers. A generous estimation is: Every
FISON contains less than 1 % of all natural numbers. There is no FISON
that contains more than 1 %. Therefore the union of all FISONs contains
less than 1 % of all natural numbers. Outside of the union of FISONs are >>> almost all natural numbers.
i.e. countable.
A relative size is very useful in order to correct the fantasy claim
that the union of all FISONs covers the infinite gap between all FISONs
and ω.
Regards, WM
On 28.12.2024 20:17, Richard Damon wrote:
On 12/28/24 11:50 AM, WM wrote:
Every Natural Number is less that almost all other natural numbers, so
its %-tile of progress is effectively 0, but together they make up the
whole infinite set.
All definable numbers (FISONs) stay below 1 %. Every union of "below 1
%" stays below 1 %.
Regards, WM
On 28.12.2024 19:48, Jim Burns wrote:
On 12/28/2024 11:50 AM, WM wrote:
A good example is the set of FISONs.
Every FISON contains only
a negligible quantity of natural numbers.
Yes.
A generous estimation is:
Every FISON contains
less than 1 % of all natural numbers.
There is no FISON that contains more than 1 %.
Yes.
Therefore the union of all FISONs contains less than 1 % of all
natural numbers.
No.
The union of FISONS contains
100% of all members of any FISON.
Of course. But that is less than 1 % of all natural numbers.
Still,
each FISON contains
less than 1% of the union of FISONs.
That is nonsense. Note that when two or 10 or many FISONs miss 99 %,
then their union misses 99 %.
Outside of the union of FISONs are almost all natural numbers.
No.
You violate straight mathematics and logic like with your Bob. There is
a upper threshold for all FISONs (99 %). This cannot be surpassed by
their union.
Regards, WM
On 28.12.2024 20:35, joes wrote:The infinite union doesn’t.
Am Sat, 28 Dec 2024 14:58:50 +0100 schrieb WM:
My theorem: If all definable numbers (FISONs) stay below a certainTherefore the union of all of them contains less than half of allWorng.
natural numbers.
threshold, then every union of those FISONs stays below that threshold.
On 12/29/24 6:01 AM, WM wrote:
All definable numbers (FISONs) stay below 1 %. Every union of "below 1
%" stays below 1 %.
Since 0 is Less than 1, you are sort of correct, but that fact doesn't
prove your claim.
The problem is that when you get to *ALL*
NBo, you "Naive" (which you call straight) mathematics just can't handle
this sort of math,
Am Sun, 29 Dec 2024 12:05:26 +0100 schrieb WM:
My theorem: If all definable numbers (FISONs) stay below a certain
threshold, then every union of those FISONs stays below that threshold.
The infinite union doesn’t.
On 29.12.2024 15:43, joes wrote:No such finite union equals N.
Am Sun, 29 Dec 2024 12:05:26 +0100 schrieb WM:
On 28.12.2024 20:35, joes wrote:
Am Sat, 28 Dec 2024 14:58:50 +0100 schrieb WM:
The union of all FISONs below a certain threshold, namely a thresholdThe infinite union doesn’t.My theorem: If all definable numbers (FISONs) stay below a certainTherefore the union of all of them contains less than half of allWorng.
natural numbers.
threshold, then every union of those FISONs stays below that
threshold.
between which and ω there exist ℵ₀ natnumbers, remains below this very threshold.
On 28.12.2024 18:31, Jim Burns wrote:
On 12/28/2024 9:03 AM, WM wrote:
[0, ω-1] = [0,ω⦆ = ℕ =/= [0, ω]
Yes,
⟦0,ω⦆ = ℕ ≠ ⟦0,ω⟧
However,
⎛ Assume k < ω
⎜ #⟦0,k⦆ ≠ #(⟦0,k⦆∪⦃k⦄)
⎜ #⟦0,k+1⦆ ≠ #(⟦0,k+1⦆∪⦃k+1⦄)
⎝ k+1 < ω
That holds for almost all natural numbers k.
However,
⎛ Assume k < ω
⎜ #⟦0,k⦆ ≠ #(⟦0,k⦆∪⦃k⦄)
⎜ #⟦0,k+1⦆ ≠ #(⟦0,k+1⦆∪⦃k+1⦄)
⎝ k+1 < ω
That holds for almost all natural numbers k.
It cannot hold for an actually infinite system
without disappearing Bob.
It cannot hold for an actually infinite system
without disappearing Bob.
On 12/29/2024 06:43 AM, joes wrote:
The infinite union doesn’t.
There is no "infinite union" in ZF,
only "pair-wise union",
according to the axiom of union.
On 12/29/2024 12:34 PM, Jim Burns wrote:
On 12/28/2024 10:54 PM, Ross Finlayson wrote:
On 12/28/2024 04:22 PM, Jim Burns wrote:
On 12/28/2024 5:36 PM, Ross Finlayson wrote:
Then it's like
"no, it's distribution is non-standard,
not-a-real-function,
with real-analytical-character".
Which is to say,
"no, it isn't what it's described to be"
You already accept
No.
You (RF) are greatly mistaken about
my (JB's) position with regard to
infinitely.many equal real.number steps
from 0 to 1
My position is and has been that they don't exist.
You already accept that the "natural/unit
equivalency function" has range with
_constant monotone strictly increasing_
has _constant_ differences, _constant_,
that as a cumulative function, for a
distribution, has that relating to
the naturals, as uniform.
My position, expressed in different ways,
is and has been that,
for each positive real x,
a finite integer n exists such that
n⋅x > 1
That conflicts with the existence of
infinitely.many equal real.number steps
from 0 to 1
Oh, you had that [0/d, 1/d, 2/d, ... oo/d]
was a thing,
So, yeah, it conflicts with yourself,
yet, that's what you said.
On 28.12.2024 18:48, joes wrote:There is no gap.
Am Sat, 28 Dec 2024 17:50:49 +0100 schrieb WM:
A relative size is very useful in order to correct the fantasy claimA good example is the set of FISONs. Every FISON contains only aA relative size is useless because it is still exactly as infinite,
negligible quantity of natural numbers. A generous estimation is:
Every FISON contains less than 1 % of all natural numbers. There is no
FISON that contains more than 1 %. Therefore the union of all FISONs
contains less than 1 % of all natural numbers. Outside of the union of
FISONs are almost all natural numbers.
i.e. countable. You seem to think there are only finitely many naturals
and therefore FISes. Of course, if you only consider fin. many, you
don’t have all.
that the union of all FISONs covers the infinite gap between all FISONs
and ω.
On 12/29/2024 12:34 PM, Jim Burns wrote:
On 12/29/2024 1:15 PM, Ross Finlayson wrote:
On 12/29/2024 06:43 AM, joes wrote:
The infinite union doesn’t.
There is no "infinite union" in ZF,
only "pair-wise union",
according to the axiom of union.
No.
https://en.wikipedia.org/wiki/Axiom_of_union
⎛ Informally, the axiom states that
⎜ for each set x there is a set y
⎜ whose elements are precisely
⎝ the elements of the elements of x.
"In-formally [naively], ...."
Am Sun, 29 Dec 2024 18:56:23 +0100 schrieb WM:
The union of all FISONs below a certain threshold, namely a thresholdNo such finite union equals N.
between which and ω there exist ℵ₀ natnumbers, remains below this very >> threshold.
Bob can disappear within a larger.enough set,
Am Sun, 29 Dec 2024 11:39:22 +0100 schrieb WM:
A relative size is very useful in order to correct the fantasy claimThere is no gap.
that the union of all FISONs covers the infinite gap between all FISONs
and ω.
On 12/29/2024 3:09 AM, WM wrote:
My theorem: Every union of FISONs which stay below a certain threshold
stays below that threshold.
On 29.12.2024 13:31, Richard Damon wrote:
NBo, you "Naive" (which you call straight) mathematics just can't
handle this sort of math,
My theorem: Every union of FISONs which stay below a certain threshold
stays below that threshold.
Find a counterexample. Don't claim it but prove it. Fail.
Contradicting this theorem without reason is outside of mathematics.
Regards, WM
On 29.12.2024 22:27, joes wrote:
Am Sun, 29 Dec 2024 11:39:22 +0100 schrieb WM:
A relative size is very useful in order to correct the fantasy claimThere is no gap.
that the union of all FISONs covers the infinite gap between all FISONs
and ω.
If all FISONs are finite and followed by infinite endsegments, then
there is an infinite gap.
My theorem: Every union of FISONs which stay below a certain threshold
stays below that threshold.
Find a counterexample. Don't claim it but prove it. Fail.
Regards, WM
On 29.12.2024 13:34, Richard Damon wrote:
On 12/29/24 6:01 AM, WM wrote:
All definable numbers (FISONs) stay below 1 %. Every union of "below
1 %" stays below 1 %.
Since 0 is Less than 1, you are sort of correct, but that fact doesn't
prove your claim.
The problem is that when you get to *ALL*
I get to all FISONs below a certain threshold, namely a threshold
between which and ω there exist ℵ₀ natnumbers.
Regards, WM
On 12/29/24 5:31 PM, WM wrote:
My theorem: Every union of FISONs which stay below a certain threshold
stays below that threshold.
Find a counterexample. Don't claim it but prove it. Fail.
But what does that prove?
The finite numbers are finite, and that is it.
Am 29.12.2024 um 23:15 schrieb Chris M. Thomasson:
On 12/29/2024 3:09 AM, WM wrote:
My theorem: Every union of FISONs which stay below a certain
threshold stays below that threshold.
Your "theorem" is false, Mückenheim. There's a simple counter example:
For each and every FISON F: F c_strict (WM: "below") IN. But the union
of all FISONs is not c_strict (WM: "below") IN but = IN.
On 12/29/24 12:53 PM, WM wrote:
On 29.12.2024 13:34, Richard Damon wrote:Which isn't "All FISONs"
On 12/29/24 6:01 AM, WM wrote:
All definable numbers (FISONs) stay below 1 %. Every union of "below
1 %" stays below 1 %.
Since 0 is Less than 1, you are sort of correct, but that fact
doesn't prove your claim.
The problem is that when you get to *ALL*
I get to all FISONs below a certain threshold, namely a threshold
between which and ω there exist ℵ₀ natnumbers.
On 30.12.2024 01:39, Richard Damon wrote:So not all. All those thresholds (and the FISONs) are finite and do not
On 12/29/24 12:53 PM, WM wrote:
On 29.12.2024 13:34, Richard Damon wrote:
On 12/29/24 6:01 AM, WM wrote:
All definable numbers (FISONs) stay below 1 %. Every union of "below >>>>> 1 %" stays below 1 %.
Since 0 is Less than 1, you are sort of correct, but that fact
doesn't prove your claim.
The problem is that when you get to *ALL*
I get to all FISONs below a certain threshold, namely a threshold
between which and ω there exist ℵ₀ natnumbers.
threshold +1Which isn't "All FISONs"Do you personally know FISONs larger than those? Please let me know
them.
On 30.12.2024 01:39, Richard Damon wrote:
On 12/29/24 12:53 PM, WM wrote:
On 29.12.2024 13:34, Richard Damon wrote:Which isn't "All FISONs"
On 12/29/24 6:01 AM, WM wrote:
All definable numbers (FISONs) stay below 1 %. Every union of
"below 1 %" stays below 1 %.
Since 0 is Less than 1, you are sort of correct, but that fact
doesn't prove your claim.
The problem is that when you get to *ALL*
I get to all FISONs below a certain threshold, namely a threshold
between which and ω there exist ℵ₀ natnumbers.
Do you personally know FISONs larger than those? Please let me know them.
Regards, WM
On 30.12.2024 01:36, Richard Damon wrote:
On 12/29/24 5:31 PM, WM wrote:
My theorem: Every union of FISONs which stay below a certain
threshold stays below that threshold.
Find a counterexample. Don't claim it but prove it. Fail.
But what does that prove?
That proves the existence of dark numbers.
The finite numbers are finite, and that is it.
FISONs do not grow by unioning them. All FISONs and their unions stay
below 1 % of ℕ.
Proof: Every FISON that is multiplied by 100 remains a FISON that can be multiplied by 100 without changing this property.
If ℕ is an actually infinite set then it is not made by FISONs.
Regards, WM
On 12/29/2024 01:41 PM, Jim Burns wrote:
On 12/29/2024 3:54 PM, Ross Finlayson wrote:
Oh, you had that [0/d, 1/d, 2/d, ... oo/d]
was a thing,
No.
What I had was that
⋃ᵢ₌᳹₀[0/i,1/i,2/i,...,i/i]
is a thing,
but the thing which it is
is not an interval of real numbers.
That's still my position.
So, yeah, it conflicts with yourself,
yet, that's what you said.
No,
it isn't [what I said].
It's also an equi-partitioning.
Here of course it IS a continuous domain
or model of real numbers, ran(EF),
because being a distribution of the
natural integers at uniform random.
Well, go ahead then and come up with
some reason why it's not an equi-partitioning
as it fulfills otherwise being a CDF.
Figuring you just can't accept it, ....
On 12/29/2024 01:48 PM, Jim Burns wrote:
On 12/29/2024 3:57 PM, Ross Finlayson wrote:
On 12/29/2024 12:34 PM, Jim Burns wrote:
On 12/29/2024 1:15 PM, Ross Finlayson wrote:
On 12/29/2024 06:43 AM, joes wrote:
There is no "infinite union" in ZF,
only "pair-wise union",
according to the axiom of union.
No.
https://en.wikipedia.org/wiki/Axiom_of_union
⎛ Informally, the axiom states that
⎜ for each set x there is a set y
⎜ whose elements are precisely
⎝ the elements of the elements of x.
"In-formally [naively], ...."
'Informaclly' isn't 'naively'.
⎛ In the formal language of the Zermelo–Fraenkel axioms,
⎜ the axiom reads:
⎝ ∀A∃B∀c(c∈B⟺∃D(c∈D∧D∈A))
-- ibid.
Yeah, "pair-wise".
I'm telling you, I am not wrong,
and for a long time, I am not under-informed,
about ZF and ZFC set theories.
Mistakes to the contrary
are wrong and/or under-informed.
On 29.12.2024 22:27, joes wrote:Nope. Infinity is just that far away - not finite.
Am Sun, 29 Dec 2024 11:39:22 +0100 schrieb WM:
If all FISONs are finite and followed by infinite endsegments, thenA relative size is very useful in order to correct the fantasy claimThere is no gap.
that the union of all FISONs covers the infinite gap between all
FISONs and ω.
there is an infinite gap.
My theorem: Every union of FISONs which stay below a certain thresholdGroundbreaking.
stays below that threshold.
On 29.12.2024 23:45, Moebius wrote:Even better, they are finite (but they do grow).
Am 29.12.2024 um 23:15 schrieb Chris M. Thomasson:That is not a proof but a silly claim. Believers of such garbage should
On 12/29/2024 3:09 AM, WM wrote:
My theorem: Every union of FISONs which stay below a certain
threshold stays below that threshold.
be exorcized from public discussions about mathematics.
Your "theorem" is false, Mückenheim. There's a simple counter example:
For each and every FISON F: F c_strict (WM: "below") IN. But the union
of all FISONs is not c_strict (WM: "below") IN but = IN.
FISONs do not grow by unioning them. All FISONs and their unions stay
below 1 % of ℕ.
Proof: Every FISON that is multiplied by 100 remains a FISON that can be multiplied by 100 without changing this property.Do you mean the set of numbers less than 100n? I thought you didn’t
Am Sun, 29 Dec 2024 23:31:46 +0100 schrieb WM:
My theorem: Every union of FISONs which stay below a certain thresholdGroundbreaking.
stays below that threshold.
Am Mon, 30 Dec 2024 09:50:26 +0100 schrieb WM:
On 30.12.2024 01:39, Richard Damon wrote:So not all. All those thresholds (and the FISONs) are finite and do not include all naturals.
On 12/29/24 12:53 PM, WM wrote:
On 29.12.2024 13:34, Richard Damon wrote:
On 12/29/24 6:01 AM, WM wrote:
All definable numbers (FISONs) stay below 1 %. Every union of "below >>>>>> 1 %" stays below 1 %.
Since 0 is Less than 1, you are sort of correct, but that fact
doesn't prove your claim.
The problem is that when you get to *ALL*
I get to all FISONs below a certain threshold, namely a threshold
between which and ω there exist ℵ₀ natnumbers.
threshold +1Which isn't "All FISONs"Do you personally know FISONs larger than those? Please let me know
them.
On 12/30/24 3:50 AM, WM wrote:
On 30.12.2024 01:39, Richard Damon wrote:
On 12/29/24 12:53 PM, WM wrote:
On 29.12.2024 13:34, Richard Damon wrote:Which isn't "All FISONs"
On 12/29/24 6:01 AM, WM wrote:
All definable numbers (FISONs) stay below 1 %. Every union of
"below 1 %" stays below 1 %.
Since 0 is Less than 1, you are sort of correct, but that fact
doesn't prove your claim.
The problem is that when you get to *ALL*
I get to all FISONs below a certain threshold, namely a threshold
between which and ω there exist ℵ₀ natnumbers.
Do you personally know FISONs larger than those? Please let me know them.
Sure, if your threshold is n,
On 12/30/24 3:44 AM, WM wrote:
On 30.12.2024 01:36, Richard Damon wrote:
On 12/29/24 5:31 PM, WM wrote:
My theorem: Every union of FISONs which stay below a certain
threshold stays below that threshold.
Find a counterexample. Don't claim it but prove it. Fail.
But what does that prove?
That proves the existence of dark numbers.
How?
It shows that there are numbers you didn't look at, but you admit to not looking at all the numbers.
The finite numbers are finite, and that is it.
FISONs do not grow by unioning them. All FISONs and their unions stay
below 1 % of ℕ.
Proof: Every FISON that is multiplied by 100 remains a FISON that can
be multiplied by 100 without changing this property.
If ℕ is an actually infinite set then it is not made by FISONs.
Who said it is made of FISONs, It could be said that it is the union of
*ALL* of the infinite set of FISONs, but the union of an infinte set of things can be bigger than any member of that infinite set.
Am Mon, 30 Dec 2024 09:40:11 +0100 schrieb WM:
FISONs do not grow by unioning them. All FISONs and their unions stayEven better, they are finite (but they do grow).
below 1 % of ℕ.
Proof: Every FISON that is multiplied by 100 remains a FISON that can beDo you mean the set of numbers less than 100n? I thought you didn’t
multiplied by 100 without changing this property.
believe in the closure of N under multiplication.
Am 29.12.2024 um 23:15 schrieb Chris M. Thomasson:
On 12/29/2024 3:09 AM, WM wrote:
My theorem: Every union of FISONs which stay below a certain
threshold stays below that threshold.
Your "theorem" is false, Mückenheim. There's a simple counter example:
For each and every FISON F: F c_strict (WM: "below") IN. But the union
of all FISONs is not c_strict (WM: "below") IN but = IN.
Mückenheim, Du bist für Mathematik zu doof und zu blöde, sorry.
.
.
.
On 30.12.2024 21:28, joes wrote:WTF? N *is* the set(!) of natural numbers, and it is infinite.
Am Mon, 30 Dec 2024 09:40:11 +0100 schrieb WM:
They do grow but will never surpass 1 % of ℕ.FISONs do not grow by unioning them. All FISONs and their unions stayEven better, they are finite (but they do grow).
below 1 % of ℕ.
The collection of natural numbers is closed under multiplication.In fact
Proof: Every FISON that is multiplied by 100 remains a FISON that canDo you mean the set of numbers less than 100n? I thought you didn’t
be multiplied by 100 without changing this property.
believe in the closure of N under multiplication.
it will ever reach let alone surpass any definable fraction of ℕ.
Am Mon, 30 Dec 2024 22:53:38 +0100 schrieb WM:
Now what is the multiplication of a FISON?
Andernfalls handelt es sich
lediglich um unbewiesene und oft genug auch falsche Behauptungen (so wie
in diesem Fall)
For each and every FISON F: F c_strict (WM: "below") IN. But the union
of all FISONs is not c_strict (WM: "below") IN but = IN.
On 30.12.2024 17:11, Richard Damon wrote:
On 12/30/24 3:50 AM, WM wrote:
On 30.12.2024 01:39, Richard Damon wrote:
On 12/29/24 12:53 PM, WM wrote:
On 29.12.2024 13:34, Richard Damon wrote:Which isn't "All FISONs"
On 12/29/24 6:01 AM, WM wrote:
All definable numbers (FISONs) stay below 1 %. Every union of
"below 1 %" stays below 1 %.
Since 0 is Less than 1, you are sort of correct, but that fact
doesn't prove your claim.
The problem is that when you get to *ALL*
I get to all FISONs below a certain threshold, namely a threshold
between which and ω there exist ℵ₀ natnumbers.
Do you personally know FISONs larger than those? Please let me know
them.
Sure, if your threshold is n,
My threshold is not a fixed number but a potentially infinite set which
never grows to more than 1 % of ℕ. Proof: Every FISON that is multiplied
by 100 remains a FISON that can be multiplied by 100 without changing
this property.
Regards, WM
On 30.12.2024 17:08, Richard Damon wrote:
On 12/30/24 3:44 AM, WM wrote:
On 30.12.2024 01:36, Richard Damon wrote:
On 12/29/24 5:31 PM, WM wrote:
My theorem: Every union of FISONs which stay below a certain
threshold stays below that threshold.
Find a counterexample. Don't claim it but prove it. Fail.
But what does that prove?
That proves the existence of dark numbers.
How?
It shows that there are numbers you didn't look at, but you admit to
not looking at all the numbers.
Neither can you look at more numbers.
The finite numbers are finite, and that is it.
FISONs do not grow by unioning them. All FISONs and their unions stay
below 1 % of ℕ.
Proof: Every FISON that is multiplied by 100 remains a FISON that can
be multiplied by 100 without changing this property.
If ℕ is an actually infinite set then it is not made by FISONs.
That is what I have been telling you for a long time.
Who said it is made of FISONs, It could be said that it is the union
of *ALL* of the infinite set of FISONs, but the union of an infinte
set of things can be bigger than any member of that infinite set.
That is a belief shared by real fools only. Every union of FISONs which
stay below a certain threshold stays below that threshold. All FISONs
stay below 1 % of ℕ. Proof: Every FISON that is multiplied by 100
remains a FISON that can be multiplied by 100 without changing this
property.
Regards, WM
On 30.12.2024 21:10, joes wrote:
Am Sun, 29 Dec 2024 23:31:46 +0100 schrieb WM:
My theorem: Every union of FISONs which stay below a certain thresholdGroundbreaking.
stays below that threshold.
And proven. My threshold is not a fixed number but a potentially
infinite set which never grows to more than 1 % of ℕ. Proof: Every FISON that is multiplied by 100 remains a FISON that can be multiplied by 100 without changing this property.
Regards, WM
On 31.12.2024 06:08, Moebius wrote:
Andernfalls handelt es sich lediglich um unbewiesene und oft genug
auch falsche Behauptungen (so wie in diesem Fall)
For each and every FISON F: F c_strict (WM: "below") IN. But the
union of all FISONs is not c_strict (WM: "below") IN but = IN.
Yes. Only a real fool could claim that by unioning the gap between less
than 1 % and ℕ could be closed.
Every FISON is less than 1 % of ℕ because by expanding it by a factor
100 the situations remains the same - forever.
Gruß, WM
On 29.12.2024 21:09, Jim Burns wrote:
Bob can disappear within a larger.enough set,
because
no finite is last of the finites,
because
swapping two values of a one.to.one map
leaves another one.to.one map.
No.
Exchange of two entities does never result in
only one of them.
Every FISON is less than 1 % of ℕ</WM>
because
by expanding it by a factor 100
the situations remains the same - forever.
Exchange of two entities does never result in
only one of them.
On 12/30/24 4:38 PM, WM wrote:
On 30.12.2024 17:08, Richard Damon wrote:
On 12/30/24 3:44 AM, WM wrote:
On 30.12.2024 01:36, Richard Damon wrote:
On 12/29/24 5:31 PM, WM wrote:
My theorem: Every union of FISONs which stay below a certain
threshold stays below that threshold.
Find a counterexample. Don't claim it but prove it. Fail.
But what does that prove?
That proves the existence of dark numbers.
How?
It shows that there are numbers you didn't look at, but you admit to
not looking at all the numbers.
Neither can you look at more numbers.
Of course I can look at "more" numbers, as I can look at the one after
where you stopped.
If ℕ is an actually infinite set then it is not made by FISONs.
That is what I have been telling you for a long time.
But there is no Natural Number that isn't in a FISON, which is NOT what
you have been saying.
The problem is your logic can't have "Every" FISON, because "every
FISON" is an infinite set of them,
All you have proved is that any finite set has a maximum value, while
the INFINITE set of Natural Numbers (or FISONs) do not, and thus your
logic can't talk about "all".
On 12/29/2024 5:27 PM, WM wrote:
Every FISON is less than 1 % of ℕ</WM>
because
by expanding it by a factor 100
the situations remains the same - forever.
Date: Tue, 31 Dec 2024 10:39:12 +0100
Message-ID: <vl0e3v$25vs8$1@dont-email.me>
For each ordinal k, there is
an ordinal k+1 which is fuller.by.one.
⎛ k = ⟦0,k⦆
⎝ k+1 = ⟦0,k⦆∪⦃k⦄
For each room k, there is
swap k⇄k+1
Before any swap,
Bob is in room 0
After swaps 0⇄1 1⇄2 ... k-1⇄k and
before swap k⇄k+1
Bob is in room k
⎛ If Bob is in room k
⎜ them k⇄k+1 is not swapped
⎜
⎜ If Bob is in any room
⎜ then not all swaps are swapped
⎜Bob comes to rest within the dark part.
⎜ For any claims P and Q
⎜ P⇒¬Q and Q⇒¬P are equally.true and equally.false
⎜
⎜ If all swaps are swapped
⎝ them Bob is not in any room.
Exchange of two entities does never result in
only one of them.
Two is finite.
How.many there are of
all swaps, all rooms, or all finite.ordinals
is more than any finite.ordinal,
is more.than.finite.
After all swaps,
⎛ Bob has never swapped into any darkᵂᴹ room,
⎝ Bob has swapped out of any visibleᵂᴹ room,
On 31.12.2024 18:25, Jim Burns wrote:
<WM>
Every FISON is less than 1 % of ℕ</WM>
because
by expanding it by a factor 100
the situations remains the same - forever.
Date: Tue, 31 Dec 2024 10:39:12 +0100
Message-ID: <vl0e3v$25vs8$1@dont-email.me>
For each ordinal k, there is
an ordinal k+1 which is fuller.by.one.
⎛ k = ⟦0,k⦆
⎝ k+1 = ⟦0,k⦆∪⦃k⦄
For each finite.ordinal k, there is
a finite.ordinal k+1 which is larger.by.one.
⎛ finite k
⎜ #⟦0,k⦆ < #⟦0,k+1⦆
⎜ ¬∃f one.to.one: ⟦0,k+1⦆ ⇉ ⟦0,k⦆
⎜ ¬∃g one.to.one: ⟦0,k+2⦆ ⇉ ⟦0,k+1⦆
⎜ #⟦0,k+1⦆ < #⟦0,k+2⦆
⎝ finite k+1
For each room k, there is
swap k⇄k+1
That does not disprove the theorem:
Every union of FISONs which
stay below a certain threshold
stays below that threshold.
Note: Every FISON stays below 1 % of ℕ.
Before any swap,
Bob is in room 0
After swaps 0⇄1 1⇄2 ... k-1⇄k and
before swap k⇄k+1
Bob is in room k
⎛ If Bob is in room k
⎜ them k⇄k+1 is not swapped
⎜
⎜ If Bob is in any room
⎜ then not all swaps are swapped
Bob is placed within
the dark parts of the matrix.
⎜ For any claims P and Q
⎜ P⇒¬Q and Q⇒¬P are equally.true and equally.false
⎜
⎜ If all swaps are swapped
⎝ them Bob is not in any room.
Bob comes to rest within the dark part.
Exchange of two entities does never result in
only one of them.
Two is finite.
How.many there are of
all swaps, all rooms, or all finite.ordinals
is more than any finite.ordinal,
is more.than.finite.
Every exchange happens between two sites.
After all swaps,
⎛ Bob has never swapped into any darkᵂᴹ room,
⎝ Bob has swapped out of any visibleᵂᴹ room,
Bob is nowhere.
But Bob is never out of the matrix
because
there is no exchange at all
between outside and inside.
On 31.12.2024 16:37, Richard Damon wrote:
[...]
I don't stop
but know that
every FISON is finite as its name says
and covers at most 1 % of ℕ.
If you don't agree
find a counterexample.
Richard Damon was thinking very hard :
On 12/30/24 4:38 PM, WM wrote:
On 30.12.2024 17:08, Richard Damon wrote:
On 12/30/24 3:44 AM, WM wrote:
On 30.12.2024 01:36, Richard Damon wrote:
On 12/29/24 5:31 PM, WM wrote:
My theorem: Every union of FISONs which stay below a certain
threshold stays below that threshold.
Find a counterexample. Don't claim it but prove it. Fail.
But what does that prove?
That proves the existence of dark numbers.
How?
It shows that there are numbers you didn't look at, but you admit to
not looking at all the numbers.
Neither can you look at more numbers.
Of course I can look at "more" numbers, as I can look at the one after
where you stopped.
The finite numbers are finite, and that is it.
FISONs do not grow by unioning them. All FISONs and their unions
stay below 1 % of ℕ.
Proof: Every FISON that is multiplied by 100 remains a FISON that
can be multiplied by 100 without changing this property.
If ℕ is an actually infinite set then it is not made by FISONs.
That is what I have been telling you for a long time.
But there is no Natural Number that isn't in a FISON, which is NOT
what you have been saying.
You don't seem to understand the difference between a set and its
members.
He has in the past explicitly stated such. Something like 'a set is
nothing more than its elements' -- I don't think he has changed his tune.
On 31.12.2024 16:37, Richard Damon wrote:
On 12/30/24 4:38 PM, WM wrote:
On 30.12.2024 17:08, Richard Damon wrote:
On 12/30/24 3:44 AM, WM wrote:
On 30.12.2024 01:36, Richard Damon wrote:
On 12/29/24 5:31 PM, WM wrote:
My theorem: Every union of FISONs which stay below a certain
threshold stays below that threshold.
Find a counterexample. Don't claim it but prove it. Fail.
But what does that prove?
That proves the existence of dark numbers.
How?
It shows that there are numbers you didn't look at, but you admit to
not looking at all the numbers.
Neither can you look at more numbers.
Of course I can look at "more" numbers, as I can look at the one after
where you stopped.
I don't stop but know that every FISON is finite as its name says and
covers at most 1 % of ℕ. If you don't agree find a counterexample.
If ℕ is an actually infinite set then it is not made by FISONs.
That is what I have been telling you for a long time.
But there is no Natural Number that isn't in a FISON, which is NOT
what you have been saying.
You cannot find a natural number that isn't in a FISON. But there must
be more because FISONs cover less than 1 % of ℕ.
The problem is your logic can't have "Every" FISON, because "every
FISON" is an infinite set of them,
Every mathematician knows that every FISON is finite as its name says
and covers at most 1 % of ℕ.
Regardes, WM
On 31.12.2024 16:39, Richard Damon wrote:omgwtf
All you have proved is that any finite set has a maximum value, whileThe set of natural numbers in FISONs is finite because it is the union
the INFINITE set of Natural Numbers (or FISONs) do not, and thus your
logic can't talk about "all".
of FISONs.
On 31.12.2024 09:56, joes wrote:In fact it is infinite.
Am Mon, 30 Dec 2024 22:53:38 +0100 schrieb WM:
Correction:
The collection of definable natural numbers (FISONs) is closed under multiplication. In fact it will never reach let alone surpass any
definable fraction of ℕ.
Are you talking about extending a FISON?Now what is the multiplication of a FISON?Every FISON is less than 1 % of ℕ because by expanding it by a factor
100 the situations remains the same - forever.
On 31.12.2024 16:37, Richard Damon wrote:Yes, you stop at some finite threshold.
On 12/30/24 4:38 PM, WM wrote:I don't stop but know that every FISON is finite as its name says and
On 30.12.2024 17:08, Richard Damon wrote:Of course I can look at "more" numbers, as I can look at the one after
On 12/30/24 3:44 AM, WM wrote:Neither can you look at more numbers.
On 30.12.2024 01:36, Richard Damon wrote:How?
On 12/29/24 5:31 PM, WM wrote:That proves the existence of dark numbers.
My theorem: Every union of FISONs which stay below a certainBut what does that prove?
threshold stays below that threshold.
Find a counterexample. Don't claim it but prove it. Fail.
It shows that there are numbers you didn't look at, but you admit to
not looking at all the numbers.
where you stopped.
covers at most 1 % of ℕ.
No??? There are infinitely many of them, and together they form N.You cannot find a natural number that isn't in a FISON. But there mustBut there is no Natural Number that isn't in a FISON, which is NOT whatIf ℕ is an actually infinite set then it is not made by FISONs.
you have been saying.
be more because FISONs cover less than 1 % of ℕ.
--The problem is your logic can't have "Every" FISON, because "every
FISON" is an infinite set of them,
On 12/31/2024 1:20 PM, WM wrote:
Every union of FISONs which
stay below a certain threshold
stays below that threshold.
Note: Every FISON stays below 1 % of ℕ.
For each finite ⟦0,j⦆, there are
more.than.#⟦0,j⦆.many finite ⟦0,k⦆
The set {⟦0,i⦆:#⟦0,i⦆<#⟦0,i+1⦆} of
all the finite ⟦0,k⦆
is not a finite set.
Bob is placed within
the dark parts of the matrix.
Then not all swaps are swapped
Bob comes to rest within the dark part.
Then not all swaps are swapped
Every exchange happens between two sites.
Yes.
Always between two visibleᵂᴹ sites.
And no single swap disappears Bob.
Infinitely.many is different from finitely.many.
For each finitely.many, there are more.
After all swaps,
⎛ Bob has never swapped into any darkᵂᴹ room,
⎝ Bob has swapped out of any visibleᵂᴹ room,
Bob is nowhere.
But Bob is never out of the matrix
because
there is no exchange at all between outside and inside.
Yes, Bob never exits the matrix.
However, after all swaps, Bob isn't in the matrix.
Therefore, infinite is different.
On 12/31/2024 1:05 PM, WM wrote:
On 31.12.2024 16:37, Richard Damon wrote:
[...]
I don't stop
but know that
every FISON is finite as its name says
and covers at most 1 % of ℕ.
If you don't agree
find a counterexample.
No FISON is smaller than 1% of the maximum FISON.
Each FISON is a counter.example --
NOT because any of them is larger, but
because no maximum FISON exists, darkᵂᴹ or visibleᵂᴹ.
On 12/31/24 1:05 PM, WM wrote:
Every mathematician knows that every FISON is finite as its name says
and covers at most 1 % of ℕ.
But, when you actually take into accout EVERY FISON, the full INFINITE
set, you get to them all
Yes, SOME of the properties of the set come from properties of its
members,
Am Tue, 31 Dec 2024 19:08:38 +0100 schrieb WM:
There are infinitely many FISONs, one for every natural. N is the
infinite union.
Am Tue, 31 Dec 2024 19:05:24 +0100 schrieb WM:
I don't stop but know that every FISON is finite as its name says andYes, you stop at some finite threshold.
covers at most 1 % of ℕ.
You cannot find a natural number that isn't in a FISON. But there mustNo??? There are infinitely many of them, and together they form N.
be more because FISONs cover less than 1 % of ℕ.
WM wrote :
The collection of definable natural numbers (FISONs) is closed under
multiplication
How can that be?
On 01.01.2025 03:03, Richard Damon wrote:
On 12/31/24 1:05 PM, WM wrote:
Every mathematician knows that every FISON is finite as its name says
and covers at most 1 % of ℕ.
But, when you actually take into accout EVERY FISON, the full INFINITE
set, you get to them all
Every union of FISONs including them all which stay below a certain
threshold stays below that threshold. Every FISON stays below 1 % of ℕ. What can't you understand?
Regards. WM
On 01.01.2025 03:03, Richard Damon wrote:
Yes, SOME of the properties of the set come from properties of its
members,
What property does not?
Regards, WM
On 01.01.2025 15:47, FromTheRafters wrote:
WM wrote :
The collection of definable natural numbers (FISONs) is closed under
multiplication
How can that be?
Potential infinity.
Regards, WM
On 1/2/25 6:36 AM, WM wrote:
On 01.01.2025 03:03, Richard Damon wrote:
Yes, SOME of the properties of the set come from properties of its
members,
What property does not?
Like being infinite.
On 1/2/25 6:40 AM, WM wrote:
Every union of FISONs including them all which stay below a certain
threshold stays below that threshold. Every FISON stays below 1 % of ℕ.
What can't you understand?
No, becuase "all which stay below a certain threshol" is not EVERY FISON.
When you actually take the union of ALL that infinite set of FISONs, you
will get the full infinite set of N.
On 1/2/25 6:52 AM, WM wrote:
On 01.01.2025 15:47, FromTheRafters wrote:
WM wrote :
The collection of definable natural numbers (FISONs) is closed under
multiplication
How can that be?
Potential infinity.
Whicn means it never ends, and doesn't have a highest member.
WM was thinking very hard :
On 01.01.2025 03:03, Richard Damon wrote:
Yes, SOME of the properties of the set come from properties of its
members,
What property does not?
Size, for one thing.
WM has brought this to us :
On 01.01.2025 14:51, joes wrote:
Am Tue, 31 Dec 2024 19:05:24 +0100 schrieb WM:
I don't stop but know that every FISON is finite as its name says andYes, you stop at some finite threshold.
covers at most 1 % of ℕ.
Go farther. Find a FISON that is not below 1 %.
One percent of infinitely many is nonsense.
On 01.01.2025 01:45, Jim Burns wrote:
On 12/31/2024 1:05 PM, WM wrote:
I don't stop
but know that
every FISON is finite as its name says
and covers at most 1 % of ℕ.
If you don't agree
find a counterexample.
No FISON is smaller than 1% of the maximum FISON.
{1} is smaller than {101} which is
smaller than all larger FISONs.
Each FISON is a counter.example --
NOT because any of them is larger, but
because no maximum FISON exists, darkᵂᴹ or visibleᵂᴹ.
Maximum FISONs cannot be grasped
because they are potentially infinite.
On 31.12.2024 21:26, Jim Burns wrote:
On 12/31/2024 1:20 PM, WM wrote:
Every union of FISONs which
stay below a certain threshold
stays below that threshold.
Note: Every FISON stays below 1 % of ℕ.
For each finite ⟦0,j⦆, there are
more.than.#⟦0,j⦆.many finite ⟦0,k⦆
So it is for all definable natural numbers.
But all natural numbers are finite,
and nevertheless there are
natural numbers which have only few successors
natural numbers which have only few successors
because when they are removed nothing remains
ℕ \ {1, 2, 3, ...} = { } .
because when they are removed nothing remains
ℕ \ {1, 2, 3, ...} = { } .
The set {⟦0,i⦆:#⟦0,i⦆<#⟦0,i+1⦆} of
all the finite ⟦0,k⦆
is not a finite set.
It is not a set
but a potentially infinite collection.
On 02.01.2025 14:22, Richard Damon wrote:
On 1/2/25 6:36 AM, WM wrote:
On 01.01.2025 03:03, Richard Damon wrote:
Yes, SOME of the properties of the set come from properties of its
members,
What property does not?
Like being infinite.
Mistake. An infinite set needs infinitely many elements.
Regards, WM
On 1/2/2025 5:43 AM, FromTheRafters wrote:
WM used his keyboard to write :
On 01.01.2025 15:47, FromTheRafters wrote:
WM wrote :
The [set] of [...] natural numbers [...] is closed under multiplication. (WM)
How can that be? (From the After)
P<bla> (WM)
Easy, he stood on a chair. (From the After)
lol! ;^)
On 1/2/2025 9:33 AM, WM wrote:
On 02.01.2025 14:40, FromTheRafters wrote:
"One percent" of infinitely many is nonsense.
It is only for people without extraordinaly mathematical skills. But
it is not false. In fact all FISONs are limited to less than
1/10^10^10^10^10^10 of ℵ₀.
Huh?
On 1/2/2025 11:47 AM, Richard Damon wrote:
On 1/2/25 12:23 PM, WM wrote:
On 02.01.2025 14:25, Richard Damon wrote:
On 1/2/25 6:40 AM, WM wrote:
Every union of FISONs including them all which stay below a certain
threshold stays below that threshold.
What can't you understand?
No, because "all which stay below a certain threshold" is not EVERY
FISON.
When you actually take the union of ALL [...] FISONs,
you will get the full infinite set of IN.
Stupid belief.
Sorry, your brain is dead.
I think so. Well, shit happens! ;^o
On 1/2/2025 9:18 AM, WM wrote:
On 02.01.2025 14:22, Richard Damon wrote:
On 1/2/25 6:36 AM, WM wrote:
On 01.01.2025 03:03, Richard Damon wrote:
Yes, SOME of the properties of the set come from properties of its
members, (RD)
What property does not? (WM)
[the property of] being infinite. (RD)
Am 03.01.2025 um 03:14 schrieb Chris M. Thomasson:
On 1/2/2025 5:57 PM, Moebius wrote:
Am 03.01.2025 um 02:40 schrieb Chris M. Thomasson:
On 1/2/2025 9:33 AM, WM wrote:
On 02.01.2025 14:40, FromTheRafters wrote:
"One percent" of infinitely many is nonsense.
Indeed. (Sorry, but I have to agree.)
It is only for people without extraordinaly mathematical skills.
But it is not false. In fact all FISONs are limited to less than
1/10^10^10^10^10^10 of ℵ₀.
Huh?
No sane person knows what he's talking about.
Hint: He's insane. Suffering from delusion.
Trying to "understand" his "ideas" (delusions) is hopeless.
.
.
.
See: https://en.wikipedia.org/wiki/Delusion
I think so... I hope he is faking it. Sigh...
You think?!
.
.
.
On 1/2/2025 5:57 PM, Moebius wrote:
Am 03.01.2025 um 02:40 schrieb Chris M. Thomasson:
On 1/2/2025 9:33 AM, WM wrote:
On 02.01.2025 14:40, FromTheRafters wrote:
"One percent" of infinitely many is nonsense.
Indeed. (Sorry, but I have to agree.)
It is only for people without extraordinaly mathematical skills. But
it is not false. In fact all FISONs are limited to less than
1/10^10^10^10^10^10 of ℵ₀.
Huh?
No sane person knows what he's talking about.
Hint: He's insane. Suffering from delusion.
Trying to "understand" his "ideas" (delusions) is hopeless.
.
.
.
See: https://en.wikipedia.org/wiki/Delusion
I think so... I hope he is faking it. Sigh...
On 1/2/2025 6:35 AM, WM wrote:
On 01.01.2025 01:45, Jim Burns wrote:
On 12/31/2024 1:05 PM, WM wrote:
I don't stop
but know that
every FISON is finite as its name says
and covers at most 1 % of ℕ.
If you don't agree
find a counterexample.
No FISON is smaller than 1% of the maximum FISON.
{1} is smaller than {101} which is
smaller than all larger FISONs.
...larger FISONs _which exist_
something which frequently goes without saying.
"Larger FISONs which exist" excludes
a maximum FISON.
How we know it's excluded is
NOT
because we can see all FISONs
but
because we can see a description of all FISONs,
A description of FISONs can be grasped.
Nope. For each and every FISON F: F c IN.
But UNION(Set_of_FISONs) = IN.
On 1/2/25 12:28 PM, WM wrote:
Hint first try to understand 50 %, then 10 %, then 1 %, then 1 ppm,
then 1 ppb, then 1/10^10^10^10^10. And then go on. Then you understand
the properties of FISONs.
And ALL finite fractional ratios of Aleph_0, are Aleph_0, and thus
infinite so
On 1/2/2025 6:31 AM, WM wrote:
On 31.12.2024 21:26, Jim Burns wrote:
On 12/31/2024 1:20 PM, WM wrote:
Every union of FISONs which
stay below a certain threshold
stays below that threshold.
Note: Every FISON stays below 1 % of ℕ.
For each finite ⟦0,j⦆, there are
more.than.#⟦0,j⦆.many finite ⟦0,k⦆
So it is for all definable natural numbers.
So it is for each finite.ordinal.
natural numbers which have only few successors
because when they are removed nothing remains
ℕ \ {1, 2, 3, ...} = { } .
... the non.finite.ordinal natural.numbers.
We define natural.numbers to be
only finite.ordinals, of which
there are more.than.any.finite.many.
If you want to know how Bob disappears,
On 1/2/25 12:30 PM, WM wrote:
On 02.01.2025 14:37, FromTheRafters wrote:
WM was thinking very hard :
On 01.01.2025 03:03, Richard Damon wrote:
Yes, SOME of the properties of the set come from properties of its
members,
What property does not?
Size, for one thing.
Size is the number of elements.
WHich is a property of the set, but not the individual members.
On 1/2/25 12:23 PM, WM wrote:
On 02.01.2025 14:25, Richard Damon wrote:
On 1/2/25 6:40 AM, WM wrote:
Every union of FISONs including them all which stay below a certain
threshold stays below that threshold. Every FISON stays below 1 % of ℕ. >>>> What can't you understand?
No, becuase "all which stay below a certain threshol" is not EVERY
FISON.
Which one is more than 1 % of ℵ₀?
There isn't one, but anytime you CHOSE what you limit is, I can go past it.
Which isn't a valid question,When you actually take the union of ALL that infinite set of FISONs,
you will get the full infinite set of N.
Stupid belief. Which FISON is more than 1 % of ℵ₀?
On 1/2/25 12:33 PM, WM wrote:
On 02.01.2025 14:40, FromTheRafters wrote:
WM has brought this to us :
On 01.01.2025 14:51, joes wrote:
Am Tue, 31 Dec 2024 19:05:24 +0100 schrieb WM:
I don't stop but know that every FISON is finite as its name says and >>>>>> covers at most 1 % of ℕ.Yes, you stop at some finite threshold.
Go farther. Find a FISON that is not below 1 %.
One percent of infinitely many is nonsense.
It is only for people without extraordinaly mathematical skills. But
it is not false. In fact all FISONs are limited to less than
1/10^10^10^10^10^10 of ℵ₀.
Which is still infinite,
On 02.01.2025 14:22, Richard Damon wrote:…which don’t need to be infinite themselves. QED.
On 1/2/25 6:36 AM, WM wrote:Mistake. An infinite set needs infinitely many elements.
On 01.01.2025 03:03, Richard Damon wrote:Like being infinite.
Yes, SOME of the properties of the set come from properties of itsWhat property does not?
members,
On 02.01.2025 14:40, FromTheRafters wrote:HAHAHAHAHA
WM has brought this to us :It is only for people without extraordinaly mathematical skills.
On 01.01.2025 14:51, joes wrote:One percent of infinitely many is nonsense.
Am Tue, 31 Dec 2024 19:05:24 +0100 schrieb WM:Go farther. Find a FISON that is not below 1 %.
I don't stop but know that every FISON is finite as its name saysYes, you stop at some finite threshold.
and covers at most 1 % of ℕ.
In fact all FISONs are limited to less thanThey are, as the name suggests, less than ANY fraction of infinity.
1/10^10^10^10^10^10 of ℵ₀.
On 02.01.2025 20:47, Richard Damon wrote:No, the (infinite) set of all FISONs does cover N by dint of being
On 1/2/25 12:28 PM, WM wrote:
the set of FISONs covers less than any fractional ratio of ℕ. This isHint first try to understand 50 %, then 10 %, then 1 %, then 1 ppm,And ALL finite fractional ratios of Aleph_0, are Aleph_0, and thus
then 1 ppb, then 1/10^10^10^10^10. And then go on. Then you understand
the properties of FISONs.
infinite so
true for their union too. Otherwise there must be a first FISON going
beyond this boundary.
On 02.01.2025 20:47, Richard Damon wrote:Fuck no. The size of {1, 2, 4} is not a member.
On 1/2/25 12:30 PM, WM wrote:A property of its members.
On 02.01.2025 14:37, FromTheRafters wrote:WHich is a property of the set, but not the individual members.
WM was thinking very hard :Size is the number of elements.
On 01.01.2025 03:03, Richard Damon wrote:Size, for one thing.
Yes, SOME of the properties of the set come from properties of its >>>>>> members,What property does not?
On 02.01.2025 19:19, Jim Burns wrote:It is very obvious there are infinitely many FISONs.
On 1/2/2025 6:31 AM, WM wrote:That is impossible, because all finite ordinals can be subtracted from ℕ without infinitely many remaining. If you don't understand that, it is useless to go on.
On 31.12.2024 21:26, Jim Burns wrote:So it is for each finite.ordinal.
On 12/31/2024 1:20 PM, WM wrote:
So it is for all definable natural numbers.Every union of FISONs which stay below a certain threshold staysFor each finite ⟦0,j⦆, there are more.than.#⟦0,j⦆.many finite ⟦0,k⦆
below that threshold.
Note: Every FISON stays below 1 % of ℕ.
Infinitely many can be removed without remainder. But only finitely manynatural numbers which have only few successors because when they are... the non.finite.ordinal natural.numbers.
removed nothing remains ℕ \ {1, 2, 3, ...} = { } .
We define natural.numbers to be only finite.ordinals, of which there
are more.than.any.finite.many.
can be defined by FISONs.
On 03.01.2025 02:52, Moebius wrote:
Nope. For each and every FISON F: F c IN.
But UNION(Set_of_FISONs) = IN.
Every union of FISONs including them all which stay below a certain
threshold stays below that threshold.
Regards, WM
On 02.01.2025 20:47, Richard Damon wrote:
On 1/2/25 12:30 PM, WM wrote:
On 02.01.2025 14:37, FromTheRafters wrote:
WM was thinking very hard :
On 01.01.2025 03:03, Richard Damon wrote:
Yes, SOME of the properties of the set come from properties of its >>>>>> members,
What property does not?
Size, for one thing.
Size is the number of elements.
WHich is a property of the set, but not the individual members.
A property of its members.
Regards, WM
On 02.01.2025 20:47, Richard Damon wrote:
On 1/2/25 12:33 PM, WM wrote:
On 02.01.2025 14:40, FromTheRafters wrote:
WM has brought this to us :
On 01.01.2025 14:51, joes wrote:
Am Tue, 31 Dec 2024 19:05:24 +0100 schrieb WM:
I don't stop but know that every FISON is finite as its name says >>>>>>> andYes, you stop at some finite threshold.
covers at most 1 % of ℕ.
Go farther. Find a FISON that is not below 1 %.
One percent of infinitely many is nonsense.
It is only for people without extraordinaly mathematical skills. But
it is not false. In fact all FISONs are limited to less than
1/10^10^10^10^10^10 of ℵ₀.
Which is still infinite,
when reached. But FISONs don't reach it.
Regards, WM
On 02.01.2025 20:47, Richard Damon wrote:
On 1/2/25 12:28 PM, WM wrote:
Hint first try to understand 50 %, then 10 %, then 1 %, then 1 ppm,
then 1 ppb, then 1/10^10^10^10^10. And then go on. Then you
understand the properties of FISONs.
And ALL finite fractional ratios of Aleph_0, are Aleph_0, and thus
infinite so
the set of FISONs covers less than any fractional ratio of ℕ. This is
true for their union too. Otherwise there must be a first FISON going
beyond this boundary.
Regards, WM
On 1/3/2025 3:39 AM, WM wrote:
all finite ordinals can be subtracted from ℕ without infinitely many
remaining.
If
Am Thu, 02 Jan 2025 18:33:24 +0100 schrieb WM:
In fact all FISONs are limited to less thanThey are, as the name suggests, less than ANY fraction of infinity.
1/10^10^10^10^10^10 of ℵ₀.
On 02.01.2025 19:19, Jim Burns wrote:
On 1/2/2025 6:31 AM, WM wrote:
On 31.12.2024 21:26, Jim Burns wrote:
On 12/31/2024 1:20 PM, WM wrote:
Every union of FISONs which
stay below a certain threshold
stays below that threshold.
Note: Every FISON stays below 1 % of ℕ.
For each finite ⟦0,j⦆, there are
more.than.#⟦0,j⦆.many finite ⟦0,k⦆
So it is for all definable natural numbers.
So it is for each finite.ordinal.
That is impossible,
That is impossible, because
all finite ordinals can be subtracted from ℕ
without infinitely many remaining.
If you don't understand that,
it is useless to go on.
No, the (infinite) set of all FISONs does cover N by dint of being
infinite.
Am Fri, 03 Jan 2025 09:39:01 +0100 schrieb WM:
Infinitely many can be removed without remainder. But only finitely manyIt is very obvious there are infinitely many FISONs.
can be defined by FISONs.
But the members taken collectively *IS* the set.
On 1/3/25 3:52 AM, WM wrote:
On 03.01.2025 02:52, Moebius wrote:
Nope. For each and every FISON F: F c IN.
But UNION(Set_of_FISONs) = IN.
Every union of FISONs including them all which stay below a certain
threshold stays below that threshold.
Which is different from *ALL* of them,
On 03.01.2025 13:26, joes wrote:No, the elements are numbers, none of which has size 3.
Am Fri, 03 Jan 2025 09:55:34 +0100 schrieb WM:
There are three elements. That is the property of the elements.Fuck no. The size of {1, 2, 4} is not a member.WHich is a property of the set, but not the individual members.A property of its members.
On 1/3/25 4:00 AM, WM wrote:
When you combine *ALL* FISONs, you get a property that none of them have individually,
Am Fri, 03 Jan 2025 18:04:16 +0100 schrieb WM:
On 03.01.2025 13:26, joes wrote:No, the elements are numbers, none of which has size 3.
Am Fri, 03 Jan 2025 09:55:34 +0100 schrieb WM:There are three elements. That is the property of the elements.
Fuck no. The size of {1, 2, 4} is not a member.WHich is a property of the set, but not the individual members.A property of its members.
On 03.01.2025 17:51, Jim Burns wrote:
On 1/3/2025 3:39 AM, WM wrote:
all finite ordinals can be subtracted from ℕ
without infinitely many remaining.
If
No if.
Is it true or false?
On 1/3/2025 12:00 PM, WM wrote:
On 03.01.2025 17:51, Jim Burns wrote:
On 1/3/2025 3:39 AM, WM wrote:
all finite ordinals can be subtracted from ℕ
without infinitely many remaining.
If
No if.
Is it true or false?
All finite.ordinals removed from
the set of each and only finite.ordinals
leaves the empty set.
There is no largest finite.ordinal.
On 03.01.2025 19:46, Jim Burns wrote:No, N is exactly the set of those numbers.
On 1/3/2025 12:00 PM, WM wrote:But removing every ordinal that you can define (and all its
On 03.01.2025 17:51, Jim Burns wrote:All finite.ordinals removed from the set of each and only
On 1/3/2025 3:39 AM, WM wrote:
No if.all finite ordinals can be subtracted from ℕ without infinitely many >>>>> remaining.If
Is it true or false?
finite.ordinals leaves the empty set.
predecessors) from ℕ leaves almost all ordinals in ℕ.
You’re almost there.There is no largest finite.ordinal.There is no definable largest finite ordinal.
On 03.01.2025 19:46, Jim Burns wrote:
There is no largest finite.ordinal.
There is no definable largest finite ordinal.
On 03.01.2025 15:49, Richard Damon wrote:
But the members taken collectively *IS* the set.
Yes. The members taken individually are lacking great parts of the set.
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
ℕ \ {1, 2, 3, ...} = { }
Regards, WM
On 1/3/2025 2:48 PM, WM wrote:
On 03.01.2025 19:46, Jim Burns wrote:
All finite.ordinals removed from
the set of each and only finite.ordinals
leaves the empty set.
Am Fri, 03 Jan 2025 20:48:57 +0100 schrieb WM:
But removing every ordinal that you can define (and all itsNo, N is exactly the set of those numbers.
predecessors) from ℕ leaves almost all ordinals in ℕ.
On 1/3/25 12:15 PM, WM wrote:
Every union of FISONs which stay below a certain threshold stays below
that threshold.
Every union of a finite number of FISONs is just an admssion that you
can't do the actual union of *ALL* FISONs.
On 1/3/2025 9:09 AM, WM wrote:
On 03.01.2025 13:35, joes wrote:
Am Fri, 03 Jan 2025 09:39:01 +0100 schrieb WM:Obvious but only potentially infinite.
Infinitely many can be removed without remainder. But only finitelyIt is very obvious there are infinitely many FISONs.
many
can be defined by FISONs.
There are infinitely many FISONs. What in the heck do you mean by using
the word, "potentially"? It's as if you don't think infinity exists?
No. One can „define” infinitely many numbers.On 1/3/2025 2:48 PM, WM wrote:But removing every ordinal that you can define (and all its
On 03.01.2025 19:46, Jim Burns wrote:
All finite.ordinals removed from the set of each and only
finite.ordinals leaves the empty set.
predecessors) from ℕ leaves almost all ordinals in ℕ.∀n ∈ ℕ_def: |ℕ \
{1, 2, 3, ..., n}| = ℵo
On 04.01.2025 05:06, Richard Damon wrote:But not for the union.
On 1/3/25 12:15 PM, WM wrote:
For all FISONs:|ℕ \ {1, 2, 3, ..., n}| = ℵo.Every union of FISONs which stay below a certain threshold stays belowEvery union of a finite number of FISONs is just an admssion that you
that threshold.
can't do the actual union of *ALL* FISONs.
On 1/3/2025 2:48 PM, WM wrote:
On 03.01.2025 19:46, Jim Burns wrote:
All finite.ordinals removed from
the set of each and only finite.ordinals
leaves the empty set.
But removing every ordinal that you can define (and all its
predecessors) from ℕ leaves almost all ordinals in ℕ.∀n ∈ ℕ_def: |ℕ \
{1, 2, 3, ..., n}| = ℵo
Regards, WM
On 03.01.2025 21:29, joes wrote:Where is the contradiction? N is not a finite set with a maximum.
Am Fri, 03 Jan 2025 20:48:57 +0100 schrieb WM:
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo contradicts your opinion.But removing every ordinal that you can define (and all itsNo, N is exactly the set of those numbers.
predecessors) from ℕ leaves almost all ordinals in ℕ.
On 1/3/2025 3:56 PM, Jim Burns wrote:
All finite.ordinals removed from
the set of each and only finite.ordinals
leaves the empty set.
But removing
every ordinal that you can define
(and all its predecessors) from ℕ leaves
almost all ordinals in ℕ.
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
On 1/3/2025 6:50 AM, Richard Damon wrote:
On 1/3/25 3:52 AM, WM wrote:
On 03.01.2025 02:52, Moebius wrote:
Nope. For each and every FISON F: F c_proper IN.
But UNION(Set_of_FISONs) = IN.
Every union of FISONs including them all which stay below a certain threshold stays below that threshold.
Regards, WM
Which is different from *ALL* of them, which your logic can't do, and
thus shows you are just a stupid idiot that just doesn't understand
the infinite.
Yet he teaches? wow.
On 1/2/2025 2:42 PM, FromTheRafters wrote:
WM was thinking very hard :
On 02.01.2025 14:40, FromTheRafters wrote:
One percent of infinitely many is nonsense.
[...] But it is not false. (WM)
On 1/4/2025 12:48 AM, WM wrote:
An interval of natural numbers without any prime number is called a
prime gap. The sequence of prime gaps assumes arbitrarily large
intervals but it cannot become actually infinite.
None of the numbers n! + 2, n! + 3, n! + 4, ..., n! + n can be prime because n! =
123... n contains 2, 3, ..., n as factors already. Therefore the >> set of gaps has no upper bound.
It is potentially infinite. It is not actually infinite however, because <bla bla bla>
On 1/4/25 3:42 AM, WM wrote:
And what keep you from "defining" the rest of the Natural Numbers.
On 1/3/2025 2:48 PM, WM wrote:
On 03.01.2025 19:46, Jim Burns wrote:
All finite.ordinals removed from
the set of each and only finite.ordinals
leaves the empty set.
But removing every ordinal that you can define (and all its
predecessors) from ℕ leaves almost all ordinals in ℕ.∀n ∈ ℕ_def: |ℕ \
{1, 2, 3, ..., n}| = ℵo
Am Sat, 04 Jan 2025 09:42:11 +0100 schrieb WM:
No. One can „define” infinitely many numbers.On 1/3/2025 2:48 PM, WM wrote:But removing every ordinal that you can define (and all its
On 03.01.2025 19:46, Jim Burns wrote:
All finite.ordinals removed from the set of each and only
finite.ordinals leaves the empty set.
predecessors) from ℕ leaves almost all ordinals in ℕ.∀n ∈ ℕ_def: |ℕ \
{1, 2, 3, ..., n}| = ℵo
On 1/4/25 3:52 AM, WM wrote:
On 04.01.2025 05:06, Richard Damon wrote:Just like every natural number is finite,
On 1/3/25 12:15 PM, WM wrote:
Every union of FISONs which stay below a certain threshold stays
below that threshold.
Every union of a finite number of FISONs is just an admssion that you
can't do the actual union of *ALL* FISONs.
For all FISONs:|ℕ \ {1, 2, 3, ..., n}| = ℵo.
That means there is no exception.
but the set of *ALL* of them
is infinite, each FISON is finite,
but the union of *ALL* of them
creates the full infinite set of Natural Numbers.
Every union of FISONs including them all which stay below a certain
threshold stays below that threshold.
I just mentioned a counterexample to your claim,
| For each and every FISON F: F c_proper IN.
But UNION(Set_of_FISONs) = IN.
So the union of (the set of) _all_ FISONs does NOT "stay below" IN,
though each and every FISON does.
On 1/4/2025 3:42 AM, WM wrote:
On 1/3/2025 3:56 PM, Jim Burns wrote:
All finite.ordinals removed from
the set of each and only finite.ordinals
leaves the empty set.
But removing
every ordinal that you can define
(and all its predecessors) from ℕ leaves
almost all ordinals in ℕ.
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
ℕ is the set of each and only finite.ordinals.
|ℕ| := ℵ₀ = |ℕ\{0}| = |ℕ\{0,1}| = ... =
|ℕ\{0,1,...,n}| = ...
The sequence of end.segments of ℕ
grows emptier.one.by.one but
it doesn't grow smaller.one.by.one.
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
ℕ is the set of each and only finite.ordinals.
Each finite.ordinal is not weird.
Even an absurdly.large one like Avogadroᴬᵛᵒᵍᵃᵈʳᵒ
is not weird.
Am Sat, 04 Jan 2025 09:17:16 +0100 schrieb WM:
On 03.01.2025 21:29, joes wrote:Where is the contradiction?
Am Fri, 03 Jan 2025 20:48:57 +0100 schrieb WM:∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo contradicts your opinion.
But removing every ordinal that you can define (and all itsNo, N is exactly the set of those numbers.
predecessors) from ℕ leaves almost all ordinals in ℕ.
For me, there are infinitely many natural numbers, period... Do you
totally disagree?
Am Sat, 04 Jan 2025 09:52:08 +0100 schrieb WM:
On 04.01.2025 05:06, Richard Damon wrote:But not for the union.
On 1/3/25 12:15 PM, WM wrote:For all FISONs:|ℕ \ {1, 2, 3, ..., n}| = ℵo.
Every union of FISONs which stay below a certain threshold stays below >>>> that threshold.Every union of a finite number of FISONs is just an admssion that you
can't do the actual union of *ALL* FISONs.
WM wrote :
On 04.01.2025 19:26, Moebius wrote:
That is not an example but a silly claim contradicted by my theorem.
My theorem is proved by the fact that nothing is in the union which
could make it larger than all elements in the union.
Every union of FISONs which stay below a certain threshold stays below
that threshold.
So the union of (the set of) _all_ FISONs does NOT "stay below" IN,
though each and every FISON does.
Your matheologial belief is outside of mathematics and does not
deserve further discussion.
Then STFU and go away.
Am 04.01.2025 um 21:38 schrieb Chris M. Thomasson:
On 1/4/2025 12:48 AM, WM wrote:
An interval of natural numbers without any prime number is called a
prime gap. The sequence of prime gaps assumes arbitrarily large
intervals but it cannot become actually infinite.
Yeah, it cannot "become" actually infinite because it already _is_
infinite,
None of the numbers n! + 2, n! + 3, n! + 4, ..., n! + n can be prime
because n! = 123... n contains 2, 3, ..., n as factors already. >>> Therefore the set of gaps has no upper bound.
DAS stimmt sogar.
Eine Menge ist ENTWEDER endlich, ODER unendlich (=nicht endlich).
Eine andere Möglichkeit gibt es nicht.
On 03.01.2025 22:38, Chris M. Thomasson wrote:
On 1/3/2025 9:09 AM, WM wrote:
On 03.01.2025 13:35, joes wrote:
Am Fri, 03 Jan 2025 09:39:01 +0100 schrieb WM:
Infinitely many can be removed without remainder. But only finitely >>>>> manyIt is very obvious there are infinitely many FISONs.
can be defined by FISONs.
Obvious but only potentially infinite.
There are infinitely many FISONs. What in the heck do you mean by using
the word, "potentially"? It's as if you don't think infinity exists?
"We introduce numbers for counting. This does not at all imply the
infinity of numbers. For, in what way should we ever arrive at infinitely-many countable things? [...] In philosophical terminology we
say that the infinite of the number sequence is only potential, i.e., existing only as a possibility." [P. Lorenzen: "Das Aktual-Unendliche in
der Mathematik", Philosophia naturalis 4 (1957) p. 4f]
"Until then, no one envisioned the possibility that infinities come in different sizes, and moreover, mathematicians had no use for 'actual infinity'. The arguments using infinity, including the Differential
Calculus of Newton and Leibniz, do not require the use of infinite sets. [...] Cantor observed that many infinite sets of numbers are countable:
the set of all integers, the set of all rational numbers, and also the
set of all algebraic numbers. Then he gave his ingenious diagonal
argument that proves, by contradiction, that the set of all real numbers
is not countable. A consequence of this is that there exists a multitude
of transcendental numbers, even though the proof, by contradiction, does
not produce a single specific example." [T. Jech: "Set theory", Stanford Encyclopedia of Philosophy (2002)]
"Numerals constitute a potential infinity. Given any numeral, we can construct a new numeral by prefixing it with S. Now imagine this
potential infinity to be completed. Imagine the inexhaustible process of constructing numerals somehow to have been finished, and call the result
the set of all numbers, denoted by . Thus is thought to be an actual infinity or a completed infinity. This is curious terminology, since the etymology of 'infinite' is 'not finished'." [E. Nelson: "Hilbert's
mistake" (2007) p. 3]
According to (Gödel's) Platonism, objects of mathematics have the same status of reality as physical objects. "Views to the effect that
Platonism is correct but only for certain relatively 'concrete'
mathematical 'objects'. Other mathematical 'objects' are man made, and
are not part of an external reality. Under such a view, what is to be
made of the part of mathematics that lies outside the scope of
Platonism? An obvious response is to reject it as utterly meaningless." [H.M. Friedman: "Philosophical problems in logic" (2002) p. 9]
"A potential infinity is a quantity which is finite but indefinitely
large. For instance, when we enumerate the natural numbers as 0, 1, 2,
..., n, n+1, ..., the enumeration is finite at any point in time, but it grows indefinitely and without bound. [...] An actual infinity is a completed infinite totality. Examples: , , C[0, 1], L2[0, 1], etc. Other examples: gods, devils, etc." [S.G. Simpson: "Potential versus
actual infinity: Insights from reverse mathematics" (2015)]
"Potential infinity refers to a procedure that gets closer and closer
to, but never quite reaches, an infinite end. For instance, the sequence
of numbers 1, 2, 3, 4, ... gets higher and higher, but it has no end; it never gets to infinity. Infinity is just an indication of a direction – it's 'somewhere off in the distance'. Chasing this kind of infinity is
like chasing a rainbow or trying to sail to the edge of the world – you may think you see it in the distance, but when you get to where you
thought it was, you see it is still further away. Geometrically, imagine
an infinitely long straight line; then 'infinity' is off at the 'end' of
the line. Analogous procedures are given by limits in calculus, whether
they use infinity or not. For example, limx0(sinx)/x = 1. This means
that when we choose values of x that are closer and closer to zero, but never quite equal to zero, then (sinx)/x gets closer and closer to one."
[E. Schechter: "Potential versus completed infinity: Its history and controversy" (5 Dec 2009)]
The sequence of increasing circumferences (or diameters, or areas) of circles is potentially infinite because the circumference of a circle
can become arbitrarily long, but it cannot be actually infinite because
then it would not belong to a circle. An infinite "circumference" would
have curvature zero, i.e., no curvature, and it could not be
distinguished what is the inner side and what is the outer side of the circle.
The length of periods of decimal representations of rational numbers is potentially infinite. The length is always finite although it has no
upper bound. The decimal representation is equal to a geometric series,
like 0.abcabcabc... = abc(10-3 + 10-6 + 10-9 + ...) which converges to the limit . A never repeating decimal sequence has an irrational limit.
An interval of natural numbers without any prime number is called a
prime gap. The sequence of prime gaps assumes arbitrarily large
intervals but it cannot become actually infinite. None of the numbers n!
+ 2, n! + 3, n! + 4, ..., n! + n can be prime because n! = 123... n
contains 2, 3, ..., n as factors already. Therefore the set of gaps has
no upper bound. It is potentially infinite. It is not actually infinite however, because there does not exist a gap with no closing prime number because there is no last prime number.
Finally, the most familiar example is this: The (magnitudes of) natural numbers are potentially infinite because, although there is no upper
bound, there is no infinite (magnitude of a) natural number.
Regards, WM
On 04.01.2025 14:17, Richard Damon wrote:That's a less sharp bound, as it is still infinite.
On 1/4/25 3:52 AM, WM wrote:Not only finite but below 1 % of |ℕ|.
On 04.01.2025 05:06, Richard Damon wrote:Just like every natural number is finite,
On 1/3/25 12:15 PM, WM wrote:For all FISONs:|ℕ \ {1, 2, 3, ..., n}| = ℵo.
Every union of FISONs which stay below a certain threshold staysEvery union of a finite number of FISONs is just an admssion that you
below that threshold.
can't do the actual union of *ALL* FISONs.
That means there is no exception.
but the set of *ALL* of them is infinite, each FISON is finite,
On 04.01.2025 11:59, joes wrote:Whatever do you mean? The union is of course larger than every single
Am Sat, 04 Jan 2025 09:52:08 +0100 schrieb WM:What should make the union larger than all FISONs?
On 04.01.2025 05:06, Richard Damon wrote:But not for the union.
On 1/3/25 12:15 PM, WM wrote:For all FISONs:|ℕ \ {1, 2, 3, ..., n}| = ℵo.
Every union of FISONs which stay below a certain threshold staysEvery union of a finite number of FISONs is just an admssion that you
below that threshold.
can't do the actual union of *ALL* FISONs.
Every union of FISONs which stay below a certain threshold stays belownThe union of all of them doesn't stay below any threshold.
that threshold.
WM <wolfgang.mueckenheim@tha.de> wrote:
The set of these prime gaps is infinite, without qualification. Euclid
could have told you that.
Finally, the most familiar example is this: The (magnitudes of) natural
numbers are potentially infinite because, although there is no upper
bound, there is no infinite (magnitude of a) natural number.
There are no "actual" and "potential" infinity in mathematics.
The
notions are fully unneeded, and add nothing to any mathematical proof.
There is finite and infinite, and that's it.
When I did my maths degree, several decades ago, "potential infinity" and "actual infinity" didn't get a look in. They weren't mentioned a single time.
The only people who talk about "potential" and "actual" infinity are non-mathematicians who lack understanding, and pioneer mathematicians
early on in the development of set theory who were still grasping after precise notions.
On 04.01.2025 12:05, joes wrote:There, you did it.
Am Sat, 04 Jan 2025 09:42:11 +0100 schrieb WM:Define all such that none remains - like in ℕ \ {1, 2, 3, ...} = { }.
No. One can „define” infinitely many numbers.On 1/3/2025 2:48 PM, WM wrote:But removing every ordinal that you can define (and all its
On 03.01.2025 19:46, Jim Burns wrote:
All finite.ordinals removed from the set of each and only
finite.ordinals leaves the empty set.
predecessors) from ℕ leaves almost all ordinals in ℕ.∀n ∈ ℕ_def: |ℕ \
{1, 2, 3, ..., n}| = ℵo
Am Sun, 05 Jan 2025 11:40:58 +0100 schrieb WM:
On 04.01.2025 14:17, Richard Damon wrote:That's a less sharp bound, as it is still infinite.
On 1/4/25 3:52 AM, WM wrote:Not only finite but below 1 % of |ℕ|.
On 04.01.2025 05:06, Richard Damon wrote:Just like every natural number is finite,
On 1/3/25 12:15 PM, WM wrote:For all FISONs:|ℕ \ {1, 2, 3, ..., n}| = ℵo.
Every union of FISONs which stay below a certain threshold staysEvery union of a finite number of FISONs is just an admssion that you >>>>> can't do the actual union of *ALL* FISONs.
below that threshold.
That means there is no exception.
but the set of *ALL* of them is infinite, each FISON is finite,
Am Sun, 05 Jan 2025 11:31:01 +0100 schrieb WM:
On 04.01.2025 11:59, joes wrote:Whatever do you mean? The union is of course larger than every single
Am Sat, 04 Jan 2025 09:52:08 +0100 schrieb WM:What should make the union larger than all FISONs?
On 04.01.2025 05:06, Richard Damon wrote:But not for the union.
On 1/3/25 12:15 PM, WM wrote:For all FISONs:|ℕ \ {1, 2, 3, ..., n}| = ℵo.
Every union of FISONs which stay below a certain threshold staysEvery union of a finite number of FISONs is just an admssion that you >>>>> can't do the actual union of *ALL* FISONs.
below that threshold.
FISON, because there are infinitely many.
Every union of FISONs which stay below a certain threshold stays belownThe union of all of them doesn't stay below any threshold.
that threshold.
Am Sun, 05 Jan 2025 11:32:23 +0100 schrieb WM:
On 04.01.2025 12:05, joes wrote:There, you did it.
Am Sat, 04 Jan 2025 09:42:11 +0100 schrieb WM:Define all such that none remains - like in ℕ \ {1, 2, 3, ...} = { }.
No. One can „define” infinitely many numbers.On 1/3/2025 2:48 PM, WM wrote:But removing every ordinal that you can define (and all its
On 03.01.2025 19:46, Jim Burns wrote:
All finite.ordinals removed from the set of each and only
finite.ordinals leaves the empty set.
predecessors) from ℕ leaves almost all ordinals in ℕ.∀n ∈ ℕ_def: |ℕ \
{1, 2, 3, ..., n}| = ℵo
On 04.01.2025 14:17, Richard Damon wrote:
On 1/4/25 3:52 AM, WM wrote:
On 04.01.2025 05:06, Richard Damon wrote:Just like every natural number is finite,
On 1/3/25 12:15 PM, WM wrote:
Every union of FISONs which stay below a certain threshold stays
below that threshold.
Every union of a finite number of FISONs is just an admssion that
you can't do the actual union of *ALL* FISONs.
For all FISONs:|ℕ \ {1, 2, 3, ..., n}| = ℵo.
That means there is no exception.
but the set of *ALL* of them is infinite, each FISON is finite,
Not only finite but below 1 % of |ℕ|.
but the union of *ALL* of them creates the full infinite set of
Natural Numbers.
Every union of FISONs which stay below a certain threshold stays below
that threshold.
Regards, WM
On 04.01.2025 14:17, Richard Damon wrote:
On 1/4/25 3:42 AM, WM wrote:
And what keep you from "defining" the rest of the Natural Numbers.
On 1/3/2025 2:48 PM, WM wrote:
On 03.01.2025 19:46, Jim Burns wrote:
All finite.ordinals removed from
the set of each and only finite.ordinals
leaves the empty set.
But removing every ordinal that you can define (and all its
predecessors) from ℕ leaves almost all ordinals in ℕ.∀n ∈ ℕ_def: |ℕ \
{1, 2, 3, ..., n}| = ℵo
Try it yourself. Then you will see it.
Regards, WM
On 05.01.2025 12:28, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
Finally, the most familiar example is this: The (magnitudes of)
natural numbers are potentially infinite because, although there is
no upper bound, there is no infinite (magnitude of a) natural number.
There are no "actual" and "potential" infinity in mathematics.
It has been exorcized by those matheologians who were afraid of the
problems introduced to matheology by these precise definitions.
The notions are fully unneeded, and add nothing to any mathematical
proof. There is finite and infinite, and that's it.
When I did my maths degree, several decades ago, "potential infinity"
and "actual infinity" didn't get a look in. They weren't mentioned a
single time.
That has opened the abyss of nonsense to engulf mathematics with such
silly results as: A union of FISONs which stay below a certain
threshold can surpass that threshold.
The only people who talk about "potential" and "actual" infinity are
non-mathematicians who lack understanding, and pioneer mathematicians
early on in the development of set theory who were still grasping
after precise notions.
All mathematicians whom you have disqualified above are genuine mathematicians.
What you
The only people who talk about "potential" and "actual" infinity are non-mathematicians who lack understanding, and pioneer mathematicians
early on in the development of set theory who were still grasping after precise notions.
On 1/5/25 5:40 AM, WM wrote:
For all FISONs:|ℕ \ {1, 2, 3, ..., n}| = ℵo.Just like every natural number is finite,
That means there is no exception.
but the set of *ALL* of them is infinite, each FISON is finite,
Not only finite but below 1 % of |ℕ|.
So? Until you can handle *ALL* of them together, your claim is just a
lie about the whole.
On 1/5/25 5:37 AM, WM wrote:
On 04.01.2025 14:17, Richard Damon wrote:
On 1/4/25 3:42 AM, WM wrote:
And what keep you from "defining" the rest of the Natural Numbers.
On 1/3/2025 2:48 PM, WM wrote:
On 03.01.2025 19:46, Jim Burns wrote:
All finite.ordinals removed from
the set of each and only finite.ordinals
leaves the empty set.
But removing every ordinal that you can define (and all its
predecessors) from ℕ leaves almost all ordinals in ℕ.∀n ∈ ℕ_def: |ℕ
\ {1, 2, 3, ..., n}| = ℵo
Try it yourself. Then you will see it.
But I CAN define any of the Natual Numbers, the whole infinite set of them.
On 1/5/25 5:31 AM, WM wrote:
On 04.01.2025 11:59, joes wrote:
Am Sat, 04 Jan 2025 09:52:08 +0100 schrieb WM:
On 04.01.2025 05:06, Richard Damon wrote:But not for the union.
On 1/3/25 12:15 PM, WM wrote:For all FISONs:|ℕ \ {1, 2, 3, ..., n}| = ℵo.
Every union of FISONs which stay below a certain threshold staysEvery union of a finite number of FISONs is just an admssion that you >>>>> can't do the actual union of *ALL* FISONs.
below
that threshold.
What should make the union larger than all FISONs?
Every union of FISONs which stay below a certain threshold stays
belown that threshold.
Because you never actually USED *ALL* FISONs,
On 05.01.2025 12:28, Alan Mackenzie wrote:
The only people who talk about "potential" and "actual" infinity are
non-mathematicians who lack understanding, and pioneer mathematicians
early on in the development of set theory who were still grasping after
precise notions.
All mathematicians whom you have disqualified above are genuine mathematicians.
What you perhaps could understand is this simple example: As Cantor said actual infinity, for instance omegas and alephs and ℕ are fixed
quantities. The set ℕ is invariable. But all finite initial segments of natural numbers FISONs {1, 2, 3, ..., n} cover less than 1 % of ℕ.
Proof: {1, 2, 3, ..., 100n} is less than ℕ. That means the set of FISONs will never cover ℕ, nor will its union reach the invariable quantity.
The set of FISONs is only potentially infinite, not a fixed quantity but growing over all finite bounds.
Regards, WM
On 05.01.2025 13:47, Richard Damon wrote:
On 1/5/25 5:37 AM, WM wrote:
On 04.01.2025 14:17, Richard Damon wrote:
On 1/4/25 3:42 AM, WM wrote:
And what keep you from "defining" the rest of the Natural Numbers.
On 1/3/2025 2:48 PM, WM wrote:
On 03.01.2025 19:46, Jim Burns wrote:
All finite.ordinals removed from
the set of each and only finite.ordinals
leaves the empty set.
But removing every ordinal that you can define (and all its
predecessors) from ℕ leaves almost all ordinals in ℕ.∀n ∈ ℕ_def: |ℕ
\ {1, 2, 3, ..., n}| = ℵo
Try it yourself. Then you will see it.
But I CAN define any of the Natual Numbers, the whole infinite set of
them.
No. Every defined number is far from the complete set.
Regards, WM
On 05.01.2025 12:28, Alan Mackenzie wrote:
The only people who talk about "potential" and "actual" infinity are
non-mathematicians who lack understanding, and pioneer mathematicians
early on in the development of set theory who were still grasping after
precise notions.
All mathematicians whom you have disqualified above are genuine mathematicians.
What you perhaps could understand is this simple example: As Cantor said actual infinity, for instance omegas and alephs and ℕ are fixed quantities. The set ℕ is invariable.
But all finite initial segments of natural numbers FISONs {1, 2, 3,
..., n} cover less than 1 % of ℕ.
Proof:
{1, 2, 3, ..., 100n} is less than ℕ. That means the set of FISONs will never cover ℕ, nor will its union reach the invariable quantity.
The set of FISONs is only potentially infinite, not a fixed quantity
but growing over all finite bounds.
Regards, WM
On 05.01.2025 13:47, Richard Damon wrote:
On 1/5/25 5:31 AM, WM wrote:
On 04.01.2025 11:59, joes wrote:
Am Sat, 04 Jan 2025 09:52:08 +0100 schrieb WM:
On 04.01.2025 05:06, Richard Damon wrote:But not for the union.
On 1/3/25 12:15 PM, WM wrote:For all FISONs:|ℕ \ {1, 2, 3, ..., n}| = ℵo.
Every union of FISONs which stay below a certain threshold stays >>>>>>> belowEvery union of a finite number of FISONs is just an admssion that you >>>>>> can't do the actual union of *ALL* FISONs.
that threshold.
What should make the union larger than all FISONs?
Every union of FISONs which stay below a certain threshold stays
belown that threshold.
Because you never actually USED *ALL* FISONs,
All FISONs are smaller than 1 % of |ℕ|.
Find a FISON {1, 2, 3, ..., n} such that {1, 2, 3, ..., 100n} is a
superset of ℕ.
Regards, WM
WM <wolfgang.mueckenheim@tha.de> wrote:
Maybe Cantor did say this, as a pioneer early on in the development of
set theory. Things have developed since then, and we see that there is nothing to be gained by construing infinite sets as "fixed quantities";
there is no mathematical proof where such a concept makes the slightest difference.
But all finite initial segments of natural numbers FISONs {1, 2, 3,
..., n} cover less than 1 % of ℕ.
That is a thoroughly unmathematical statement. To talk about 1% of an infinite set is meaningless.
Finally, it is
wrong, absurdly wrong. The union of all FISONs _is_ N.
No, not a mathematical proof.
{1, 2, 3, ..., 100n} is less than ℕ. That means the set of FISONs will
never cover ℕ, nor will its union reach the invariable quantity.
No, it doesn't mean that at all. The set of FISONs does indeed "cover"
N, in the sense that their union is equal to N. A proof of this is
trivial, well within the understanding of a school student studying
maths.
On 04.01.2025 21:38, Chris M. Thomasson wrote:Well yes, the size of N is itself not a natural number. Big surprise.
For me, there are infinitely many natural numbers, period... Do youNo. There are actually infinitely many natural numbers. All can be
totally disagree?
removed from ℕ, but only collectively ℕ \ {1, 2, 3, ...} = { }.
It is impossible to remove the numbers individually ∀n ∈ ℕ_def: |ℕ \ {1,
2, 3, ..., n}| = ℵo.
Am Sun, 05 Jan 2025 12:14:47 +0100 schrieb WM:
On 04.01.2025 21:38, Chris M. Thomasson wrote:Well yes, the size of N is itself not a natural number. Big surprise.
For me, there are infinitely many natural numbers, period... Do youNo. There are actually infinitely many natural numbers. All can be
totally disagree?
removed from ℕ, but only collectively ℕ \ {1, 2, 3, ...} = { }.
It is impossible to remove the numbers individually ∀n ∈ ℕ_def: |ℕ \ {1,
2, 3, ..., n}| = ℵo.
On 1/5/25 12:26 PM, WM wrote:
On 05.01.2025 13:47, Richard Damon wrote:
On 1/5/25 5:31 AM, WM wrote:
On 04.01.2025 11:59, joes wrote:
Am Sat, 04 Jan 2025 09:52:08 +0100 schrieb WM:
On 04.01.2025 05:06, Richard Damon wrote:But not for the union.
On 1/3/25 12:15 PM, WM wrote:For all FISONs:|ℕ \ {1, 2, 3, ..., n}| = ℵo.
Every union of FISONs which stay below a certain threshold stays >>>>>>>> belowEvery union of a finite number of FISONs is just an admssion that >>>>>>> you
that threshold.
can't do the actual union of *ALL* FISONs.
What should make the union larger than all FISONs?
Every union of FISONs which stay below a certain threshold stays
belown that threshold.
Because you never actually USED *ALL* FISONs,
All FISONs are smaller than 1 % of |ℕ|.
Find a FISON {1, 2, 3, ..., n} such that {1, 2, 3, ..., 100n} is a
superset of ℕ.
I never said there was,
but your claim doesn't match your conclusion, as
the union of *ALL* the FISIONs will reach the size of the Natural
Numbers, even though no finite subset reaches a measurable percentage of
it.
On 04.01.2025 17:20, Jim Burns wrote:
On 1/4/2025 3:42 AM, WM wrote:
On 1/3/2025 3:56 PM, Jim Burns wrote:
All finite.ordinals removed from
the set of each and only finite.ordinals
leaves the empty set.
But removing
every ordinal that you can define
(and all its predecessors) from ℕ leaves
almost all ordinals in ℕ.
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
ℕ is the set of each and only finite.ordinals.
Yes.
|ℕ| := ℵ₀ = |ℕ\{0}| = |ℕ\{0,1}| = ... =
|ℕ\{0,1,...,n}| = ...
The sequence of end.segments of ℕ
grows emptier.one.by.one but
it doesn't grow smaller.one.by.one.
It does
but you cannot give the numbers
because they are dark.
The sequence of end.segments of ℕ
grows emptier.one.by.one but
it doesn't grow smaller.one.by.one.
It does
but you cannot give the numbers
because they are dark.
A precise measure must detect
the loss of one element.
ℵo is no precise measure but only
another expression for infinitely many.
WM <wolfgang.mueckenheim@tha.de> wrote: <nonsense>
The union of all FISONs _is_ N.
The plain fact is that the set of FISONs is infinite
The set of FISONs
WM <wolfgang.mueckenheim@tha.de> wrote: <nonsense>
The union of all FISONs _is_ N.
The plain fact is that the set of FISONs is infinite
The set of FISONs
On 1/5/2025 11:47 AM, WM wrote:
On 05.01.2025 18:39, Richard Damon wrote:
On 1/5/25 12:26 PM, WM wrote:
On 05.01.2025 13:47, Richard Damon wrote:
On 1/5/25 5:31 AM, WM wrote:
On 04.01.2025 11:59, joes wrote:
Am Sat, 04 Jan 2025 09:52:08 +0100 schrieb WM:
For all FISONs: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
But not for the union.
What should make the union larger than all FISONs?
Every union of FISONs which stay below a certain threshold stays
belown that threshold.
Because you never actually USED *ALL* FISONs [for the union(s)],
You said that I never used all FISONs. But I do.
All are insufficient.
but your claim doesn't match your conclusion, as the union of *ALL*
FISONs will reach the size of the [set of all] Natural Numbers,
even though no [union of] finite[ly many FISONs] reaches [...] it.
Do one or more FISONs grow during the union process? (WM)
WM <wolfgang.mueckenheim@tha.de> wrote:
On 05.01.2025 12:28, Alan Mackenzie wrote:
The only people who talk about "potential" and "actual" infinity are
non-mathematicians who lack understanding, and [...]
All mathematicians whom you have disqualified above are genuine
mathematicians.
[...] all finite initial segments of natural numbers FISONs {1, 2, 3,
..., n} cover less than 1 % of ℕ.
That is a thoroughly unmathematical statement. To talk about 1% of an infinite set is meaningless. To say "cover" in the context of set
theory rather than topological spaces is inappropriate. Above all, to
say "all finite initial segments" is unmathematical, since what is meant
is not the set of FISONs, but the union of FISONs. Finally, it is
wrong, absurdly wrong. The union of all FISONs _is_ N.
Proof:
No, not a mathematical proof. You have never studied maths to degree
level, and have no idea what a mathematical proof looks like. [...]
[...] The set of FISONs does indeed "cover"
N, in the sense that their union is equal to N. A proof of this is
trivial, well within the understanding of a school student studying
maths.
The set of FISONs is only potentially infinite, not <bla>
This "potentially" and "actually" infinite has led you astray, away from
the truth. They are solely historical notions, with no place in modern [classical] mathematics [i. e. set theory + classical logic --moebius].
The plain fact is that the set of FISONs is infinite [...]
"We introduce numbers for counting. This does not at all imply the
infinity of numbers. For, in what way should we ever arrive at
infinitely-many countable things? [...] In philosophical terminology we
say that the infinite of the number sequence is only potential, i.e.,
existing only as a possibility." [P. Lorenzen: "Das Aktual-Unendliche in
der Mathematik", Philosophia naturalis 4 (1957) p. 4f]
Philosopy.
"Until then, no one envisioned the possibility that infinities come in
different sizes, and moreover, mathematicians had no use for 'actual
infinity'. The arguments using infinity, including the Differential
Calculus of Newton and Leibniz, do not require the use of infinite sets.
[...] Cantor observed that many infinite sets of numbers are countable:
the set of all integers, the set of all rational numbers, and also the
set of all algebraic numbers. Then he gave his ingenious diagonal
argument that proves, by contradiction, that the set of all real numbers
is not countable. A consequence of this is that there exists a multitude
of transcendental numbers, even though the proof, by contradiction, does
not produce a single specific example." [T. Jech: "Set theory", Stanford
Encyclopedia of Philosophy (2002)]
Also philosophy.
"Numerals constitute a potential infinity. Given any numeral, we can
construct a new numeral by prefixing it with S. Now imagine this
potential infinity to be completed. Imagine the inexhaustible process of
constructing numerals somehow to have been finished, and call the result
the set of all numbers, denoted by . Thus is thought to be an actual >> infinity or a completed infinity. This is curious terminology, since the
etymology of 'infinite' is 'not finished'." [E. Nelson: "Hilbert's
mistake" (2007) p. 3]
E. Nelson is clearly not a mathematician.
According to (Gödel's) Platonism, objects of mathematics have the same
status of reality as physical objects. "Views to the effect that
Platonism is correct but only for certain relatively 'concrete'
mathematical 'objects'. Other mathematical 'objects' are man made, and
are not part of an external reality. Under such a view, what is to be
made of the part of mathematics that lies outside the scope of
Platonism? An obvious response is to reject it as utterly meaningless."
[H.M. Friedman: "Philosophical problems in logic" (2002) p. 9]
Possibly philosophy, more likely complete nonsense.
"A potential infinity is a quantity which is finite but indefinitely
large. For instance, when we enumerate the natural numbers as 0, 1, 2,
..., n, n+1, ..., the enumeration is finite at any point in time, but it
grows indefinitely and without bound. [...] An actual infinity is a
completed infinite totality. Examples: , , C[0, 1], L2[0, 1], etc.
Other examples: gods, devils, etc." [S.G. Simpson: "Potential versus
actual infinity: Insights from reverse mathematics" (2015)]
Another philosopher?
"Potential infinity refers to a procedure that gets closer and closer
to, but never quite reaches, an infinite end. For instance, the sequence
of numbers 1, 2, 3, 4, ... gets higher and higher, but it has no end; it
never gets to infinity. Infinity is just an indication of a direction –
it's 'somewhere off in the distance'. Chasing this kind of infinity is
like chasing a rainbow or trying to sail to the edge of the world – you
may think you see it in the distance, but when you get to where you
thought it was, you see it is still further away. Geometrically, imagine
an infinitely long straight line; then 'infinity' is off at the 'end' of
the line. Analogous procedures are given by limits in calculus, whether
they use infinity or not. For example, limx0(sinx)/x = 1. This means
that when we choose values of x that are closer and closer to zero, but
never quite equal to zero, then (sinx)/x gets closer and closer to one."
[E. Schechter: "Potential versus completed infinity: Its history and
controversy" (5 Dec 2009)]
There may be a history to it, but there is no controversy, at least not
in mathematical circles.
There are no "actual" and "potential" infinity in mathematics.
When I did my maths degree, several decades ago, "potential infinity" and "actual infinity" didn't get a look in. They weren't mentioned a single time. Instead, precise definitions were given to "finite" and
"infinite", and we learnt how to use these definitions and what could be
done with them.
The only people who talk about "potential" and "actual" infinity are non-mathematicians who lack understanding
WM <wolfgang.mueckenheim@tha.de> wrote:
On 05.01.2025 12:28, Alan Mackenzie wrote:
The only people who talk about "potential" and "actual" infinity are
non-mathematicians who lack understanding, and [...]
All mathematicians whom you have disqualified above are genuine
mathematicians.
[...] all finite initial segments of natural numbers FISONs {1, 2, 3,
..., n} cover less than 1 % of ℕ.
That is a thoroughly unmathematical statement. To talk about 1% of an infinite set is meaningless. To say "cover" in the context of set
theory rather than topological spaces is inappropriate. Above all, to
say "all finite initial segments" is unmathematical, since what is meant
is not the set of FISONs, but the union of FISONs. Finally, it is
wrong, absurdly wrong. The union of all FISONs _is_ N.
Proof:
No, not a mathematical proof. You have never studied maths to degree
level, and have no idea what a mathematical proof looks like. [...]
[...] The set of FISONs does indeed "cover"
N, in the sense that their union is equal to N. A proof of this is
trivial, well within the understanding of a school student studying
maths.
The set of FISONs is only potentially infinite, not <bla>
This "potentially" and "actually" infinite has led you astray, away from
the truth. They are solely historical notions, with no place in modern [classical] mathematics [i. e. set theory + classical logic --moebius].
The plain fact is that the set of FISONs is infinite [...]
On 05.01.2025 19:03, joes wrote:
Am Sun, 05 Jan 2025 12:14:47 +0100 schrieb WM:
On 04.01.2025 21:38, Chris M. Thomasson wrote:Well yes, the size of N is itself not a natural number. Big surprise.
For me, there are infinitely many natural numbers, period... Do youNo. There are actually infinitely many natural numbers. All can be
totally disagree?
removed from ℕ, but only collectively ℕ \ {1, 2, 3, ...} = { }.
It is impossible to remove the numbers individually ∀n ∈ ℕ_def: |ℕ \ {1,
2, 3, ..., n}| = ℵo.
ℕ cannot be covered by FISONs, neither by many nor by their union. If ℕ could be covered by FISONs then one would be sufficient. But for all we
have: Extension by 100 is insufficient. Every union of FISONs which stay below 1 % stays below 1 %.
1 % is an abbreviation for: Extension by a factor of 100 does not cover ℕ.
Regards, WM
Regards, WM
On 1/5/2025 11:47 AM, WM wrote:
What should make the union larger than all FISONs?
A reasonable question might be:
"What should make the union larger than each and every FISON?"
That fact that the union of ALL FISONs "comprises" ALL natural numbers,
while each and every FISON only "comprises" finitely many numbers.
Every union of FISONs which stay below a certain threshold stays >>>>>>> below that threshold.
Nope.
On 1/5/2025 6:07 AM, WM wrote:
On 04.01.2025 17:20, Jim Burns wrote:
On 1/4/2025 3:42 AM, WM wrote:
On 1/3/2025 3:56 PM, Jim Burns wrote:
All finite.ordinals removed from
the set of each and only finite.ordinals
leaves the empty set.
But removing
every ordinal that you can define
(and all its predecessors) from ℕ leaves
almost all ordinals in ℕ.
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
ℕ is the set of each and only finite.ordinals.
Yes.
Anything which is a finite.ordinal,
darkᵂᴹ and visibleᵂᴹ, is in ℕ
Q. What is a finite.ordinal?
⎛ It is an element of ℕ
⎜
⎜ It is an ordinal,
⎜ one of the (well.ordered) ordinals.
⎜
⎜ It is
⎜ larger.by.one than emptier.by.one
⎜ ( #⟦0,k⦆ > #(⟦0,k⦆\⦃0⦄)
⎜ or it is emptiest.
⎝ ( k = 0 = ⦃⦄
We know that
all of that is true.without.exception
among the finite.ordinals,
whether.or.not I can giveᵂᴹ the finite.ordinal,
whether.or.not it's darkᵂᴹ.
The sequence of end.segments of ℕ
grows emptier.one.by.one but
it doesn't grow smaller.one.by.one.
It does
but you cannot give the numbers
because they are dark.
Not.giving numbers doesn't prevent us from
making claims which we know are without.exception.
A precise measure must detect
the loss of one element.
ℵo is no precise measure but only
another expression for infinitely many.
An accurate measure must recognize that
a set larger.than any.finite.set
is not any.finite.set.
https://en.wikipedia.org/wiki/Accuracy_and_precision
For each finite.set, there is
a finite.ordinal of the same size.
For each finite.ordinal, there is
a larger.by.one finite.ordinal,
and
it and its priors are a subset of
the set of all finite.ordinals.
For each finite.set, there is
a larger.than.that subset of
the set of all finite.ordinals.
For each finite.set, that set is not
the set of all finite.ordinals.
The set of all finite.ordinals is not
any.finite.set.
Q. What is a finite set?
On 1/5/25 2:47 PM, WM wrote:
You said that I never used all FISONs. But I do. All are insufficient.
No, you DON'T use all. Because ALL has an infinite number of members,
and you need to process them one by one,
but your claim doesn't match your conclusion, as the union of *ALL*How do they do it? Do one or more FISONs grow during the union
the FISIONs will reach the size of the Natural Numbers, even though
no finite subset reaches a measurable percentage of it.
process? (One would be sufficient.)
Nope, they are just infinite in number.
WM pretended :
0 is a cardinal but not an ordinal.
Wrong.
On 05.01.2025 19:03, joes wrote:
Am Sun, 05 Jan 2025 12:14:47 +0100 schrieb WM:
On 04.01.2025 21:38, Chris M. Thomasson wrote:
For me,
there are infinitely many natural numbers, period...
Do you totally disagree?
No.
There are actually infinitely many natural numbers.
All can be removed from ℕ, but only collectively
ℕ \ {1, 2, 3, ...} = { }.
It is impossible to remove the numbers individually
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
Well yes,
the size of N is itself
not a natural number.
Big surprise.
ℕ cannot be covered by FISONs,
neither by many nor by their union.
If ℕ could be covered by FISONs
then one would be sufficient.
If ℕ could be covered by FISONs
then one would be sufficient.
ℕ cannot be covered by FISONs,
neither by many nor by their union.
But for all we have:
Extension by 100 is insufficient.
On 1/5/2025 1:14 PM, WM wrote:
ℕ cannot be covered by FISONs,
neither by many nor by their union.
If ℕ could be covered by FISONs
then one would be sufficient.
ℕ is the set of finite.ordinals.
ℕ holds each finite ordinal.
ℕ holds only finite.ordinals.
k ∈ ℕ ⇒ k+1 ∈ ℕ
is true for both the darkᵂᴹ and the visibleᵂᴹ.
If ℕ could be covered by FISONs
then one would be sufficient.
ℕ is the set of finite.ordinals.
A FISON is a set of finite.ordinals
up to that FISON's maximum (finite.ordinal) element.
If one FISON covered ℕ,
that FISON.cover would equal ℕ,
and the maximum of that FISON.cover
would be the maximum.of.all finite.ordinal.
However,
no finite.ordinal k is the maximum.of.all.
k ∈ ℕ ⇒ k+1 ∈ ℕ
That is true for both the darkᵂᴹ and the visibleᵂᴹ.
Contradiction.
No one FISON covers ℕ.
ℕ is the set of finite ordinals.
Each finite.ordinal k is in
at least one FISON: ⟦0,k⟧
Each finite.ordinal is in
the union of FISONs
The union of FISONs covers
the set ℕ of finite.ordinals
But for all we have:
Extension by 100 is insufficient.
Correct.
Which is weird, but accurate.
The source of that weird result is lemma 1.
WM presented the following explanation :
On 06.01.2025 19:23, FromTheRafters wrote:
WM pretended :
0 is a cardinal but not an ordinal.
Wrong.
Who is the zeroest clown?
Okay, maybe not with "your" definition of the ordinal numbers.
This is
mathematics,
He is hyper finite, and that's the way it is.
On 01/06/2025 02:43 PM, Jim Burns wrote:
It would be great if you (WM) did NOT
find lemma 1 weird,
but it is what it is.
The inductive set being covered by
initial segments is an _axiom_ of ZF.
WM has brought this to us :
On 06.01.2025 23:43, Jim Burns wrote:
k ∈ ℕ ⇒ k+1 ∈ ℕ
is true for both the darkᵂᴹ and the visibleᵂᴹ.
One exception exists: ω-1.
Which remains undefined.
First several von Neumann ordinals
0 = {} = ∅
1 = {0} = {∅}
2 = {0,1} = {∅,{∅}}
3 = {0,1,2} = {∅,{∅},{∅,{∅}}} 4 = {0,1,2,3} = {∅,{∅},{∅,{∅}},{∅,{∅},{∅,{∅}}}}
=====================================
Notice they start at zero (emptyset)
If there were such
things as "potential" and "actual" infinity in maths,
make a difference to some mathematical result.
Am 05.01.2025 um 12:28 schrieb Alan Mackenzie:
"We introduce numbers for counting. This does not at all imply the
infinity of numbers. For, in what way should we ever arrive at
infinitely-many countable things? [...] In philosophical terminology we
say that the infinite of the number sequence is only potential, i.e.,
existing only as a possibility." [P. Lorenzen: "Das Aktual-Unendliche in >>> der Mathematik", Philosophia naturalis 4 (1957) p. 4f]
Philosopy.
Sure, though P. Lorenzen was an eminent mathematician who developed a
form of constructive mathematics (constructive analysis) and dialogical logic.
See: https://en.wikipedia.org/wiki/Paul_Lorenzen
and: https://en.wikipedia.org/wiki/Dialogical_logic
Note that we can't be sure if Mückenheim's translation is accurate.
"Until then, no one envisioned the possibility that infinities come in
different sizes, and moreover, mathematicians had no use for 'actual
infinity'. The arguments using infinity, including the Differential
Calculus of Newton and Leibniz, do not require the use of infinite sets. >>> [...] Cantor observed that many infinite sets of numbers are countable:
the set of all integers, the set of all rational numbers, and also the
set of all algebraic numbers. Then he gave his ingenious diagonal
argument that proves, by contradiction, that the set of all real numbers >>> is not countable. A consequence of this is that there exists a multitude >>> of transcendental numbers, even though the proof, by contradiction, does >>> not produce a single specific example." [T. Jech: "Set theory", Stanford >>> Encyclopedia of Philosophy (2002)]
Also philosophy.
Sure. But T. Jech is a leading set theorist.
https://en.wikipedia.org/wiki/Thomas_Jech
"Numerals constitute a potential infinity. Given any numeral, we can
construct a new numeral by prefixing it with S. Now imagine this
potential infinity to be completed. Imagine the inexhaustible process of >>> constructing numerals somehow to have been finished, and call the result >>> the set of all numbers, denoted by . Thus is thought to be an actual >>> infinity or a completed infinity. This is curious terminology, since the >>> etymology of 'infinite' is 'not finished'." [E. Nelson: "Hilbert's
mistake" (2007) p. 3]
E. Nelson is clearly not a mathematician.
Holy shit!
"Edward Nelson (May 4, 1932 – September 10, 2014) was an American mathematician. He was professor in the Mathematics Department at
Princeton University. He was known for his work on mathematical physics
and mathematical logic. In mathematical logic, he was noted especially
for his internal set theory, and views on ultrafinitism and the
consistency of arithmetic. In philosophy of mathematics he advocated the view of formalism rather than platonism or intuitionism."
https://en.wikipedia.org/wiki/Edward_Nelson
See: https://en.wikipedia.org/wiki/Internal_set_theory
According to (Gödel's) Platonism, objects of mathematics have the same
status of reality as physical objects. "Views to the effect that
Platonism is correct but only for certain relatively 'concrete'
mathematical 'objects'. Other mathematical 'objects' are man made, and
are not part of an external reality. Under such a view, what is to be
made of the part of mathematics that lies outside the scope of
Platonism? An obvious response is to reject it as utterly meaningless."
[H.M. Friedman: "Philosophical problems in logic" (2002) p. 9]
Possibly philosophy, more likely complete nonsense.
*sigh*
"Harvey Friedman (born 23 September 1948)[1] is an American mathematical logician at Ohio State University in Columbus, Ohio. He has worked on reverse mathematics, a project intended to derive the axioms of
mathematics from the theorems considered to be necessary. In recent
years, this has advanced to a study of Boolean relation theory, which attempts to justify large cardinal axioms by demonstrating their
necessity for deriving certain propositions considered "concrete"."
https://en.wikipedia.org/wiki/Harvey_Friedman
"This chapter focuses on the work of mathematical logician Harvey
Friedman, who was recently awarded the National Science Foundation's
annual Waterman Prize, honoring the most outstanding American scientist under thirty-five years of age in all fields of science and engineering. Friedman's contributions span all branches of mathematical logic
(recursion theory, proof theory, model theory, set theory, and theory of computation). He is a generalist in an age of specialization, yet his theorems often require extraordinary technical virtuosity, of which only
a few selected highlights are discussed. Friedman's ideas have yielded radically new kinds of independence results. The kinds of statements
that were proved to be independent before Friedman were mostly disguised properties of formal systems (such as Gödel's theorem on unprovability
of consistency) or assertions about abstract sets (such as the continuum hypothesis or Souslin's hypothesis). In contrast, Friedman's
independence results are about questions of a more concrete nature involving, for example, Borel functions or the Hilbert cube."
Source: https://www.sciencedirect.com/science/article/abs/pii/S0049237X09701545 (1985)
"A potential infinity is a quantity which is finite but indefinitely
large. For instance, when we enumerate the natural numbers as 0, 1, 2,
..., n, n+1, ..., the enumeration is finite at any point in time, but it >>> grows indefinitely and without bound. [...] An actual infinity is a
completed infinite totality. Examples: , , C[0, 1], L2[0, 1], etc. >>> Other examples: gods, devils, etc." [S.G. Simpson: "Potential versus
actual infinity: Insights from reverse mathematics" (2015)]
Another philosopher?
"Stephen George Simpson (born September 8, 1945) is an American mathematician whose research concerns the foundations of mathematics, including work in mathematical logic, recursion theory, and Ramsey
theory. He is known for his extensive development of the field of
reverse mathematics founded by Harvey Friedman, in which the goal is to determine which axioms are needed to prove certain mathematical
theorems.[1] He has also argued for the benefits of finitistic
mathematical systems, such as primitive recursive arithmetic, which do
not include actual infinity."
Source: https://en.wikipedia.org/wiki/Steve_Simpson_(mathematician)
"Potential infinity refers to a procedure that gets closer and closer
to, but never quite reaches, an infinite end. For instance, the sequence >>> of numbers 1, 2, 3, 4, ... gets higher and higher, but it has no end; it >>> never gets to infinity. Infinity is just an indication of a direction – >>> it's 'somewhere off in the distance'. Chasing this kind of infinity is
like chasing a rainbow or trying to sail to the edge of the world – you >>> may think you see it in the distance, but when you get to where you
thought it was, you see it is still further away. Geometrically, imagine >>> an infinitely long straight line; then 'infinity' is off at the 'end' of >>> the line. Analogous procedures are given by limits in calculus, whether
they use infinity or not. For example, limx0(sinx)/x = 1. This means
that when we choose values of x that are closer and closer to zero, but
never quite equal to zero, then (sinx)/x gets closer and closer to one." >>> [E. Schechter: "Potential versus completed infinity: Its history and
controversy" (5 Dec 2009)]
There may be a history to it, but there is no controversy, at least not
in mathematical circles.
You are not familiar with "foundations of mathematics", right?
Of course: "In practice, most mathematicians either do not work from axiomatic systems, or if they do, do not doubt the consistency of ZFC, generally their preferred axiomatic system."
Source: https://en.wikipedia.org/wiki/Foundations_of_mathematics
There are no "actual" and "potential" infinity in mathematics.
If you say so. :-P
When I did my maths degree, several decades ago, "potential infinity" and
"actual infinity" didn't get a look in. They weren't mentioned a single
time. Instead, precise definitions were given to "finite" and
"infinite", and we learnt how to use these definitions and what could be
done with them.
Sure. But this needs a context; usually (some sort of) set theory.
The only people who talk about "potential" and "actual" infinity are
non-mathematicians who lack understanding
Like those mentioned above?
On 07.01.2025 10:40, FromTheRafters wrote:
WM has brought this to us :
On 06.01.2025 23:43, Jim Burns wrote:
k ∈ ℕ ⇒ k+1 ∈ ℕ
is true for both the darkᵂᴹ and the visibleᵂᴹ.
One exception exists: ω-1.
Which remains undefined.
Like all dark numbers.
Regards, WM
On 1/7/25 5:20 AM, WM wrote:
On 07.01.2025 02:36, Ross Finlayson wrote:
On 01/06/2025 02:43 PM, Jim Burns wrote:
It would be great if you (WM) did NOT
find lemma 1 weird,
but it is what it is.
It is not weird. But your conclusions are weird.
The inductive set being covered by
initial segments is an _axiom_ of ZF.
And the existence of the set ℕ is also an axiom of ZF. Therefore ZF is
incompatible with mathematics.
No, ZF doesn't have as an axiom that the set of Natural Numbers exist.
All FISONs stay below the threshold |ℕ|/100, or in other words,
multiplication of any FISON by 100 is insufficient to cover |ℕ|.
So? The union of an infinite set of them can have properites different
that any set that is a union of only a finite number of them. That is a nature of infinity.
Every union of FISONs {1, 2, 3, ..., n} which stay below this
threshold stays below this threshold too.
But not the union of *EVERY* FISON, the FULL INFINITE set of them.
SOething your "logic" can't handle,
WM was thinking very hard :
On 07.01.2025 02:36, Ross Finlayson wrote:
On 01/06/2025 02:43 PM, Jim Burns wrote:
It would be great if you (WM) did NOT
find lemma 1 weird,
but it is what it is.
It is not weird. But your conclusions are weird.
The inductive set being covered by
initial segments is an _axiom_ of ZF.
And the existence of the set ℕ is also an axiom of ZF.
Which axiom?
WM wrote :
On 07.01.2025 10:51, FromTheRafters wrote:
First several von Neumann ordinals
v. Neumann was bright but not bright enough.
Why should we use the nomenclature of his disproven theory?
Disproven?
0 = {} = ∅
1 = {0} = {∅}
2 = {0,1} = {∅,{∅}}
3 = {0,1,2} = {∅,{∅},{∅,{∅}}}
4 = {0,1,2,3} = {∅,{∅},{∅,{∅}},{∅,{∅},{∅,{∅}}}}
=====================================
Notice they start at zero (emptyset)
It would be more important to reach the full set.
It does, in the infinite union.
WM was thinking very hard :
On 07.01.2025 15:05, FromTheRafters wrote:
WM wrote :
On 07.01.2025 10:51, FromTheRafters wrote:
First several von Neumann ordinals
v. Neumann was bright but not bright enough.
Why should we use the nomenclature of his disproven theory?
Disproven?
0 = {} = ∅
1 = {0} = {∅}
2 = {0,1} = {∅,{∅}}
3 = {0,1,2} = {∅,{∅},{∅,{∅}}} >>>>> 4 = {0,1,2,3} = {∅,{∅},{∅,{∅}},{∅,{∅},{∅,{∅}}}}
=====================================
Notice they start at zero (emptyset)
It would be more important to reach the full set.
It does, in the infinite union.
It does not because the infinite union cannot be larger than every
unioned FISON.
Just because you saya so?
Every unioned FISON is smaller than any definable fraction of the full
set ℕ.
So what?
On 07.01.2025 15:05, FromTheRafters wrote:
WM wrote :
On 07.01.2025 10:51, FromTheRafters wrote:
First several von Neumann ordinals
v. Neumann was bright but not bright enough.
Why should we use the nomenclature of his disproven theory?
Disproven?
0 = {} = ∅
1 = {0} = {∅}
2 = {0,1} = {∅,{∅}}
3 = {0,1,2} = {∅,{∅},{∅,{∅}}}
4 = {0,1,2,3} = {∅,{∅},{∅,{∅}},{∅,{∅},{∅,{∅}}}}
=====================================
Notice they start at zero (emptyset)
It would be more important to reach the full set.
It does, in the infinite union.
It does not because the infinite union cannot be larger than every
unioned FISON. Every unioned FISON is smaller than any definable
fraction of the full set ℕ.
Regards, WM
On 06.01.2025 23:43, Jim Burns wrote:
On 1/5/2025 1:14 PM, WM wrote:
ℕ cannot be covered by FISONs,
neither by many nor by their union.
If ℕ could be covered by FISONs
then one would be sufficient.
ℕ is the set of finite.ordinals.
ℕ holds each finite ordinal.
ℕ holds only finite.ordinals.
|ℕ| is by definition
the smallest transfinite number,
On 1/6/2025 3:28 AM, FromTheRafters wrote:
Chris M. Thomasson laid this down on his screen :
If a tree falls in a forest, does it make a sound?
If something 'hears' it, yes.
On 06.01.2025 23:43, Jim Burns wrote:
k ∈ ℕ ⇒ k+1 ∈ ℕ
is true for both the darkᵂᴹ and the visibleᵂᴹ.
One exception exists: ω-1.
(0, ω)*2 = {2, 4, 6, ..., ω, ω+2, ω+4, ...}.
Except you don't use *EVERY* FISON, only those below some limit n that
is below a faction of infinity.
On 1/7/25 7:51 AM, WM wrote:
No, ZF doesn't have as an axiom that the set of Natural Numbers exist.
AoI: There exists an infinite set S.
Which isn't that the NATURAL NUMBERS are an infinite set.
Every union of FISONs {1, 2, 3, ..., n} which stay below this
threshold stays below this threshold too.
But not the union of *EVERY* FISON, the FULL INFINITE set of them.
All are below 1 %.
No,
So, all you prove is that a finite subset of an infinite set doesn't
cover all of the infinite set.
Your logic can't handle *EVERY* FISON at once.
SOething your "logic" can't handle,
Luckily.
So, you think you are "lucky" to be ignorant?
I guess that just shows how your logic works.
Regards, WM
On 1/7/2025 4:13 AM, WM wrote:
The cardinal:ordinal distinction
-- which does not matter in the finite domain
matters in the infinite domain.
On 1/7/2025 4:13 AM, WM wrote:
On 06.01.2025 23:43, Jim Burns wrote:
k ∈ ℕ ⇒ k+1 ∈ ℕ
is true for both the darkᵂᴹ and the visibleᵂᴹ.
One exception exists: ω-1.
No.
ω-1 does not exist, darkᵂᴹ or visibleᵂᴹ,
because
⦃k: k < ω ≤ k+1⦄ = ⦃⦄
⎛ Assume otherwise.
⎜ Assume ω-1 exists.
⎜ ω-1 < ω
⎜ ¬∃ᵒʳᵈψ: ω-1 < ψ < ω
⎜
⎜ However,
⎜ ω-1 ∉ ⦃⦄ = ⦃k: k < ω ≤ k+1⦄
On 08.01.2025 00:30, Richard Damon wrote:
On 1/7/25 7:51 AM, WM wrote:
No, ZF doesn't have as an axiom that the set of Natural Numbers exist.
AoI: There exists an infinite set S.
Which isn't that the NATURAL NUMBERS are an infinite set.
The infinite set has been designed by Zermelo according to Dedekind's definition of the natural numbers, as Zermelo noted. https:// gdz.sub.uni-goettingen.de/id/PPN235181684_0065? tify=%7B%22pages%22%3A%5B276%5D%2C%22pan%22%3A%7B%22x%22%3A0.461%2C%22y%22%3A1.103%7D%2C%22view%22%3A%22info%22%2C%22zoom%22%3A0.884%7D
Every union of FISONs {1, 2, 3, ..., n} which stay below this
threshold stays below this threshold too.
But not the union of *EVERY* FISON, the FULL INFINITE set of them.
All are below 1 %.
No,
Show one FISON that is larger than 1 %.
So, all you prove is that a finite subset of an infinite set doesn't
cover all of the infinite set.
Your logic can't handle *EVERY* FISON at once.
Show one FISON that is larger than 1 %.
Regards, WM
SOething your "logic" can't handle,
Luckily.
So, you think you are "lucky" to be ignorant?
I guess that just shows how your logic works.
Regards, WM
On 08.01.2025 06:45, Jim Burns wrote:
On 1/7/2025 4:13 AM, WM wrote:
On 06.01.2025 23:43, Jim Burns wrote:
k ∈ ℕ ⇒ k+1 ∈ ℕ
is true for both the darkᵂᴹ and the visibleᵂᴹ.
One exception exists: ω-1.
No.
ω-1 does not exist, darkᵂᴹ or visibleᵂᴹ,
because
⦃k: k < ω ≤ k+1⦄ = ⦃⦄
If ω exists, then ω-1 exists.
Then your claim is wrong.
⎛ Assume otherwise.
⎜ Assume ω-1 exists.
⎜ ω-1 < ω
⎜ ¬∃ᵒʳᵈψ: ω-1 < ψ < ω
⎜
⎜ However,
⎜ ω-1 ∉ ⦃⦄ = ⦃k: k < ω ≤ k+1⦄
⦃⦄ = ⦃k: k < ω ≤ k+1⦄
is a wrong presupposition.
⎜ ¬(w-1 < w ≤ (w-1)+1)
⎜ w-1 < (w-1)+1 < w
⎜ ∃ᵒʳᵈψ: ω-1 < ψ < ω
⎝ Contradiction.
⦃k: k < ω ≤ k+1⦄ has one element.
It is dark like ω.
WM formulated on Wednesday :
If ω exists, then ω-1 exists.
Wrong.
⦃k: k < ω ≤ k+1⦄ = ⦃⦄
ω-1 does not exist.
On 1/8/25 4:06 AM, WM wrote:
On 08.01.2025 00:30, Richard Damon wrote:There isn't one, but doesn't need to be.
Except you don't use *EVERY* FISON, only those below some limit n
that is below a faction of infinity.
Show a FISON that expanded by a factor of 100 or more covers ℕ. Fail!
On 01/07/2025 03:36 AM, Alan Mackenzie wrote:
Moebius <invalid@example.invalid> wrote:
Am 05.01.2025 um 12:28 schrieb Alan Mackenzie:
None of the above extracts is about mathematics. If there were such
things as "potential" and "actual" infinity in maths, then they would
make a difference to some mathematical result. There would be some
theorem provable given the existence of PI and AI which would be false
or unprovable with just plain infinite, or vice versa. Or something
like that. Nobody in this discussion has so far attempted to cite
such a result.
When I did my maths degree, several decades ago, "potential
infinity" and "actual infinity" didn't get a look in. They weren't
mentioned a single time. Instead, precise definitions were given to
"finite" and "infinite", and we learnt how to use these definitions
and what could be done with them.
Sure. But this needs a context; usually (some sort of) set theory.
The only people who talk about "potential" and "actual" infinity are
non-mathematicians who lack understanding
Like those mentioned above?
OK, and mathematicians in their time off. ;-)
Oh, how about yin-yang ad-infinitum,
a simple and graphical example of
something that is, in the infinite,
not what it is, in the limit,
that it's actual infinite limit,
differs its potential infinite limit.
On 07.01.2025 12:36, Alan Mackenzie wrote:
If there were such
things as "potential" and "actual" infinity in maths,
Your comments about my quotes show that you have lost all contact with mathematics.
then they would make a difference to some mathematical result.
Of course. Here is a simple example, accessible to every student who is
not yet stultified by matheology.
For the inclusion-monotonic sequence of endsegments of natural numbers
E(k) = {k+1, k+2, k+3, ...} the intersection of all terms is empty. But
if every number k has infinitely many successors, as ZF claims, then the intersection is not empty.
Therefore set theory, claiming both, is false.
Inclusion monotonic sequences can only have an empty intersection if
they have an empty term.
Therefore the empty intersection of all requires the existence of
finite terms which must be dark.
Further there are not infinitely many infinite endsegments possible
because the indices of an actually infinite set of endsegements without
gaps must be all natural numbers.
Regards, WM
On 08.01.2025 00:50, Jim Burns wrote:
The cardinal:ordinal distinction
-- which does not matter in the finite domain
matters in the infinite domain.
The reason is that
the infinite cardinal ℵ₀ is based on
the mapping of
the potentially infinite collection of
natural numbers n,
all of which have
infinitely many successors.
The cardinal ℵ₀ is not based on
the mapping of
the actually infinite set ℕ where
ℕ \ {1, 2, 3, ...} = { }.
On 08.01.2025 12:04, FromTheRafters wrote:
WM formulated on Wednesday :
If ω exists, then ω-1 exists.
Wrong.
A set like ℕ has a fixed number of elements.
If ω-1 does not exist,
what is the fixed border of existence?
WM <wolfgang.mueckenheim@tha.de> wrote:
On 07.01.2025 12:36, Alan Mackenzie wrote:
If there were such
things as "potential" and "actual" infinity in maths,
Your comments about my quotes show that you have lost all contact with
mathematics.
then they would make a difference to some mathematical result.
Of course. Here is a simple example, accessible to every student who is
not yet stultified by matheology.
For the inclusion-monotonic sequence of endsegments of natural numbers
E(k) = {k+1, k+2, k+3, ...} the intersection of all terms is empty. But
if every number k has infinitely many successors, as ZF claims, then the
intersection is not empty.
That is false. The intersection of even just two infinite sets can be
empty.
As for the intersection of all endsegments of natural numbers, this is obviously empty.
Therefore set theory, claiming both, is false.
Set theory doesn't "claim" both. Set theory doesn't "claim" at all. It
has axioms and theorems derived from those axioms. If one accepts the axioms, and nearly all mathematicians do, then one is logically forced to accept the theorems, too.
Inclusion monotonic sequences can only have an empty intersection if
they have an empty term.
False. Where do you get such an idea from?
Such sequences have an empty
intersection if there is no element which is a member of each set in the sequence. This is trivially true for the sequence of endsegments of the natural numbers.
Therefore the empty intersection of all requires the existence of
finite terms which must be dark.
That isn't mathematics. Jim proved some while ago that there are no dark numbers, in as far as he could get a definition of them out of you.
Further there are not infinitely many infinite endsegments possible
because the indices of an actually infinite set of endsegements without
gaps must be all natural numbers.
That's meaningless gobbledegook.
On 1/8/2025 9:35 AM, WM wrote:
On 08.01.2025 12:04, FromTheRafters wrote:
WM formulated on Wednesday :
If ω exists, then ω-1 exists.
Wrong.
A set like ℕ has a fixed number of elements.
Yes.
Our sets do not change.
Our set ℕ does not change.
If ω-1 does not exist,
what is the fixed border of existence?
Membership in ℕ is determined by ℕ.rule.compliance,
not by position relative to a border.element.
Each object complying with the ℕ.rule is in ℕ
Each object not.complying is not.in ℕ
Compliance and non-compliance do not change.
Membership does not change.
No element is the border.element
because
each element is smaller.than another, fuller element,
and so, not on the border.
Which elements are in ℕ doesn't change.
On 1/8/2025 4:16 AM, WM wrote:
On 08.01.2025 00:50, Jim Burns wrote:
The cardinal:ordinal distinction
-- which does not matter in the finite domain
matters in the infinite domain.
The reason is that
the infinite cardinal ℵ₀ is based on
the mapping of the potentially infinite collection of
natural numbers n,
all of which have
infinitely many successors.
The cardinal ℵ₀ is not based on
the mapping of
the actually infinite set ℕ where
ℕ \ {1, 2, 3, ...} = { }.
For each set smaller.than a fuller.by.one set,
the cardinal:ordinal distinction doesn't matter.
Cardinals and ordinals always go together.
For each set smaller.than a fuller.by.one set
there is an ordinal of its size in
the set ℕ of all finite ordinals.
Each set for which
there is NOT an ordinal of its size in
the set ℕ of all finite ordinals
is NOT a set smaller.than a fuller.by.one set.
WM explained :
On 08.01.2025 12:04, FromTheRafters wrote:
WM formulated on Wednesday :
If ω exists, then ω-1 exists.
Wrong.
A set like ℕ has a fixed number of elements. If ω-1 does not exist,
what is the fixed border of existence?
Who says that there has to be a fixed border of existence?
Omega is a
limit ordinal not a successor.
On 1/8/2025 6:35 AM, WM wrote:
ℕ has a fixed number of elements. If ω-1 does not exist,
what is the fixed border of existence?
On 1/8/2025 6:35 AM, WM wrote:
ℕ has a fixed number of elements. If ω-1 does not exist,
what is the fixed border of existence?
On 08.01.2025 14:31, Jim Burns wrote:Because there is no "last".
⦃k: k < ω ≤ k+1⦄ = ⦃⦄Let us accept this result.
ω-1 does not exist.
Then the sequence of endsegments loses every natnumber but not a last
one.
Then the empty intersection of infinite but inclusion monotonicNo, this isn't even the case. The infinite(!) intersection is empty.
endsegments is violating basic logic. (Losing all numbers but keeping infinitely many can only be possible if new numbers are acquired.)
On 08.01.2025 12:04, FromTheRafters wrote:It has an infinite number of elements, and that number happens to be
WM formulated on Wednesday :
A set like ℕ has a fixed number of elements. If ω-1 does not exist, what is the fixed border of existence?If ω exists, then ω-1 exists.Wrong.
On 08.01.2025 20:19, Jim Burns wrote:You can't talk about size without using |abs|. Yes, it is a subset.
On 1/8/2025 4:16 AM, WM wrote:
On 08.01.2025 00:50, Jim Burns wrote:
The cardinal:ordinal distinction -- which does not matter in the
finite domain matters in the infinite domain.
The reason is that the infinite cardinal ℵ₀ is based on the mapping of >>> the potentially infinite collection of natural numbers n,
all of which have infinitely many successors.
The cardinal ℵ₀ is not based on the mapping of the actually infinite >>> set ℕ where ℕ \ {1, 2, 3, ...} = { }.
For each set smaller.than a fuller.by.one set, the cardinal:ordinal
distinction doesn't matter.
Cardinals and ordinals always go together.
For each set smaller.than a fuller.by.one set there is an ordinal of
its size in the set ℕ of all finite ordinals.
Each set for which there is NOT an ordinal of its size in the set ℕ of
all finite ordinals is NOT a set smaller.than a fuller.by.one set.
The set {1, 2, 3, ...} is smaller by one element than the set {0, 1, 2,
3, ...}. Proof: {0, 1, 2, 3, ...} \ {1, 2, 3, ...} = {0}. Cardinality
cannot describe this difference because it covers only mappings of
elements which have almost all elements as successors.
On 08.01.2025 21:07, Jim Burns wrote:No, |N| is not a member of itself and does not behave like it.
On 1/8/2025 9:35 AM, WM wrote:|ℕ| is invariable. |ℕ| = |ℕ|/2 is wrong.
On 08.01.2025 12:04, FromTheRafters wrote:Yes.
WM formulated on Wednesday :
A set like ℕ has a fixed number of elements.If ω exists, then ω-1 exists.Wrong.
They don't cease. They simply aren't in the same league, if you will.If ω-1 does not exist,
what is the fixed border of existence?
Membership in ℕ is determined by ℕ.rule.compliance,
not by position relative to a border.element.
Each object complying with the ℕ.rule is in ℕ Each object not.complying >> is not.in ℕ
The rule is for n there is n+1. But the successor is not created but
does exist. How far do successors reach? Why do they not reach to ω-1?
Where do they cease before?
Exactly omega is the "border", but this suggests a wrong mental image,Compliance and non-compliance do not change. Membership does notThen tell me if n = 7 exists and n = ω-1 does not exist where the border lies.
change.
No, it is a finite expression of an infinity.No element is the border.element because each element is smaller.than"And so on" can only happen, when the elements are created. Potential infinity.
another, fuller element, and so, not on the border.
--Which elements are in ℕ doesn't change.
On 08.01.2025 16:23, Alan Mackenzie wrote:Only valid for finite sets.
WM <wolfgang.mueckenheim@tha.de> wrote:
On 07.01.2025 12:36, Alan Mackenzie wrote:
Of course. But do you know what inclusion-monotony means? E(n+1) is aThat is false. The intersection of even just two infinite sets can bethen they would make a difference to some mathematical result.For the inclusion-monotonic sequence of endsegments of natural numbers
E(k) = {k+1, k+2, k+3, ...} the intersection of all terms is empty.
But if every number k has infinitely many successors, as ZF claims,
then the intersection is not empty.
empty.
proper subset of E(n): {n+2, n+3, n+4, ...} c {n+1, n+2, n+3, n+4, ...}.
Here the intersection cannot be empty unless there is an empty
endsegment.
And nobody said otherwise, since there are infinitely many segments.As for the intersection of all endsegments of natural numbers, this isBut for all definable endsegments the intersection is infinite, and from endsegmnet to endsegment only one number is lost, never more!
obviously empty.
the general law of mathematics ∀k ∈ ℕ : E(k+1) = E(k) \ {k+1}On the contrary: for every (finite!) k, E(k) is not empty.
or ∀k ∈ ℕ : ∩{E(0), E(1), E(2), ..., E(k+1)} = ∩{E(0), E(1), E(2), ...,
E(k)} \ {k+1}
proves that the empty intersection requires finite intersections
preceding it.
Unless you claim that the general law does not hold for ∀k ∈ ℕ.It does not hold for the infinite intersection.
No such thing.But the theorems contradict the general law of mathematics.Therefore set theory, claiming both, is false.Set theory doesn't "claim" both. Set theory doesn't "claim" at all.
It has axioms and theorems derived from those axioms. If one accepts
the axioms, and nearly all mathematicians do, then one is logically
forced to accept the theorems, too.
Which noone contradicted.It is trivially true that only one element can vanish with eachInclusion monotonic sequences can only have an empty intersection ifFalse.
they have an empty term.
Such sequences have an empty
intersection if there is no element which is a member of each set in
the sequence. This is trivially true for the sequence of endsegments
of the natural numbers.
endsegment.
Wrong. The limit of the harmonic series is zero, even though none of theJim "proved" that when exchanging two elements O and X, one of them can disappear. His "proofs" violate logic which says that lossless exchangeTherefore the empty intersection of all requires the existence ofThat isn't mathematics. Jim proved some while ago that there are no
finite terms which must be dark.
dark numbers, in as far as he could get a definition of them out of
you.
will never suffer losses.
Why should it?That's a simple fact. The sequence of natural numbers 1, 2, 3, ..., n,Further there are not infinitely many infinite endsegments possibleThat's meaningless gobbledegook.
because the indices of an actually infinite set of endsegements
without gaps must be all natural numbers.
n+1, ...
cannot be cut into two actually infinite sequences, namely indices and contents.
When all contents is appearing as an infinite sequence of indices thenYes, there are no more numbers after the naturals???
no number can remain in the contents.
On 1/8/2025 3:20 PM, Moebius wrote:
Am 08.01.2025 um 23:31 schrieb Chris M. Thomasson:
On 1/8/2025 6:35 AM, WM wrote:
ℕ has a fixed number of elements. If ω-1 does not exist, what is the >>>> fixed border of existence?
What is the (fixed) "border of existence"?
Afaict, WM thinks that way because he is hell bent on saying infinity
does not exist in any form, any where... ? ;^o
Mücke, Du bist reif für die Klapse.
.
.
.
On 1/8/2025 3:28 PM, Moebius wrote:
Am 08.01.2025 um 23:31 schrieb Chris M. Thomasson:
On 1/8/2025 6:35 AM, WM wrote:
ℕ has a fixed number of elements. If ω-1 does not exist, what is the >>>> fixed border of existence?
What is the (fixed) "border of existence"?
One of his hyper finite thoughts? According to him, 1+2 does not exist
unless he visually and/or mentally thought of the dark number 3? Humm...
Mücke, Du bist reif für die Klapse.
.
.
.
On 1/8/2025 3:20 PM, Moebius wrote:
Am 08.01.2025 um 23:31 schrieb Chris M. Thomasson:
On 1/8/2025 6:35 AM, WM wrote:
ℕ has a fixed number of elements. If ω-1 does not exist, what is the >>>> fixed border of existence?
What is the (fixed) "border of existence"?
Afaict, WM thinks that way because he is hell bent on saying infinity
does not exist in any form, any where... ? ;^o
Mücke, Du bist reif für die Klapse.
.
.
.
On 08.01.2025 20:19, Jim Burns wrote:
On 1/8/2025 4:16 AM, WM wrote:
On 08.01.2025 00:50, Jim Burns wrote:
The cardinal:ordinal distinction
-- which does not matter in the finite domain
matters in the infinite domain.
The reason is that
the infinite cardinal ℵ₀ is based on
the mapping of the potentially infinite collection of
natural numbers n,
all of which have
infinitely many successors.
The cardinal ℵ₀ is not based on
the mapping of
the actually infinite set ℕ where
ℕ \ {1, 2, 3, ...} = { }.
For each set smaller.than a fuller.by.one set,
the cardinal:ordinal distinction doesn't matter.
Cardinals and ordinals always go together.
For each set smaller.than a fuller.by.one set
there is an ordinal of its size in
the set ℕ of all finite ordinals.
Each set for which
there is NOT an ordinal of its size in
the set ℕ of all finite ordinals
is NOT a set smaller.than a fuller.by.one set.
The set {1, 2, 3, ...} is smaller by one element than the set {0, 1, 2,
3, ...}. Proof: {0, 1, 2, 3, ...} \ {1, 2, 3, ...} = {0}. Cardinality
cannot describe this difference because it covers only mappings of
elements which have almost all elements as successors.
Regards, WM
On 1/8/2025 4:25 AM, Richard Damon wrote:
On 1/8/25 4:04 AM, WM wrote:
On 08.01.2025 00:30, Richard Damon wrote:
On 1/7/25 7:51 AM, WM wrote:
No, ZF doesn't have as an axiom that the set of Natural Numbers
exist.
AoI: There exists an infinite set S.
Which isn't that the NATURAL NUMBERS are an infinite set.
The infinite set has been designed by Zermelo according to Dedekind's
definition of the natural numbers, as Zermelo noted. https://
gdz.sub.uni-goettingen.de/id/PPN235181684_0065?
tify=%7B%22pages%22%3A%5B276%5D%2C%22pan%22%3A%7B%22x%22%3A0.461%2C%22y%22%3A1.103%7D%2C%22view%22%3A%22info%22%2C%22zoom%22%3A0.884%7D
So? it doesn't mean that ZF has made it an axiom that the set of
Natural Numbers exist, he has made his Axiom of Infinity to be
designed so that the existance of the Natural Numbers can be derived
from it. You confuse cause from effect.
IT is good to know where you are trying to go, or it can be hard to
get there.
IF you claim that the axiom of infinity is NOT valid, then why do you
keep on using the results of it in your logic? One of your problems is
you have ADMITTED that you "logic" isn't axiomized (since you admit
you can't provide a set of actual axioms to define it) and thus you
admit that your "logic" isn't actually LOGIC. Your "Theorem" can't be
actually a Theorem, as you don't have any axioms on which to prove it.
Every union of FISONs {1, 2, 3, ..., n} which stay below this
threshold stays below this threshold too.
But not the union of *EVERY* FISON, the FULL INFINITE set of them.
All are below 1 %.
No,
Show one FISON that is larger than 1 %.
There isn't one, but that doesn't matter,
Show me a Natural Number that is bigger than Aleph_0 / 100?
It doesn't exist, because Aleph_0 is infinite, and an infinite number
divided by ANY finite value is still that infinite value, and thus
there is no finite value greater than that.
This doesn't give you your "dark numbers" as "non-defined finite
numbers", but shows that your logic is just broken.
[...]
WM must think that Aleph_0 is some really big natural number.
;^)
Am Wed, 08 Jan 2025 15:31:13 +0100 schrieb WM:
(Losing all numbers but keepingNo, this isn't even the case. The infinite(!) intersection is empty.
infinitely many can only be possible if new numbers are acquired.)
Am Wed, 08 Jan 2025 15:35:44 +0100 schrieb WM:
A set like ℕ has a fixed number of elements. If ω-1 does not exist, what >> is the fixed border of existence?It has an infinite number of elements, and that number happens to be invariant under finite subtraction/addition.
Am Wed, 08 Jan 2025 23:06:27 +0100 schrieb WM:
The set {1, 2, 3, ...} is smaller by one element than the set {0, 1, 2,You can't talk about size without using |abs|.
3, ...}. Proof: {0, 1, 2, 3, ...} \ {1, 2, 3, ...} = {0}. Cardinality
cannot describe this difference because it covers only mappings of
elements which have almost all elements as successors.
Am Wed, 08 Jan 2025 22:45:26 +0100 schrieb WM:
On 08.01.2025 16:23, Alan Mackenzie wrote:
Of course. But do you know what inclusion-monotony means? E(n+1) is aOnly valid for finite sets.
proper subset of E(n): {n+2, n+3, n+4, ...} c {n+1, n+2, n+3, n+4, ...}.
Here the intersection cannot be empty unless there is an empty
endsegment.
And nobody said otherwise, since there are infinitely many segments.As for the intersection of all endsegments of natural numbers, this isBut for all definable endsegments the intersection is infinite, and from
obviously empty.
endsegmenet to endsegment only one number is lost, never more!
Unless you claim that the general law does not hold for ∀k ∈ ℕ.It does not hold for the infinite intersection.
Inclusion monotonic sequences can only have an empty intersection ifFalse.
they have an empty term.
It is trivially true that only one element can vanish with eachWhich noone contradicted.
endsegment.
Jim "proved" that when exchanging two elements O and X, one of them canWrong. The limit of the harmonic series is zero, even though none of the terms are.
disappear. His "proofs" violate logic which says that lossless exchange
will never suffer losses.
That's a simple fact. The sequence of natural numbers 1, 2, 3, ..., n,Why should it?
n+1, ...
cannot be cut into two actually infinite sequences, namely indices and
contents.
When all contents is appearing as an infinite sequence of indices thenYes, there are no more numbers after the naturals???
no number can remain in the contents.
Am Wed, 08 Jan 2025 22:57:52 +0100 schrieb WM:
The rule is for n there is n+1. But the successor is not created butThey don't cease. They simply aren't in the same league, if you will.
does exist. How far do successors reach? Why do they not reach to ω-1?
Where do they cease before?
On 09.01.2025 01:07, joes wrote:
Am Wed, 08 Jan 2025 22:57:52 +0100 schrieb WM:
The rule is for n there is n+1. But the successor is not created butThey don't cease. They simply aren't in the same league, if you will.
does exist. How far do successors reach? Why do they not reach to ω-1?
Where do they cease before?
Cantor will. Every set of numbers of the first and second number class
has a smallest element. Hence they all are on the ordinal line.
Regards, WM
On 09.01.2025 00:45, joes wrote:
Am Wed, 08 Jan 2025 23:06:27 +0100 schrieb WM:
The set {1, 2, 3, ...} is smaller by one element than the set {0, 1, 2,You can't talk about size without using |abs|.
3, ...}. Proof: {0, 1, 2, 3, ...} \ {1, 2, 3, ...} = {0}. Cardinality
cannot describe this difference because it covers only mappings of
elements which have almost all elements as successors.
I can and I do. And everybody understands it in case of subsets. This
proves, in this special case (and more is not required), that Cantor's
size is only a qualitative measure, not a quantitative one.
Regards, WM
On 1/8/25 5:06 PM, WM wrote:
The set {1, 2, 3, ...} is smaller by one element than the set {0, 1,But Alelph_0, the size of the second, is also the size of the first, as Aleph_0 - 1 is Aleph_0.
2, 3, ...}. Proof: {0, 1, 2, 3, ...} \ {1, 2, 3, ...} = {0}.
Cardinality cannot describe this difference because it covers only
mappings of elements which have almost all elements as successors.
WM wrote :
On 09.01.2025 01:07, joes wrote:
Am Wed, 08 Jan 2025 22:57:52 +0100 schrieb WM:
The rule is for n there is n+1. But the successor is not created butThey don't cease. They simply aren't in the same league, if you will.
does exist. How far do successors reach? Why do they not reach to ω-1? >>>> Where do they cease before?
Cantor will. Every set of numbers of the first and second number class
has a smallest element. Hence they all are on the ordinal line.
Zero is the smallest in the natural number class, omega is the smallest
of the infinite number class. Neither has a predecessor in its class.
On 1/8/25 9:40 AM, WM wrote:
On 08.01.2025 13:25, Richard Damon wrote:
On 1/8/25 4:06 AM, WM wrote:
On 08.01.2025 00:30, Richard Damon wrote:There isn't one, but doesn't need to be.
Except you don't use *EVERY* FISON, only those below some limit n
that is below a faction of infinity.
Show a FISON that expanded by a factor of 100 or more covers ℕ. Fail! >>>>
Read your sentence above. Think over its meaning.
WHat is wrong with it.
On 1/8/25 9:31 AM, WM wrote:
On 08.01.2025 14:31, Jim Burns wrote:
⦃k: k < ω ≤ k+1⦄ = ⦃⦄Let us accept this result.
ω-1 does not exist.
Then the sequence of endsegments loses every natnumber but not a last
one. Then the empty intersection of infinite but inclusion monotonic
endsegments is violating basic logic. (Losing all numbers but keeping
infinitely many can only be possible if new numbers are acquired.)
Then the only possible way to satisfy logic is the non-existence of ω
and of endsegments as complete sets.
It is useless to prove your claim as long as you cannot solve this
problem.
There isn't a "last one" to lose
on 1/8/2025, WM supposed :
On 08.01.2025 18:32, FromTheRafters wrote:
WM explained :
On 08.01.2025 12:04, FromTheRafters wrote:
WM formulated on Wednesday :
If ω exists, then ω-1 exists.
Wrong.
A set like ℕ has a fixed number of elements. If ω-1 does not exist, >>>> what is the fixed border of existence?
Who says that there has to be a fixed border of existence?
According to set theory every set has a fixed set of elements, not
more and not less.
This applies to only finite sets.
Omega is a limit ordinal not a successor.
But the natural numbers are invariable. For every n, there is n+1
which is not created but simply exists.
Great for every n which are elements of N but omega is not an element of
N. In the naturals, n+1 is usually axiomatic and n-1 is a theorem. In
the infinite ordinals omega+1 is axiomatic and omega-1 is undefined or
not existent.
WM explained :
The set {1, 2, 3, ...} is smaller by one element than the set {0, 1,
2, 3, ...}.
Both sets are equal in size
On 1/9/25 4:38 AM, WM wrote:
On 09.01.2025 00:45, joes wrote:
Am Wed, 08 Jan 2025 23:06:27 +0100 schrieb WM:
The set {1, 2, 3, ...} is smaller by one element than the set {0, 1, 2, >>>> 3, ...}. Proof: {0, 1, 2, 3, ...} \ {1, 2, 3, ...} = {0}. CardinalityYou can't talk about size without using |abs|.
cannot describe this difference because it covers only mappings of
elements which have almost all elements as successors.
I can and I do. And everybody understands it in case of subsets. This
proves, in this special case (and more is not required), that Cantor's
size is only a qualitative measure, not a quantitative one.
Sorry it *IS* true,
On 1/9/25 6:39 AM, WM wrote:
On 09.01.2025 01:07, joes wrote:
Am Wed, 08 Jan 2025 22:57:52 +0100 schrieb WM:
The rule is for n there is n+1. But the successor is not created butThey don't cease. They simply aren't in the same league, if you will.
does exist. How far do successors reach? Why do they not reach to ω-1? >>>> Where do they cease before?
Cantor will. Every set of numbers of the first and second number class
has a smallest element. Hence they all are on the ordinal line.
Which doesn't prove your claim,
WM wrote on 1/9/2025 :
On 09.01.2025 13:27, FromTheRafters wrote:
WM wrote :
On 09.01.2025 01:07, joes wrote:
Am Wed, 08 Jan 2025 22:57:52 +0100 schrieb WM:Cantor will. Every set of numbers of the first and second number
The rule is for n there is n+1. But the successor is not created but >>>>>> does exist. How far do successors reach? Why do they not reach toThey don't cease. They simply aren't in the same league, if you will. >>>>
ω-1?
Where do they cease before?
class has a smallest element. Hence they all are on the ordinal line.
Zero is the smallest in the natural number class, omega is the
smallest of the infinite number class. Neither has a predecessor in
its class.
Are the natural numbers fixed or do they evolve?
Neither
they are the smallest infinite set.
On 08.01.2025 14:31, Jim Burns wrote:
⦃k: k < ω ≤ k+1⦄ = ⦃⦄
ω-1 does not exist.
Let us accept this result.
Then
the sequence of endsegments
loses every natnumber but
not a last one.
Then
the empty intersection of
infinite but
inclusion monotonic endsegments
is violating basic logic.
(Losing all numbers but
keeping infinitely many
can only be possible if
new numbers are acquired.)
Then
the only possible way
to satisfy logic is
the non-existence of ω and
of endsegments as complete sets.
(Losing all numbers but
keeping infinitely many
can only be possible if
new numbers are acquired.)
It is useless to prove your claim
as long as you cannot solve this problem.
(Losing all numbers but
keeping infinitely many
can only be possible if
new numbers are acquired.)
On 1/8/2025 9:31 AM, WM wrote:
On 08.01.2025 14:31, Jim Burns wrote:
⦃k: k < ω ≤ k+1⦄ = ⦃⦄
ω-1 does not exist.
Let us accept this result.
Then
the sequence of endsegments
loses every natnumber but
not a last one.
Then
the empty intersection of
infinite but
inclusion monotonic endsegments
is violating basic logic.
(Losing all numbers but
keeping infinitely many
can only be possible if
new numbers are acquired.)
Then the only possible way
to satisfy logic is
the non-existence of ω and
of endsegments as complete sets.
(Losing all numbers but
keeping infinitely many
can only be possible if
new numbers are acquired.)
No.
Sets do not change.
Not all sets are finite.
⎛ By 'finite', I mean
⎝ 'smaller.than fuller.by.one sets'
It is useless to prove your claim
as long as you cannot solve this problem.
(Losing all numbers but
keeping infinitely many
can only be possible if
new numbers are acquired.)
No.
Sets emptier.by.one than ℕ are not smaller.
In the sequence of end.segments of ℕ
there is no number which
empties an infinite set to a finite set.
and
there is no number which
is in common with all its end.segments.
ℕ has only infinite end.segments.
The intersection of
all (infinite) end.segments of ℕ
is empty.
Sets do not change.
Not all sets are finite
On 09.01.2025 13:18, Richard Damon wrote:
On 1/9/25 6:39 AM, WM wrote:
Cantor will.
Every set of numbers of
the first and second number class
has a smallest element.
Hence they all are on the ordinal line.
Which doesn't prove your claim,
It proves that the numbers of
the first and second number class
form a linear system.
On 09.01.2025 18:52, Jim Burns wrote:
On 1/8/2025 9:31 AM, WM wrote:
(Losing all numbers but
keeping infinitely many
can only be possible if
new numbers are acquired.)
No.
Losing all numbers but keeping infinitely many
is impossible in inclusion-monotonic sequences.
Sets do not change.
But the terms (E(n))
differ from their successors by one number.
Not all sets are finite.
⎛ By 'finite', I mean
⎝ 'smaller.than fuller.by.one sets'
Spare your gobbledegook.
Finite means like a natural number.
On 1/9/2025 11:51 AM, WM wrote:
It proves that the numbers of
the first and second number class
form a linear system.
Apparently,
'linear' is yet.another term which
means something other than
what you (WM) want it to mean.
Apparently,
what you (WM) think you have concluded is that
the ordinal.line has the Archimedeanⁿᵒᵗᐧᵂᴹ property
The ordinal.line is linearⁿᵒᵗᐧᵂᴹ and
its linearityⁿᵒᵗᐧᵂᴹ does follow from its well.order,
but that's not the Archimedeanⁿᵒᵗᐧᵂᴹ property.
⎛ Because ⦃α,β⦄ holds a first ordinal,
⎜ α≠β ⇒ α<β ∨ α>β
⎜ and '<' is connected
⎜
⎜ Because ⦃α,β,γ⦄ holds a first ordinal,
⎜ α<β ∧ β<γ ⇒ α<γ
⎜ and '<' is transitive.
⎜
⎜ By definition, ¬(α<α)
⎜ and '<' is irreflexive.
⎜
⎜ A connected, transitive, irreflexive order
⎝ is a linear order.
On 09.01.2025 17:11, FromTheRafters wrote:They are neither finite in number nor do they "come into being".
WM wrote on 1/9/2025 :There is no third alternative.
On 09.01.2025 13:27, FromTheRafters wrote:Neither
WM wrote :Are the natural numbers fixed or do they evolve?
On 09.01.2025 01:07, joes wrote:Zero is the smallest in the natural number class, omega is the
Am Wed, 08 Jan 2025 22:57:52 +0100 schrieb WM:Cantor will. Every set of numbers of the first and second number
The rule is for n there is n+1. But the successor is not created >>>>>>> but does exist. How far do successors reach? Why do they not reach >>>>>>> to ω-1?They don't cease. They simply aren't in the same league, if you
Where do they cease before?
will.
class has a smallest element. Hence they all are on the ordinal
line.
smallest of the infinite number class. Neither has a predecessor in
its class.
Any special reason for that figure?they are the smallest infinite set.The set of prime numbers is infinite but smaller because it is a proper subset. It has less than 1 % content.
On 09.01.2025 13:18, Richard Damon wrote:Is the second class all infinite(!) ordinals >= omega but less than
On 1/9/25 6:39 AM, WM wrote:It proves that the numbers of the first and second number class form a
On 09.01.2025 01:07, joes wrote:Which doesn't prove your claim,
Am Wed, 08 Jan 2025 22:57:52 +0100 schrieb WM:Cantor will. Every set of numbers of the first and second number class
The rule is for n there is n+1. But the successor is not created but >>>>> does exist. How far do successors reach? Why do they not reach toThey don't cease. They simply aren't in the same league, if you will.
ω-1?
Where do they cease before?
has a smallest element. Hence they all are on the ordinal line.
linear system.
On 1/9/2025 1:25 PM, WM wrote:
On 09.01.2025 18:52, Jim Burns wrote:
On 1/8/2025 9:31 AM, WM wrote:
(Losing all numbers but
keeping infinitely many
can only be possible if
new numbers are acquired.)
No.
Losing all numbers but keeping infinitely many
is impossible in inclusion-monotonic sequences.
Sets do not change.
But the terms (E(n))
differ from their successors by one number.
Each end.segment is larger than
any ordinal smaller.than fuller.by.one sets.
ℕ is the set of such ordinals.
Finite means like a natural number.
...which doesn't define 'natural number'.
Do you (WM) disagree with
'finite' meaning
'smaller.than fuller.by.one sets'?
On 09.01.2025 18:52, Jim Burns wrote:This case doesn't occur.
On 1/8/2025 9:31 AM, WM wrote:Losing all numbers but keeping infinitely many is impossible in inclusion-monotonic sequences.
On 08.01.2025 14:31, Jim Burns wrote:No.
⦃k: k < ω ≤ k+1⦄ = ⦃⦄
ω-1 does not exist.
Let us accept this result.
Then the sequence of endsegments loses every natnumber but not a last
one.
Then the empty intersection of infinite but inclusion monotonic
endsegments is violating basic logic.
(Losing all numbers but keeping infinitely many can only be possible
if new numbers are acquired.)
Then the only possible way to satisfy logic is the non-existence of ω
and of endsegments as complete sets.
(Losing all numbers but keeping infinitely many can only be possible
if new numbers are acquired.)
Especially not of the same cardinality as n+1.Sets do not change.But the terms (E(n)) differ from their successors by one number.
Not all sets are finite.Spare your gobbledegook. Finite means like a natural number.
⎛ By 'finite', I mean ⎝ 'smaller.than fuller.by.one sets'
Much waffle deleted.Honest thanks for the note.
It is possible with infinite sets, which can't be reduced by a finiteIt is useless to prove your claim as long as you cannot solve this
problem.
Don't be silly.(Losing all numbers but keeping infinitely many can only be possibleNo.
if new numbers are acquired.)
No, they are subsets of the same cardinality. There is no contradiction.Sets emptier.by.one than ℕ are not smaller.They are. But that is irrelevant here. The sequence of endsegments loses
all numbers. If all endsegments remain infinite, we have a
contradiction.
No term of the sequence is empty, if you mean that.In the sequence of end.segments of ℕ there is no number which emptiesThen there cannot exist a sequence of endsegments obeying
an infinite set to a finite set.
∀k ∈ ℕ: E(k+1) = E(k) \ {k+1} for all k ∈ ℕ and getting empty.
WDYM "become"? There is no point at which all naturals would beand there is no number which is in common with all its end.segments.Therefore all numbers get lost from the content and become indices.
Huh? No. Then not all numbers would be "indices".ℕ has only infinite end.segments.Then it has only finitely many, because not all numbers get lost from
the content.
There is no such endsegment.The intersection of all (infinite) end.segments of ℕ is empty.What is the content if all elements of ℕ have become indices?
--Sets do not change.
On 09.01.2025 10:56, FromTheRafters wrote:Nobody said they were equal to each other.
WM explained :
No. Both sets appear equal (although everybody can see that they areThe set {1, 2, 3, ...} is smaller by one element than the set {0, 1,Both sets are equal in size
2, 3, ...}.
not) when measured by an insufficient tool.
On 09.01.2025 20:46, Jim Burns wrote:
Do you (WM) disagree with
'finite' meaning
'smaller.than fuller.by.one sets'?
That is also true for infinite sets.
On 09.01.2025 00:45, joes wrote:You have not defined any other concept of "size". How do, say,
Am Wed, 08 Jan 2025 23:06:27 +0100 schrieb WM:
I can and I do. And everybody understands it in case of subsets. ThisThe set {1, 2, 3, ...} is smaller by one element than the set {0, 1,You can't talk about size without using |abs|.
2,
3, ...}. Proof: {0, 1, 2, 3, ...} \ {1, 2, 3, ...} = {0}. Cardinality
cannot describe this difference because it covers only mappings of
elements which have almost all elements as successors.
proves, in this special case (and more is not required), that Cantor's
size is only a qualitative measure, not a quantitative one.
On 09.01.2025 01:00, joes wrote:The empty (inf.) intersection is not a segment.
Am Wed, 08 Jan 2025 22:45:26 +0100 schrieb WM:
On 08.01.2025 16:23, Alan Mackenzie wrote:
Why? What changes the basics? The intersection is only empty, when noOf course. But do you know what inclusion-monotony means? E(n+1) is aOnly valid for finite sets.
proper subset of E(n): {n+2, n+3, n+4, ...} c {n+1, n+2, n+3, n+4,
...}.
Here the intersection cannot be empty unless there is an empty
endsegment.
natural number remains in all endsegments.
If none is empty, then other numbers must be inside. Contradiction.No number is an element of all segments.
Only insofar as every number eventually "leaves" the endsegments.Infinitely many endsegments need infinitely many indices. Therefore no natural number must remain as content in the sequence of endsegments.And nobody said otherwise, since there are infinitely many segments.As for the intersection of all endsegments of natural numbers, thisBut for all definable endsegments the intersection is infinite, and
is obviously empty.
from endsegmenet to endsegment only one number is lost, never more!
Sorry, I lost track. Which law?Why not?Unless you claim that the general law does not hold for ∀k ∈ ℕ.It does not hold for the infinite intersection.
Does not follow.Simple logic. For an empty intersection, there must be infinitely many endsegments. That means no natural number can remain in the content. IfInclusion monotonic sequences can only have an empty intersection if >>>>> they have an empty term.False.
a number n remained, then it would impose an upper limit on the sequence
of indices.
Then the empty intersection is preceded by finite intersections.It is trivially true that only one element can vanish with eachWhich noone contradicted.
endsegment.
Same reason: the limit may have different properties than the terms.There is no exchange involved.Jim "proved" that when exchanging two elements O and X, one of themWrong. The limit of the harmonic series is zero, even though none of
can disappear. His "proofs" violate logic which says that lossless
exchange will never suffer losses.
the terms are.
But this never happens. There can be no infinite starting segment shortBecause an infinite sequence of indices followed by an infinite sequenceThat's a simple fact. The sequence of natural numbers 1, 2, 3, ..., n,Why should it?
n+1, ...
cannot be cut into two actually infinite sequences, namely indices and
contents.
of contentent would require two infinite sequences.
There are infinitely many naturals though. I fail to picture yourSo it is. The claim of infinitely many infinite endsegments is false.When all contents is appearing as an infinite sequence of indices thenYes, there are no more numbers after the naturals???
no number can remain in the contents.
On 09.01.2025 00:44, joes wrote:You have a faulty mental image.There is an infinite sequence of
Am Wed, 08 Jan 2025 15:31:13 +0100 schrieb WM:
Empty intersection means that all numbers of the contents have become(Losing all numbers but keeping infinitely many can only be possibleNo, this isn't even the case. The infinite(!) intersection is empty.
if new numbers are acquired.)
indices k of the endsegments E(k). What remains in the always infinite
E(k)?
On 09.01.2025 00:42, joes wrote:No, this is an entirely distinct concept and I don't even use those terms.
Am Wed, 08 Jan 2025 15:35:44 +0100 schrieb WM:
That is potential infinity, not actual infinity.A set like ℕ has a fixed number of elements. If ω-1 does not exist,It has an infinite number of elements, and that number happens to be
what is the fixed border of existence?
invariant under finite subtraction/addition.
Invariability under finite subtraction implies the impossibility to*in finitely many steps - it is infinite after all. You should recognise
empty the endsegments.
That implies the impossibility to extract allNo, you just need "extract/apply" infinitely many, which is more than
elements of contents in order to apply them as indices.
That destroys Cantor's approach. His sequences do not exist:What does this have to do with Aleph_0?
"thus we get the epitome (ω) of all real algebraic numbers [...] and
with respect to this order we can talk about the nth algebraic number
where not a single one of this epitome (ω) has been forgotten." [E.
Zermelo: "Georg Cantor – Gesammelte Abhandlungen mathematischen und philosophischen Inhalts", Springer, Berlin (1932) p. 116]
Am Thu, 09 Jan 2025 17:51:43 +0100 schrieb WM:
all ordinals have an order,
but omega still has no predecessor
Am Thu, 09 Jan 2025 19:25:19 +0100 schrieb WM:
Losing all numbers but keeping infinitely many is impossible inThis case doesn't occur.
inclusion-monotonic sequences.
If all endsegments remain infinite, we have aNo, they are subsets of the same cardinality. There is no contradiction.
contradiction.
No term of the sequence is empty, if you mean that.In the sequence of end.segments of ℕ there is no number which emptiesThen there cannot exist a sequence of endsegments obeying
an infinite set to a finite set.
∀k ∈ ℕ: E(k+1) = E(k) \ {k+1} for all k ∈ ℕ and getting empty.
WDYM "become"? There is no point at which all naturals would beand there is no number which is in common with all its end.segments.Therefore all numbers get lost from the content and become indices.
counted - N being infinite.
Huh? No. Then not all numbers would be "indices".ℕ has only infinite end.segments.Then it has only finitely many, because not all numbers get lost from
the content.
There is no such endsegment.The intersection of all (infinite) end.segments of ℕ is empty.What is the content if all elements of ℕ have become indices?
Am Thu, 09 Jan 2025 17:59:07 +0100 schrieb WM:
They are neither finite in numberThere is no third alternative.Are the natural numbers fixed or do they evolve?Neither
The set of prime numbers is infinite but smaller because it is a properAny special reason for that figure?
subset. It has less than 1 % content.
On 09.01.2025 20:46, Jim Burns wrote:
On 1/9/2025 1:25 PM, WM wrote:
On 09.01.2025 18:52, Jim Burns wrote:
Sets do not change.
But the terms (E(n))
differ from their successors by one number.
Each end.segment is larger than
any ordinal smaller.than fuller.by.one sets.
What makes it so?
Am Thu, 09 Jan 2025 17:18:16 +0100 schrieb WM:
On 09.01.2025 10:56, FromTheRafters wrote:Nobody said they were equal to each other.
WM explained :No. Both sets appear equal (although everybody can see that they are
The set {1, 2, 3, ...} is smaller by one element than the set {0, 1,Both sets are equal in size
2, 3, ...}.
not) when measured by an insufficient tool.
I wonder which cardinality you assign to the sets
{k^2 +2} and {k^2 +1}, k e N?
WM has brought this to us :
No, I simply know that all natural numbers exist and then comes ω.
No, after the natural numbers comes omega plus one. Omega *IS* the order
type of the natural numbers.
On 1/9/2025 3:23 PM, WM wrote:
On 09.01.2025 20:46, Jim Burns wrote:
Do you (WM) disagree with
'finite' meaning
'smaller.than fuller.by.one sets'?
That is also true for infinite sets.
Apparently it's true for your infiniteᵂᴹ sets.
However,
it's false for our infiniteⁿᵒᵗᐧᵂᴹ sets.
Whatever claim you (WM) make about infinite sets
is not a claim about _our_ infiniteⁿᵒᵗᐧᵂᴹ sets.
Am Thu, 09 Jan 2025 10:38:44 +0100 schrieb WM:
On 09.01.2025 00:45, joes wrote:You have not defined any other concept of "size".
Am Wed, 08 Jan 2025 23:06:27 +0100 schrieb WM:I can and I do. And everybody understands it in case of subsets. This
The set {1, 2, 3, ...} is smaller by one element than the set {0, 1,You can't talk about size without using |abs|.
2,
3, ...}. Proof: {0, 1, 2, 3, ...} \ {1, 2, 3, ...} = {0}. Cardinality
cannot describe this difference because it covers only mappings of
elements which have almost all elements as successors.
proves, in this special case (and more is not required), that Cantor's
size is only a qualitative measure, not a quantitative one.
Am Thu, 09 Jan 2025 11:54:27 +0100 schrieb WM:
If none is empty, then other numbers must be inside. Contradiction.No number is an element of all segments.
Infinitely many endsegments need infinitely many indices. Therefore noOnly insofar as every number eventually "leaves" the endsegments.
natural number must remain as content in the sequence of endsegments.
This however does not imply and empty endsegment,
since there inf. many
of both naturals and therefore endsegments.
Sorry, I lost track. Which law?Why not?Unless you claim that the general law does not hold for ∀k ∈ ℕ.It does not hold for the infinite intersection.
Does not follow.Then the empty intersection is preceded by finite intersections.It is trivially true that only one element can vanish with eachWhich noone contradicted.
endsegment.
Same reason: the limit may have different properties than the terms.There is no exchange involved.Jim "proved" that when exchanging two elements O and X, one of themWrong. The limit of the harmonic series is zero, even though none of
can disappear. His "proofs" violate logic which says that lossless
exchange will never suffer losses.
the terms are.
But this never happens.Because an infinite sequence of indices followed by an infinite sequenceThat's a simple fact. The sequence of natural numbers 1, 2, 3, ..., n, >>>> n+1, ...Why should it?
cannot be cut into two actually infinite sequences, namely indices and >>>> contents.
of contentent would require two infinite sequences.
The claim of infinitely many infinite endsegments is false.There are infinitely many naturals though.
Am Thu, 09 Jan 2025 10:34:42 +0100 schrieb WM:
There is an infinite sequence of
infinite segments.
On 1/9/2025 3:23 PM, WM wrote:
Thus, because ordinals are well.ordered,
Am Thu, 09 Jan 2025 10:30:25 +0100 schrieb WM:
On 09.01.2025 00:42, joes wrote:No, this is an entirely distinct concept and I don't even use those terms.
Am Wed, 08 Jan 2025 15:35:44 +0100 schrieb WM:That is potential infinity, not actual infinity.
A set like ℕ has a fixed number of elements. If ω-1 does not exist, >>>> what is the fixed border of existence?It has an infinite number of elements, and that number happens to be
invariant under finite subtraction/addition.
That implies the impossibility to extract allNo, you just need "extract/apply" infinitely many,
elements of contents in order to apply them as indices.
That destroys Cantor's approach. His sequences do not exist:What does this have to do with Aleph_0?
"thus we get the epitome (ω) of all real algebraic numbers [...] and
with respect to this order we can talk about the nth algebraic number
where not a single one of this epitome (ω) has been forgotten." [E.
Zermelo: "Georg Cantor – Gesammelte Abhandlungen mathematischen und
philosophischen Inhalts", Springer, Berlin (1932) p. 116]
In the context of classical mathematik, they don't "evolve".WM wrote :
Are the natural numbers fixed or do they evolve?
Am Wed, 08 Jan 2025 22:57:52 +0100 schrieb WM:
The rule is for n there is n+1. But the successor is not created but >>>>>> does exist. How far do successors reach? Why do they not reach to
ω-1?
On 1/9/2025 8:18 AM, WM wrote:
On 09.01.2025 10:56, FromTheRafters wrote:
WM explained :
The set {1, 2, 3, ...} is smaller by one element than the set {0, 1,
2, 3, ...}.
Both sets are equal in size
No. Both sets appear equal (although everybody can see that they are
not) when measured by an insufficient tool.
{ 0, 1, 2, 3, ... } = { 1 - 1, 2 - 1, 3 - 1, 4 - 1, ... }
On 1/9/2025 2:12 PM, WM wrote:
On 09.01.2025 21:46, FromTheRafters wrote:
WM has brought this to us :
No, I simply know that all natural numbers exist and then comes ω.
No, after the natural numbers comes omega plus one. Omega *IS* the
order type of the natural numbers.
ω is the first infinite ordinal number. It comes after all finite
ordinal numbers and can be considered as their limit. The order type
is another meaning of ω.
Nothing from that implies that there is a "largest natural number". Not
at all. :^)
On 1/9/2025 5:15 PM, Moebius wrote:
Am 09.01.2025 um 22:12 schrieb Chris M. Thomasson:Well, wrt [...]
On 1/9/2025 8:18 AM, WM wrote:
On 09.01.2025 10:56, FromTheRafters wrote:
WM explained :
The set {1, 2, 3, ...} is smaller by one element than the set {0,
1, 2, 3, ...}.
Both sets are equal in size
No. Both sets appear equal (although everybody can see that they are
not) when measured by an insufficient tool.
{ 0, 1, 2, 3, ... } = { 1 - 1, 2 - 1, 3 - 1, 4 - 1, ... }
Really? :-)
On 1/9/2025 5:15 PM, Moebius wrote:
Am 09.01.2025 um 22:12 schrieb Chris M. Thomasson:Well, wrt [...]
On 1/9/2025 8:18 AM, WM wrote:
On 09.01.2025 10:56, FromTheRafters wrote:
WM explained :
The set {1, 2, 3, ...} is smaller by one element than the set {0,
1, 2, 3, ...}.
Both sets are equal in size
No. Both sets appear equal (although everybody can see that they are
not) when measured by an insufficient tool.
{ 0, 1, 2, 3, ... } = { 1 - 1, 2 - 1, 3 - 1, 4 - 1, ... }
Really? :-)
On 09.01.2025 21:57, Jim Burns wrote:
On 1/9/2025 3:23 PM, WM wrote:
On 09.01.2025 20:46, Jim Burns wrote:
Do you (WM) disagree with
'finite' meaning
'smaller.than fuller.by.one sets'?
That is also true for infinite sets.
Apparently it's true for your infiniteᵂᴹ sets.
However,
it's false for our infiniteⁿᵒᵗᐧᵂᴹ sets.
It is false for potentially infinite sets.
Whatever claim you (WM) make about infinite sets
is not a claim about _our_ infiniteⁿᵒᵗᐧᵂᴹ sets.
I use actual infinity,
you use potential infinity,
best recognizable by the same cardinality of
almost all your sets.
Am 10.01.2025 um 02:19 schrieb Chris M. Thomasson:
On 1/9/2025 5:15 PM, Moebius wrote:
Am 09.01.2025 um 22:12 schrieb Chris M. Thomasson:
On 1/9/2025 8:18 AM, WM wrote:
On 09.01.2025 10:56, FromTheRafters wrote:
WM explained :
The set {1, 2, 3, ...} is smaller by one element than the set {0, >>>>>>> 1, 2, 3, ...}.
Both sets are equal in size
No. Both sets appear equal (although everybody can see that they
are not) when measured by an insufficient tool.
Am 10.01.2025 um 02:45 schrieb Moebius:
Am 10.01.2025 um 02:19 schrieb Chris M. Thomasson:
On 1/9/2025 5:15 PM, Moebius wrote:
Am 09.01.2025 um 22:12 schrieb Chris M. Thomasson:
On 1/9/2025 8:18 AM, WM wrote:
On 09.01.2025 10:56, FromTheRafters wrote:
WM explained :
The set {1, 2, 3, ...} is smaller by one element than the set
{0, 1, 2, 3, ...}.
Both sets are equal in size
No. Both sets appear equal (although everybody can see that they
are not) when measured by an insufficient tool.
Hint: WM here meant (of course): "Both sets appear equal IN SIZE ..."
.
.
.
Am 10.01.2025 um 02:48 schrieb Moebius:
Am 10.01.2025 um 02:45 schrieb Moebius:
Am 10.01.2025 um 02:19 schrieb Chris M. Thomasson:
On 1/9/2025 5:15 PM, Moebius wrote:
Am 09.01.2025 um 22:12 schrieb Chris M. Thomasson:
On 1/9/2025 8:18 AM, WM wrote:
On 09.01.2025 10:56, FromTheRafters wrote:
WM explained :
The set {1, 2, 3, ...} is smaller by one element than the set >>>>>>>>> {0, 1, 2, 3, ...}.
Both sets are equal in size
No. Both sets appear equal (although everybody can see that they >>>>>>> are not) when measured by an insufficient tool.
Hint: WM here meant (of course): "Both sets appear equal IN SIZE ..."
Hint@WM: The size of {1, 2, 3, ...} EQUALS the size of {0, 1, 2, 3, ...}
when "measured" by the "tool" /equivalence/.
See: https://www.britannica.com/science/set-theory/Equivalent-sets
____________________________________________________________________
Hint: Using Zermelo's definition of the natural numbers we have 1 = {0},
2 = {1}, 3 = {2}, 4 = {3}, ...
And hence {1, 2, 3, 4, ...} = {{0}, {1}, {2}, {3}, ...}
If we NOW compare
{{0}, {1}, {2}, {3}, ...} (= {1, 2, 3, 4, ...})
with
{ 0 , 1 , 2 , 3 , ...} ,
does ist STILL make sense to claim "everybody can see that they are not
equal in size"?
.
.
.
Am 10.01.2025 um 03:01 schrieb Moebius:
Am 10.01.2025 um 02:48 schrieb Moebius:
Am 10.01.2025 um 02:45 schrieb Moebius:
Am 10.01.2025 um 02:19 schrieb Chris M. Thomasson:
On 1/9/2025 5:15 PM, Moebius wrote:
Am 09.01.2025 um 22:12 schrieb Chris M. Thomasson:
On 1/9/2025 8:18 AM, WM wrote:
On 09.01.2025 10:56, FromTheRafters wrote:
WM explained :
The set {1, 2, 3, ...} is smaller by one element than the set >>>>>>>>>> {0, 1, 2, 3, ...}.
Both sets are equal in size
No. Both sets appear equal (although everybody can see that they >>>>>>>> are not) when measured by an insufficient tool.
Hint: WM here meant (of course): "Both sets appear equal IN SIZE ..."
Hint@WM: The size of {1, 2, 3, ...} EQUALS the size of {0, 1, 2,
3, ...} when "measured" by the "tool" /equivalence/.
See: https://www.britannica.com/science/set-theory/Equivalent-sets
____________________________________________________________________
Hint: Using Zermelo's definition of the natural numbers we have 1 =
{0}, 2 = {1}, 3 = {2}, 4 = {3}, ...
And hence {1, 2, 3, 4, ...} = {{0}, {1}, {2}, {3}, ...}
If we NOW compare
{{0}, {1}, {2}, {3}, ...} (= {1, 2, 3, 4, ...})
with
{ 0 , 1 , 2 , 3 , ...} ,
does ist STILL make sense to claim "everybody can see that they are
not equal in size"?
Again, referring to the sucessor operation s, we have
{1, 2, 3, 4, ...} = {s0, s1, s2, s3, ...} .
If we NOW compare
{s0, s1, s2, s3, ...} (= {1, 2, 3, 4, ...})
with
{ 0, 1, 2, 3, ...} ,
does ist STILL make sense to claim "everybody can see that they are not
equal in size"?
.
.
.
Am 10.01.2025 um 03:01 schrieb Moebius:
Am 10.01.2025 um 02:48 schrieb Moebius:
Am 10.01.2025 um 02:45 schrieb Moebius:
Am 10.01.2025 um 02:19 schrieb Chris M. Thomasson:
On 1/9/2025 5:15 PM, Moebius wrote:
Am 09.01.2025 um 22:12 schrieb Chris M. Thomasson:
On 1/9/2025 8:18 AM, WM wrote:
On 09.01.2025 10:56, FromTheRafters wrote:
WM explained :
The set {1, 2, 3, ...} is smaller by one element than the set >>>>>>>>>> {0, 1, 2, 3, ...}.
Both sets are equal in size
No. Both sets appear equal (although everybody can see that they >>>>>>>> are not) when measured by an insufficient tool.
Hint: WM here meant (of course): "Both sets appear equal IN SIZE ..."
Hint@WM: The size of {1, 2, 3, ...} EQUALS the size of {0, 1, 2,
3, ...} when "measured" by the "tool" /equivalence/.
See: https://www.britannica.com/science/set-theory/Equivalent-sets
____________________________________________________________________
Hint: Using Zermelo's definition of the natural numbers we have 1 =
{0}, 2 = {1}, 3 = {2}, 4 = {3}, ...
And hence {1, 2, 3, 4, ...} = {{0}, {1}, {2}, {3}, ...}
If we NOW compare
{{0}, {1}, {2}, {3}, ...} (= {1, 2, 3, 4, ...})
with
{ 0 , 1 , 2 , 3 , ...} ,
does ist STILL make sense to claim "everybody can see that they are
not equal in size"?
Again, referring to the sucessor operation s, we have
{1, 2, 3, 4, ...} = {s0, s1, s2, s3, ...} .
If we NOW compare
{s0, s1, s2, s3, ...} (= {1, 2, 3, 4, ...})
with
{ 0, 1, 2, 3, ...} ,
does ist STILL make sense to claim "everybody can see that they are not
equal in size"?
.
.
.
On 09.01.2025 22:03, joes wrote:Inside what? Inside each segment are infinitely many naturals. Mind
Am Thu, 09 Jan 2025 11:54:27 +0100 schrieb WM:
But what numbers are inside if all natural numbers are outside?If none is empty, then other numbers must be inside. Contradiction.No number is an element of all segments.
In what? The intersection is empty.But what remains?Infinitely many endsegments need infinitely many indices. Therefore noOnly insofar as every number eventually "leaves" the endsegments.
natural number must remain as content in the sequence of endsegments.
Nothing "remains". There is no end, only a limit.This however does not imply and empty endsegment,What remains?
Yes, in the limit.since there inf. many of both naturals and therefore endsegments.Infinitely many numbers leave. All elements of ℕ leave.
Well, in the limit (sigh) infinitely many numbers have been "lost".The law that the intersection gets empty but only by one element perSorry, I lost track. Which law?Why not?Unless you claim that the general law does not hold for ∀k ∈ ℕ. >>>> It does not hold for the infinite intersection.
term.
Care to write out that deduction?It follows from the law that only one element per term can leave.Does not follow.Then the empty intersection is preceded by finite intersections.It is trivially true that only one element can vanish with eachWhich noone contradicted.
endsegment.
It is an infinite "process".There is no limit involved when counting the fractions.Same reason: the limit may have different properties than the terms.There is no exchange involved.Jim "proved" that when exchanging two elements O and X, one of themWrong. The limit of the harmonic series is zero, even though none of
can disappear. His "proofs" violate logic which says that lossless
exchange will never suffer losses.
the terms are.
No, it doesn't. There is no "infinitieth" segment.An infinite set of infinite endsegments happens according to matheology.But this never happens.Because an infinite sequence of indices followed by an infiniteThat's a simple fact. The sequence of natural numbers 1, 2, 3, ...,Why should it?
n, n+1, ...
cannot be cut into two actually infinite sequences, namely indices
and contents.
sequence of contentent would require two infinite sequences.
It requires two infinite sequences.
Those are the same set N.But all must be in the set of indices, if there are infinitely many endsegments. But infinitely many must be in the set of contents, if all endsegments are infinite.The claim of infinitely many infinite endsegments is false.There are infinitely many naturals though.
On 09.01.2025 23:05, Jim Burns wrote:There is no "infinitieth" index with an empty content.
On 1/9/2025 3:23 PM, WM wrote:
Thus, because ordinals are well.ordered,an infinite set of endsegments requires all natural numbers as indices.
That means the content becomes/is empty if the set becomes/is infinite.
On 09.01.2025 21:29, joes wrote:Of course every set is bijective to itself.
Am Thu, 09 Jan 2025 17:18:16 +0100 schrieb WM:They are said to be in bijection.
On 09.01.2025 10:56, FromTheRafters wrote:Nobody said they were equal to each other.
WM explained :No. Both sets appear equal (although everybody can see that they are
The set {1, 2, 3, ...} is smaller by one element than the set {0, 1, >>>>> 2, 3, ...}.Both sets are equal in size
not) when measured by an insufficient tool.
Sorry, the sets for all k.I wonder which cardinality you assign to the sets {k^2 +2} and {k^2Cardinality is a useless tool. |{k^2 + 2}| = |{k^2 + 1}| = 1 for every
+1}, k e N?
k.
On 09.01.2025 21:19, joes wrote:Now this is gibberish. What do those terms mean? Are you referring
Am Thu, 09 Jan 2025 17:59:07 +0100 schrieb WM:
Gibberish.They are neither finite in numberThere is no third alternative.Are the natural numbers fixed or do they evolve?Neither
Are the natural numbers fixed or do they evolve?
And wrong. A good teacher should not pick an arbitrary numberIt is vivid and true - the characteristic feature of my lectures.The set of prime numbers is infinite but smaller because it is aAny special reason for that figure?
proper subset. It has less than 1 % content.
On 09.01.2025 21:17, joes wrote:...for different cases. There is no empty segment, each is infinite.
Am Thu, 09 Jan 2025 19:25:19 +0100 schrieb WM:
Loss of all numbers is proven by the empty intersection.Losing all numbers but keeping infinitely many is impossible inThis case doesn't occur.
inclusion-monotonic sequences.
Keeping infinitely many is poved by Fritsche.
You may have noticed that every segment is different.They remain infinite. But infinitely many endsegments require allIf all endsegments remain infinite, we have aNo, they are subsets of the same cardinality. There is no
contradiction.
contradiction.
natnumbers as indices. What makes up their infinite content?
Untrue. The sequence is, unfathomably, infinite.Then not all natnumbers are outside of content and inside of the set of indices.No term of the sequence is empty, if you mean that.In the sequence of end.segments of ℕ there is no number which empties >>>> an infinite set to a finite set.Then there cannot exist a sequence of endsegments obeying ∀k ∈ ℕ:
E(k+1) = E(k) \ {k+1} for all k ∈ ℕ and getting empty.
Contradiction. There are inf. many.The endsegment E(n) loses its element n+1 ad becomes E(n+1).WDYM "become"? There is no point at which all naturals would be countedand there is no number which is in common with all its endsegments.Therefore all numbers get lost from the content and become indices.
- N being infinite.
Then there are only finitely many indices.Huh? No. Then not all numbers would be "indices".ℕ has only infinite endsegments.Then it has only finitely many, because not all numbers get lost from
the content.
omega is not an element of N.What element of ℕ does not become an index?There is no such endsegment.The intersection of all (infinite) end.segments of ℕ is empty.What is the content if all elements of ℕ have become indices?
Again, referring to the sucessor operation s, we have
{1, 2, 3, 4, ...} = {s0, s1, s2, s3, ...} .
If we NOW compare
{s0, s1, s2, s3, ...} (= {1, 2, 3, 4, ...})
with
{ 0, 1, 2, 3, ...} ,
does ist STILL make sense to claim "everybody can see that they are not
equal in size"?
WM wrote :
In the context of classical mathematik, they don't "evolve".Are the natural numbers fixed or do they evolve?
Hint: The set of all natural numbers, IN, does not change.
Am Thu, 09 Jan 2025 23:27:21 +0100 schrieb WM:
Mind
your quantifiers: It is not required that the intersection be nonempty.
In what? The intersection is empty.But what remains?Infinitely many endsegments need infinitely many indices. Therefore no >>>> natural number must remain as content in the sequence of endsegments.Only insofar as every number eventually "leaves" the endsegments.
Nothing "remains". There is no end, only a limit.This however does not imply and empty endsegment,What remains?
Yes, in the limit.since there inf. many of both naturals and therefore endsegments.Infinitely many numbers leave. All elements of ℕ leave.
Am Thu, 09 Jan 2025 23:06:47 +0100 schrieb WM:
On 09.01.2025 21:29, joes wrote:Of course every set is bijective to itself.
Am Thu, 09 Jan 2025 17:18:16 +0100 schrieb WM:They are said to be in bijection.
On 09.01.2025 10:56, FromTheRafters wrote:Nobody said they were equal to each other.
WM explained :No. Both sets appear equal (although everybody can see that they are
The set {1, 2, 3, ...} is smaller by one element than the set {0, 1, >>>>>> 2, 3, ...}.Both sets are equal in size
not) when measured by an insufficient tool.
Am Thu, 09 Jan 2025 22:55:13 +0100 schrieb WM:
On 09.01.2025 21:17, joes wrote:...for different cases. There is no empty segment, each is infinite.
Am Thu, 09 Jan 2025 19:25:19 +0100 schrieb WM:Loss of all numbers is proven by the empty intersection.
Losing all numbers but keeping infinitely many is impossible inThis case doesn't occur.
inclusion-monotonic sequences.
Keeping infinitely many is poved by Fritsche.
What element of ℕ does not become an index?
Am Thu, 09 Jan 2025 23:41:40 +0100 schrieb WM:
On 09.01.2025 23:05, Jim Burns wrote:There is no "infinitieth" index with an empty content.
On 1/9/2025 3:23 PM, WM wrote:an infinite set of endsegments requires all natural numbers as indices.
Thus, because ordinals are well.ordered,
That means the content becomes/is empty if the set becomes/is infinite.
On 09.01.2025 13:17, Richard Damon wrote:
On 1/9/25 4:38 AM, WM wrote:
On 09.01.2025 00:45, joes wrote:
Am Wed, 08 Jan 2025 23:06:27 +0100 schrieb WM:
The set {1, 2, 3, ...} is smaller by one element than the set {0,You can't talk about size without using |abs|.
1, 2,
3, ...}. Proof: {0, 1, 2, 3, ...} \ {1, 2, 3, ...} = {0}. Cardinality >>>>> cannot describe this difference because it covers only mappings of
elements which have almost all elements as successors.
I can and I do. And everybody understands it in case of subsets. This
proves, in this special case (and more is not required), that
Cantor's size is only a qualitative measure, not a quantitative one.
Sorry it *IS* true,
It is true that {1, 2, 3, ...} is a set and {0, 1, 2, 3, ...} is a
greater set. Your hysteric moaning cannot change that.
Regards, WM
On 10.01.2025 10:15, joes wrote:Not true; the sequence is infinite.
Am Thu, 09 Jan 2025 22:55:13 +0100 schrieb WM:Without empty endsegment, not all numbers become indices.
On 09.01.2025 21:17, joes wrote:...for different cases. There is no empty segment, each is infinite.
Am Thu, 09 Jan 2025 19:25:19 +0100 schrieb WM:Loss of all numbers is proven by the empty intersection.
Losing all numbers but keeping infinitely many is impossible inThis case doesn't occur.
inclusion-monotonic sequences.
Keeping infinitely many is poved by Fritsche.
Note that bijections need all the indices. There is no limit accepted.An infinite bijection is not finite.
There are no such elements.Those which remain in all endsegments. If they did not remain, not all endsegments could be infinite, because of inclusion monotony.What element of ℕ does not become an index?
On 10.01.2025 02:25, Moebius wrote:There are no points without numbers.
So all natural numbers are fixed. Then for every point on the ordinalIn the context of classical mathematik, they don't "evolve".WM wrote :
Are the natural numbers fixed or do they evolve?
Hint: The set of all natural numbers, IN, does not change.
line it is determined whether there is a natural number. Although we
cannot determine it because most are dark.
On 1/9/25 11:48 AM, WM wrote:
It is true that {1, 2, 3, ...} is a set and {0, 1, 2, 3, ...} is a
greater set.
No, one may be the proper subset of the other, but it turns out that due
to the way that infinity works, they are both are the same size.
On 1/9/25 7:37 AM, WM wrote:
As I said, cardinality cannot describe this difference of one element.
Because the property that "cardinality" describes doesn't have that difference.
On 1/9/25 5:01 PM, WM wrote:
You don't know it. That does not prove its non-existence.
It has no predecessor,
Am Fri, 10 Jan 2025 11:38:49 +0100 schrieb WM:
On 10.01.2025 10:15, joes wrote:Not true; the sequence is infinite.
Am Thu, 09 Jan 2025 22:55:13 +0100 schrieb WM:Without empty endsegment, not all numbers become indices.
On 09.01.2025 21:17, joes wrote:...for different cases. There is no empty segment, each is infinite.
Am Thu, 09 Jan 2025 19:25:19 +0100 schrieb WM:Loss of all numbers is proven by the empty intersection.
Losing all numbers but keeping infinitely many is impossible inThis case doesn't occur.
inclusion-monotonic sequences.
Keeping infinitely many is poved by Fritsche.
Note that bijections need all the indices. There is no limit accepted.An infinite bijection is not finite.
Am Fri, 10 Jan 2025 10:52:46 +0100 schrieb WM:
There are no points without numbers.Hint: The set of all natural numbers, IN, does not change.So all natural numbers are fixed. Then for every point on the ordinal
line it is determined whether there is a natural number. Although we
cannot determine it because most are dark.
On 10.01.2025 14:09, joes wrote:Points don't even exist without the numbers. There is no number with
Am Fri, 10 Jan 2025 10:52:46 +0100 schrieb WM:
As I said. You can prove it when doubling all elements of the set {1, 2,There are no points without numbers.Hint: The set of all natural numbers, IN, does not change.So all natural numbers are fixed. Then for every point on the ordinal
line it is determined whether there is a natural number. Although we
cannot determine it because most are dark.
3, ..., ω}. The regular distance of next neighbours remains as a
conserved property in correct mathematics.
On 10.01.2025 13:41, Richard Damon wrote:Depends. The complement of a starting segment of N (wrt N) is infinite;
On 1/9/25 7:37 AM, WM wrote:
So it is. Cardinality cannot describe the difference of one element, notAs I said, cardinality cannot describe this difference of one element.Because the property that "cardinality" describes doesn't have that
difference.
even of infinitely many.
Infinite is simply infinite, no matter what the real size is.There is no "real size", there is only the set. Your expectations
You need not understand that. But the claim that different sets like ℕThey have the same cardinality, defined as being bijective.
and ℚ are of same size shows that you have been stultified by set
theory.
Am Fri, 10 Jan 2025 17:42:39 +0100 schrieb WM:
On 10.01.2025 14:09, joes wrote:
Am Fri, 10 Jan 2025 10:52:46 +0100 schrieb WM:As I said. You can prove it when doubling all elements of the set
There are no points without numbers.Hint: The set of all natural numbers, IN, does not change.So all natural numbers are fixed. Then for every point on the ordinal
line it is determined whether there is a natural number. Although we
cannot determine it because most are dark.
The regular distance of next neighbours remains as aPoints don't even exist without the numbers. There is no number with
conserved property in correct mathematics.
a finite distance from omega.
On 1/9/25 5:34 PM, WM wrote:
On 09.01.2025 22:15, joes wrote:
Am Thu, 09 Jan 2025 10:34:42 +0100 schrieb WM:
There is an infinite sequence of
infinite segments.
You cannot cut the set of natural numbers at any position to get two
infinite sets. Infinite sequence means no content. Infinite content
(content at all) means no infinite sequence.
Sure you can, you just need a special knife, like one that separates the
odds from the evens.
Am Fri, 10 Jan 2025 17:19:44 +0100 schrieb WM:
On 10.01.2025 13:41, Richard Damon wrote:
On 1/9/25 7:37 AM, WM wrote:
Infinite is simply infinite, no matter what the real size is.There is no "real size", there is only the set.
Your expectations
about cardinality do not match it.
You need not understand that. But the claim that different sets like ℕThey have the same cardinality, defined as being bijective.
and ℚ are of same size shows that you have been stultified by set
theory.
"Nubmers" or "Sets" don't evolve.
You seem to THINK that sets,
particularly "potentially infinite" set "evolve" in that numbers get
added to them as you move along the generator, but the set doesn't
change, only our knowledge of the set.
they are the smallest infinite set.
The set of prime numbers is infinite but smaller because it is a
proper subset. It has less than 1 % content.
It may SEEM smaller, but it turns out it is of the same countable
infinite cardinality.
On 10.01.2025 17:50, joes wrote:You mean N u {w}.
Am Fri, 10 Jan 2025 17:42:39 +0100 schrieb WM:{1, 2,3, ..., ω}. (*)
On 10.01.2025 14:09, joes wrote:
Am Fri, 10 Jan 2025 10:52:46 +0100 schrieb WM:As I said. You can prove it when doubling all elements of the set
There are no points without numbers.Hint: The set of all natural numbers, IN, does not change.So all natural numbers are fixed. Then for every point on the
ordinal line it is determined whether there is a natural number.
Although we cannot determine it because most are dark.
Why do you want to include omega? What new numbers? There are no evensSince all natural numbers existing below ω are multiplied by 2, when doubling the elements of (*), no further numbers below ω can be createdThe regular distance of next neighbours remains as a conservedPoints don't even exist without the numbers. There is no number with a
property in correct mathematics.
finite distance from omega.
- in actual infinity. What happens with the new 50 % of even numbers?
On 10.01.2025 18:01, joes wrote:No, the even numbers are twice the naturals.
Am Fri, 10 Jan 2025 17:19:44 +0100 schrieb WM:There is real size. The natural numbers have twice the size of the even numbers.
On 10.01.2025 13:41, Richard Damon wrote:There is no "real size", there is only the set.
On 1/9/25 7:37 AM, WM wrote:
Infinite is simply infinite, no matter what the real size is.
You wrongly expect this to hold in the infinite.Your expectations about cardinality do not match it.I have no expectations about cardinality. I know that for every finite initial segment the even numbers are about half of the natural numbers.
This does not change anywhere. It is true up to every natural number.
More are not available.
You are mistaken. Clearly |{1*2, 2*2, 3*2, ...}| = |N|Already the bijection between even and natural numbers has beenYou need not understand that. But the claim that different sets like ℕ >>> and ℚ are of same size shows that you have been stultified by setThey have the same cardinality, defined as being bijective.
theory.
disproved above.
On 10.01.2025 13:41, Richard Damon wrote:They either are members, or not.
"Nubmers" or "Sets" don't evolve.
Fine. Then the set of natural numbers is completed. Multiply all natural numbers by 2. The set of even numbers then doubles.Very wrong. The set {2*k for k e N} = G = 2N = {2*1, 2*2, 2*3, ...} =
The domain below ωNone are "created". The multiples of 4 are ALSO in the original set.
is unable to absorb new numbers. What happens to the newly created even numbers?
Then don't talk about "creation".You seem to THINK that sets,The multiplication above concerns the set, not only the numbers we know.
particularly "potentially infinite" set "evolve" in that numbers get
added to them as you move along the generator, but the set doesn't
change, only our knowledge of the set.
*sizes of the setsIt turns out that countable cardinality is not able to distinguish theIt may SEEM smaller, but it turns out it is of the same countablethey are the smallest infinite set.The set of prime numbers is infinite but smaller because it is a
proper subset. It has less than 1 % content.
infinite cardinality.
sets* of natural numbers and of even numbers.
But mathematics. Every setCould you formalise what you mean here?
{1, 2, 3, 4, 5, ..., n} contains roughly twice the even numbers.
This holds for all n. Hence it holds for the infinite set.No. The limit of the ratio is different from the ratio of the limit.
Am Fri, 10 Jan 2025 18:21:43 +0100 schrieb WM:
I have no expectations about cardinality. I know that for every finiteYou wrongly expect this to hold in the infinite.
initial segment the even numbers are about half of the natural numbers.
This does not change anywhere. It is true up to every natural number.
More are not available.
Am Fri, 10 Jan 2025 18:12:15 +0100 schrieb WM:
On 10.01.2025 17:50, joes wrote:
Am Fri, 10 Jan 2025 17:42:39 +0100 schrieb WM:
On 10.01.2025 14:09, joes wrote:
Am Fri, 10 Jan 2025 10:52:46 +0100 schrieb WM:
You mean N u {w}.{1, 2,3, ..., ω}. (*)As I said. You can prove it when doubling all elements of the set
Why do you want to include omega?Since all natural numbers existing below ω are multiplied by 2, whenThe regular distance of next neighbours remains as a conservedPoints don't even exist without the numbers. There is no number with a
property in correct mathematics.
finite distance from omega.
doubling the elements of (*), no further numbers below ω can be created
- in actual infinity. What happens with the new 50 % of even numbers?
What new numbers? There are no evens
above omega.
Am Fri, 10 Jan 2025 18:33:31 +0100 schrieb WM:
On 10.01.2025 13:41, Richard Damon wrote:They either are members, or not.
"Nubmers" or "Sets" don't evolve.
Fine. Then the set of natural numbers is completed. Multiply all naturalVery wrong. The set {2*k for k e N} = G = 2N = {2*1, 2*2, 2*3, ...} =
numbers by 2. The set of even numbers then doubles.
{2, 4, 6, ...} is a proper subset of N.
I have no idea what you think.
The domain below ωNone are "created". The multiples of 4 are ALSO in the original set.
is unable to absorb new numbers. What happens to the newly created even
numbers?
{1, 2, 3, 4, 5, ..., n} contains roughly twice the even numbers.Could you formalise what you mean here?
This holds for all n. Hence it holds for the infinite set.No. The limit of the ratio is different from the ratio of the limit.
The limit of the ratio is different from the ratio of the limit.
On 10.01.2025 19:28, joes wrote:But it is true for every natural (if you formalise it correctly)!
Am Fri, 10 Jan 2025 18:21:43 +0100 schrieb WM:
No, I expect it is true for all natural numbers, none of which isI have no expectations about cardinality. I know that for every finiteYou wrongly expect this to hold in the infinite.
initial segment the even numbers are about half of the natural
numbers.
This does not change anywhere. It is true up to every natural number.
infinite.
On 10.01.2025 19:36, joes wrote:Your notation misleadingly implies predecessors to omega.
Am Fri, 10 Jan 2025 18:12:15 +0100 schrieb WM:
On 10.01.2025 17:50, joes wrote:
Am Fri, 10 Jan 2025 17:42:39 +0100 schrieb WM:
On 10.01.2025 14:09, joes wrote:
Am Fri, 10 Jan 2025 10:52:46 +0100 schrieb WM:
Yes.You mean N u {w}.{1, 2,3, ..., ω}. (*)As I said. You can prove it when doubling all elements of the set
It is not a natural number and destroys the symmetry.Why not?Why do you want to include omega?Since all natural numbers existing below ω are multiplied by 2, whenThe regular distance of next neighbours remains as a conservedPoints don't even exist without the numbers. There is no number with
property in correct mathematics.
a finite distance from omega.
doubling the elements of (*), no further numbers below ω can be
created - in actual infinity. What happens with the new 50 % of even
numbers?
The set N u {w}, for example, is not equal to N. You can't callWhat new numbers? There are no evens above omega.If actual infinity is invariable, then we can take the set ℕ and
consider it as invariable. When we add further elements, it grows.
Afterwards it contains further elements. Same can be accomplished by multiplying all natural numbers by 2.There is no natural n such that 2*n is not natural.
On 10.01.2025 13:41, Richard Damon wrote:
On 1/9/25 11:48 AM, WM wrote:
It is true that
{1, 2, 3, ...} is a set and
{0, 1, 2, 3, ...} is a greater set.
No,
one may be the proper subset of the other,
but it turns out that
due to the way that infinity works,
they are both are the same size.
This has nothing to do with
"how infinity works".
It simply is a result of an insufficient method
to measure infinite sets.
Do you (WM) disagree with
'finite' meaning
'smaller.than fuller.by.one sets'?
That is also true for infinite sets.
On 09.01.2025 22:22, joes wrote:In particular it means there is no largest one.
Am Thu, 09 Jan 2025 10:30:25 +0100 schrieb WM:
On 09.01.2025 00:42, joes wrote:
Am Wed, 08 Jan 2025 15:35:44 +0100 schrieb WM:
A set like ℕ has a fixed number of elements. If ω-1 does not exist, >>>>> what is the fixed border of existence?It has an infinite number of elements, and that number happens to be
invariant under finite subtraction/addition.
which means all natural numbers. Not even one must be missing from theThat implies the impossibility to extract all elements of contents inNo, you just need "extract/apply" infinitely many,
order to apply them as indices.
set of indices.
Yes they can, because there are an infinity of them.It means that no limits are involved but that all not yet used contentThat destroys Cantor's approach. His sequences do not exist:What does this have to do with Aleph_0?
"thus we get the epitome (ω) of all real algebraic numbers [...] and
with respect to this order we can talk about the nth algebraic number
where not a single one of this epitome (ω) has been forgotten." [E.
Zermelo: "Georg Cantor – Gesammelte Abhandlungen mathematischen und
philosophischen Inhalts", Springer, Berlin (1932) p. 116]
of endsegments must become indices. Not all endsegments can be infinite.
Am Fri, 10 Jan 2025 20:22:44 +0100 schrieb WM:
If actual infinity is invariable, then we can take the set ℕ andThe set N u {w}, for example, is not equal to N. You can't call
consider it as invariable. When we add further elements, it grows.
that invariable.
Afterwards it contains further elements. Same can be accomplished byThere is no natural n such that 2*n is not natural.
multiplying all natural numbers by 2.
Am Fri, 10 Jan 2025 20:17:37 +0100 schrieb WM:
On 10.01.2025 19:28, joes wrote:But it is true for every natural
Am Fri, 10 Jan 2025 18:21:43 +0100 schrieb WM:No, I expect it is true for all natural numbers, none of which is
I have no expectations about cardinality. I know that for every finite >>>> initial segment the even numbers are about half of the naturalYou wrongly expect this to hold in the infinite.
numbers.
This does not change anywhere. It is true up to every natural number.
infinite.
(if you formalise it correctly)!
That doesn't make it true for N and G.
Where OUR infinityⁿᵒᵗᐧᵂᴹ "doesn't work",
it's you who's saying it doesn't work,
Am Thu, 09 Jan 2025 23:39:21 +0100 schrieb WM:
On 09.01.2025 22:22, joes wrote:
Am Thu, 09 Jan 2025 10:30:25 +0100 schrieb WM:
On 09.01.2025 00:42, joes wrote:
Am Wed, 08 Jan 2025 15:35:44 +0100 schrieb WM:
A set like ℕ has a fixed number of elements. If ω-1 does not exist, >>>>>> what is the fixed border of existence?It has an infinite number of elements, and that number happens to be >>>>> invariant under finite subtraction/addition.
In particular it means there is no largest one.which means all natural numbers. Not even one must be missing from theThat implies the impossibility to extract all elements of contents inNo, you just need "extract/apply" infinitely many,
order to apply them as indices.
set of indices.
Yes they can, because there are an infinity of them.It means that no limits are involved but that all not yet used contentThat destroys Cantor's approach. His sequences do not exist:What does this have to do with Aleph_0?
"thus we get the epitome (ω) of all real algebraic numbers [...] and
with respect to this order we can talk about the nth algebraic number
where not a single one of this epitome (ω) has been forgotten." [E.
Zermelo: "Georg Cantor – Gesammelte Abhandlungen mathematischen und
philosophischen Inhalts", Springer, Berlin (1932) p. 116]
of endsegments must become indices. Not all endsegments can be infinite.
On 10.01.2025 21:06, joes wrote:Good. It is not true for the infinite sets.
Am Fri, 10 Jan 2025 20:17:37 +0100 schrieb WM:Of course. Otherwise you would have to find a counterexample.
On 10.01.2025 19:28, joes wrote:But it is true for every natural
Am Fri, 10 Jan 2025 18:21:43 +0100 schrieb WM:No, I expect it is true for all natural numbers, none of which is
I have no expectations about cardinality. I know that for everyYou wrongly expect this to hold in the infinite.
finite initial segment the even numbers are about half of the
natural numbers.
This does not change anywhere. It is true up to every natural
number.
infinite.
Mathematics is all about formalising.(if you formalise it correctly)!Irrelevant.
∀n ∈ ℕ: |{1, 2, 3, ..., 2n}|/|{2, 4, 6, ..., 2n}| = 2.Those are not N and E.
No. For n->oo, G is both the set {2, 4, ..., 2n} and {2, 4, ..., 42n};That doesn't make it true for N and G.I am not interested in these letters but only in all natural numbers.
All natural numbers are twice as many as all even natural numbers. If
your N and G denote all natural numbers and all even numbers, then 2 is
true also for them.
On 10.01.2025 21:03, joes wrote:Right. N stays N even when you construct a new set N u {k} (which is
Am Fri, 10 Jan 2025 20:22:44 +0100 schrieb WM:
Every set in ZF is invariable.If actual infinity is invariable, then we can take the set ℕ andThe set N u {w}, for example, is not equal to N. You can't call that
consider it as invariable. When we add further elements, it grows.
invariable.
Bzzt wrong. It proves it's true for infinitely many numbers,That it true for every definable number and proves potential infinity.Afterwards it contains further elements. Same can be accomplished byThere is no natural n such that 2*n is not natural.
multiplying all natural numbers by 2.
It contradicts actual infinity. If ***all*** natural numbers are
doubled, then you get more than you started with or you did not start
with ***all*** natural numbers.
On 10.01.2025 21:30, joes wrote:The naturals *are* the indices of the sequence. And it is infinite.
Am Thu, 09 Jan 2025 23:39:21 +0100 schrieb WM:Relevant is only that none remains outside of the set of indices. It
On 09.01.2025 22:22, joes wrote:In particular it means there is no largest one.
Am Thu, 09 Jan 2025 10:30:25 +0100 schrieb WM:
On 09.01.2025 00:42, joes wrote:
Am Wed, 08 Jan 2025 15:35:44 +0100 schrieb WM:
A set like ℕ has a fixed number of elements. If ω-1 does notIt has an infinite number of elements, and that number happens to
exist,
what is the fixed border of existence?
be invariant under finite subtraction/addition.
which means all natural numbers. Not even one must be missing from theThat implies the impossibility to extract all elements of contentsNo, you just need "extract/apply" infinitely many,
in order to apply them as indices.
set of indices.
would make the set finite.
That is wrong. Infinitely many of them can only exist when no naturalYes they can, because there are an infinity of them.It means that no limits are involved but that all not yet used contentThat destroys Cantor's approach. His sequences do not exist:What does this have to do with Aleph_0?
"thus we get the epitome (ω) of all real algebraic numbers [...] and >>>>> with respect to this order we can talk about the nth algebraic
number where not a single one of this epitome (ω) has been
forgotten." [E. Zermelo: "Georg Cantor – Gesammelte Abhandlungen
mathematischen und philosophischen Inhalts", Springer, Berlin (1932) >>>>> p. 116]
of endsegments must become indices. Not all endsegments can be
infinite.
natural number is missing an an index.
Therefore none can remain in theThe limit is indeed empty.
content. Therefore your argument is fools crap.
On 10.01.2025 21:08, Jim Burns wrote:
Where OUR infinityⁿᵒᵗᐧᵂᴹ "doesn't work",
it's you who's saying it doesn't work,
You are inconsistent. You claim that all natural numbers are an
invariable set. But when all elements are doubled then your set grows, showing it is not inc´variable. That is nonsense.
Regards, WM
On 1/10/2025 1:52 AM, WM wrote:
Hint: The set of all natural numbers, IN, does not change. [moebius]
So all natural numbers are fixed.
Then for every [ordinal number] it is determined whether [it] is a natural number [or not].
Although we cannot determine it because <bla>
On 1/10/2025 4:41 AM, Richard Damon wrote:
On 1/9/25 5:01 PM, WM wrote:
On 09.01.2025 21:24, joes wrote:
Am Thu, 09 Jan 2025 17:51:43 +0100 schrieb WM:
all ordinals have an order, but omega still has no predecessor
You <bla bla bla>
It has no predecessor, just like 0 has no predecessor [...]
0 has no predecessor in the unsigned integers.
Well, we can go into the signed integers where a predecessor of 0 is 0 -
1 ?
Am 10.01.2025 um 23:18 schrieb Chris M. Thomasson:
On 1/10/2025 1:52 AM, WM wrote:
Then for every [ordinal number] it is determined whether [it] is a
natural number [or not].
Indeed!
Although we cannot determine it because <bla>
Well, there's a difference between mathematical facts and what we KNOW
about these facts (or even if they a r e facts).
There are many mathematical facts we cannot "determine" (at least not yet).
.
.
.
Am 10.01.2025 um 23:18 schrieb Chris M. Thomasson:
On 1/10/2025 1:52 AM, WM wrote:
Then for every [ordinal number] it is determined whether [it] is a
natural number [or not].
Indeed!
Am 11.01.2025 um 04:51 schrieb Moebius:
Am 10.01.2025 um 23:18 schrieb Chris M. Thomasson:
On 1/10/2025 1:52 AM, WM wrote:
Then for every [ordinal number] it is determined whether [it] is a
natural number [or not].
Indeed!
There is even a simple criteria for this (i.e. if an ordinal number is a natural number or not):
For each and every ordinal number x:
x is a natural number iff x < omega.
.
.
.
Am 10.01.2025 um 23:18 schrieb Chris M. Thomasson:
On 1/10/2025 1:52 AM, WM wrote:
Then for every [ordinal number] it is determined whether [it] is a
natural number [or not].
Indeed!
Although we cannot determine it because <bla>
Well, there's a difference between mathematical facts and what we KNOW
about these facts (or even if they a r e facts).
There are many mathematical facts we cannot "determine" (at least not yet).
.
.
.
Am Fri, 10 Jan 2025 22:30:52 +0100 schrieb WM:
On 10.01.2025 21:03, joes wrote:Right. N stays N even when you construct a new set N u {k} (which is
Am Fri, 10 Jan 2025 20:22:44 +0100 schrieb WM:Every set in ZF is invariable.
If actual infinity is invariable, then we can take the set ℕ andThe set N u {w}, for example, is not equal to N. You can't call that
consider it as invariable. When we add further elements, it grows.
invariable.
a different set only if k !e N}. So N doesn't "grow".
Am Fri, 10 Jan 2025 22:38:51 +0100 schrieb WM:
On 10.01.2025 21:06, joes wrote:Good. It is not true for the infinite sets.
Am Fri, 10 Jan 2025 20:17:37 +0100 schrieb WM:Of course. Otherwise you would have to find a counterexample.
On 10.01.2025 19:28, joes wrote:But it is true for every natural
Am Fri, 10 Jan 2025 18:21:43 +0100 schrieb WM:No, I expect it is true for all natural numbers, none of which is
I have no expectations about cardinality. I know that for everyYou wrongly expect this to hold in the infinite.
finite initial segment the even numbers are about half of the
natural numbers.
This does not change anywhere. It is true up to every natural
number.
infinite.
Mathematics is all about formalising.(if you formalise it correctly)!Irrelevant.
∀n ∈ ℕ: |{1, 2, 3, ..., 2n}|/|{2, 4, 6, ..., 2n}| = 2.Those are not N and E.
No. For n->oo,That doesn't make it true for N and G.I am not interested in these letters but only in all natural numbers.
All natural numbers are twice as many as all even natural numbers. If
your N and G denote all natural numbers and all even numbers, then 2 is
true also for them.
G is both the set {2, 4, ..., 2n} and {2, 4, ..., 42n};
indeed, {2, 4, ..., 2kn} for every k e N.
Am Fri, 10 Jan 2025 22:44:31 +0100 schrieb WM:
The naturals *are* the indices of the sequence. And it is infinite.Yes they can, because there are an infinity of them.That is wrong. Infinitely many of them can only exist when no natural
natural number is missing an an index.
Therefore none can remain in theThe limit is indeed empty.
content. Therefore your argument is fools crap.
On 10.01.2025 22:51, joes wrote:But not for omega, which is not a natural.
Am Fri, 10 Jan 2025 22:38:51 +0100 schrieb WM:The natural numbers are an infinite set. For all of them it is true,
On 10.01.2025 21:06, joes wrote:Good. It is not true for the infinite sets.
Am Fri, 10 Jan 2025 20:17:37 +0100 schrieb WM:Of course. Otherwise you would have to find a counterexample.
On 10.01.2025 19:28, joes wrote:But it is true for every natural
Am Fri, 10 Jan 2025 18:21:43 +0100 schrieb WM:No, I expect it is true for all natural numbers, none of which is
I have no expectations about cardinality. I know that for everyYou wrongly expect this to hold in the infinite.
finite initial segment the even numbers are about half of the
natural numbers.
This does not change anywhere. It is true up to every natural
number.
infinite.
Informal reasoning gets you nowhere, see the centuries before that.No, that is only a habit of the last century.Mathematics is all about formalising.(if you formalise it correctly)!Irrelevant.
Not what I said. Every natural is finite, and so are theFind an element of N or E that is not covered by the equation.∀n ∈ ℕ: |{1, 2, 3, ..., 2n}|/|{2, 4, 6, ..., 2n}| = 2.Those are not N and E.
The *set* of all of them isn't.Every n is finite.No. For n->oo,That doesn't make it true for N and G.I am not interested in these letters but only in all natural numbers.
All natural numbers are twice as many as all even natural numbers. If
your N and G denote all natural numbers and all even numbers, then 2
is true also for them.
All what?G is both the set {2, 4, ..., 2n} and {2, 4, ..., 42n};And all of them can be denoted by n.
indeed, {2, 4, ..., 2kn} for every k e N.
If these sets are not fixed, then there is no bijection possible.They are fixed for every single k and n->oo, they are the same even.
"thus we get the epitome (ω) of all real algebraic numbers [...] andThere is no "after" an infinity. No extension is going on. No
with respect to this order we can talk about the th algebraic number
where not a single one of this epitome () has been forgotten." [E. Zermelo: "Georg Cantor – Gesammelte Abhandlungen mathematischen und philosophischen Inhalts", Springer, Berlin (1932) p. 116]
Afterwards no extension by 42 is allowed.
On 10.01.2025 23:42, joes wrote:The limit is empty, no natural "remains" in every endsegment, the
Am Fri, 10 Jan 2025 22:44:31 +0100 schrieb WM:
Not if any natural is missing because it remains in the content.The naturals *are* the indices of the sequence. And it is infinite.Yes they can, because there are an infinity of them.That is wrong. Infinitely many of them can only exist when no natural
natural number is missing an an index.
Yes it does, the (infinite) sequence of (infinite) end segmentsThere is no limit! All indices are required for bijections. Bijections concern all elements, not limits. In particular a sequence of infiniteTherefore none can remain in the content. Therefore your argument isThe limit is indeed empty.
fools crap.
sets has no empty "limit".
A union of FISONs which stay below a certain threshold can surpass that threshold.Only a finite threshold. Every FISON is (surprise) finite, and every
On 1/10/25 4:48 PM, WM wrote:
On 10.01.2025 21:08, Jim Burns wrote:But the set doesn't grow.
Where OUR infinityⁿᵒᵗᐧᵂᴹ "doesn't work",
it's you who's saying it doesn't work,
You are inconsistent. You claim that all natural numbers are an
invariable set. But when all elements are doubled then your set grows,
showing it is not invariable. That is nonsense.
Which element is in the doubled set that wasn't there in the first place?
Am 10.01.2025 um 23:18 schrieb Chris M. Thomasson:
On 1/10/2025 1:52 AM, WM wrote:
Hint: The set of all natural numbers, IN, does not change. [moebius]
So all natural numbers are fixed.
Was immer das auch bedeuten soll.
Well, there's a difference between mathematical facts and what we KNOW
about these facts (or even if they a r e facts).
On 11.01.2025 01:28, Richard Damon wrote:Nope. This is all just your conception of Aleph_0 as finite. It does
On 1/10/25 4:48 PM, WM wrote:The number of elements remains constant. All odd numbers of ℕ are
On 10.01.2025 21:08, Jim Burns wrote:But the set doesn't grow.
Where OUR infinityⁿᵒᵗᐧᵂᴹ "doesn't work",You are inconsistent. You claim that all natural numbers are an
it's you who's saying it doesn't work,
invariable set. But when all elements are doubled then your set grows,
showing it is not invariable. That is nonsense.
Which element is in the doubled set that wasn't there in the first
place?
deleted. That implies that new even numbers are added.
"The infinite sequence thus defined has the peculiar property to containThis just means it is a bijection to N, which has an order.
the positive rational numbers completely, and each of them only once at
a determined place." [G. Cantor, letter to R. Lipschitz (19 Nov 1883)]
If Cantor has constructed a sequence containing all even numbers of the original set ℕ, then the doubled even numbers are missing.What? Doubled even numbers are also even numbers.
Am Sat, 11 Jan 2025 10:03:01 +0100 schrieb WM:
On 10.01.2025 23:42, joes wrote:
Am Fri, 10 Jan 2025 22:44:31 +0100 schrieb WM:
The limit is empty,Not if any natural is missing because it remains in the content.The naturals *are* the indices of the sequence. And it is infinite.Yes they can, because there are an infinity of them.That is wrong. Infinitely many of them can only exist when no natural
natural number is missing an an index.
no natural "remains" in every endsegment, the
sequence of naturals diverges beyond any (finite) bound. Furthermore,
every one of the inf.many naturals has a corresponding endsegment,
none is "missing"; the sequence is infinitely long
Yes it does, the (infinite) sequence of (infinite) end segmentsThere is no limit! All indices are required for bijections. BijectionsTherefore none can remain in the content. Therefore your argument isThe limit is indeed empty.
fools crap.
concern all elements, not limits. In particular a sequence of infinite
sets has no empty "limit".
converges to the empty set,
because no element can be in every
segment.
A union of FISONs which stay below a certain threshold can surpass thatOnly a finite threshold. Every FISON is (surprise) finite, and every
threshold.
finite union of consecutive ones is again finite and equal to the
largest one. Every infinite union of (not necessarily consecutive)
FISONs has no largest natural.
Am Sat, 11 Jan 2025 09:50:02 +0100 schrieb WM:
On 10.01.2025 22:51, joes wrote:But not for omega, which is not a natural.
Am Fri, 10 Jan 2025 22:38:51 +0100 schrieb WM:The natural numbers are an infinite set. For all of them it is true,
On 10.01.2025 21:06, joes wrote:Good. It is not true for the infinite sets.
Am Fri, 10 Jan 2025 20:17:37 +0100 schrieb WM:Of course. Otherwise you would have to find a counterexample.
On 10.01.2025 19:28, joes wrote:But it is true for every natural
Am Fri, 10 Jan 2025 18:21:43 +0100 schrieb WM:No, I expect it is true for all natural numbers, none of which is
I have no expectations about cardinality. I know that for every >>>>>>>> finite initial segment the even numbers are about half of theYou wrongly expect this to hold in the infinite.
natural numbers.
This does not change anywhere. It is true up to every natural
number.
infinite.
Informal reasoning gets you nowhere, see the centuries before that.No, that is only a habit of the last century.Mathematics is all about formalising.(if you formalise it correctly)!Irrelevant.
Not what I said. Every natural is finite, and so are theFind an element of N or E that is not covered by the equation.∀n ∈ ℕ: |{1, 2, 3, ..., 2n}|/|{2, 4, 6, ..., 2n}| = 2.Those are not N and E.
starting segments of N and E.
The whole sets (which can be seen
as the limits) are not finite.
The *set* of all of them isn't.Every n is finite.No. For n->oo,That doesn't make it true for N and G.I am not interested in these letters but only in all natural numbers.
All natural numbers are twice as many as all even natural numbers. If
your N and G denote all natural numbers and all even numbers, then 2
is true also for them.
All what?G is both the set {2, 4, ..., 2n} and {2, 4, ..., 42n};And all of them can be denoted by n.
indeed, {2, 4, ..., 2kn} for every k e N.
"thus we get the epitome (ω) of all real algebraic numbers [...] andThere is no "after" an infinity.
with respect to this order we can talk about the th algebraic number
where not a single one of this epitome () has been forgotten." [E.
Zermelo: "Georg Cantor – Gesammelte Abhandlungen mathematischen und
philosophischen Inhalts", Springer, Berlin (1932) p. 116]
Afterwards no extension by 42 is allowed.
Am Sat, 11 Jan 2025 11:04:56 +0100 schrieb WM:
On 11.01.2025 01:28, Richard Damon wrote:Nope. This is all just your conception of Aleph_0 as finite.
On 1/10/25 4:48 PM, WM wrote:The number of elements remains constant. All odd numbers of ℕ are
On 10.01.2025 21:08, Jim Burns wrote:But the set doesn't grow.
Where OUR infinityⁿᵒᵗᐧᵂᴹ "doesn't work",You are inconsistent. You claim that all natural numbers are an
it's you who's saying it doesn't work,
invariable set. But when all elements are doubled then your set grows, >>>> showing it is not invariable. That is nonsense.
Which element is in the doubled set that wasn't there in the first
place?
deleted. That implies that new even numbers are added.
It does
not behave like that. All countable sets are bijective to each other.
"The infinite sequence thus defined has the peculiar property to containThis just means it is a bijection to N, which has an order.
the positive rational numbers completely, and each of them only once at
a determined place." [G. Cantor, letter to R. Lipschitz (19 Nov 1883)]
If Cantor has constructed a sequence containing all even numbers of theWhat? Doubled even numbers are also even numbers.
original set ℕ, then the doubled even numbers are missing.
On 11.01.2025 11:25, joes wrote:Infinite things don't have an end.
Am Sat, 11 Jan 2025 11:04:56 +0100 schrieb WM:Not finite but complete.
On 11.01.2025 01:28, Richard Damon wrote:Nope. This is all just your conception of Aleph_0 as finite.
On 1/10/25 4:48 PM, WM wrote:The number of elements remains constant. All odd numbers of ℕ are
On 10.01.2025 21:08, Jim Burns wrote:But the set doesn't grow.
Where OUR infinityⁿᵒᵗᐧᵂᴹ "doesn't work",You are inconsistent. You claim that all natural numbers are an
it's you who's saying it doesn't work,
invariable set. But when all elements are doubled then your set
grows,
showing it is not invariable. That is nonsense.
Which element is in the doubled set that wasn't there in the first
place?
deleted. That implies that new even numbers are added.
"Doubling" IS multiplication by 2, turning the naturals into the evens.It does not behave like that. All countable sets are bijective to eachIt appears so. But it is wrong. Construct a bijection between natural
other.
numbers and even natural numbers. f(n) = 2n. Then double the even
numbers by multiplication. Many are not in the bijection.
All even numbers are naturals. Multiplication by any natural numberBut not natural numbers, since *all* natural numbers have been doubled.If Cantor has constructed a sequence containing all even numbers ofWhat? Doubled even numbers are also even numbers.
the original set ℕ, then the doubled even numbers are missing.
When we mark all natural numbers with an astrisk, then none remains.It is not. The sequence 2*n does not go beyond infinity.
When wie double them, then none remains. The odd numbers leave. All even numbers remain. But the same number of of even numbers is larger than ω.
1, 2, 3, 4, 5, ..., ω becomes 2, 4, 6, 8, 10, ..., ω, ω+2, ω+4, ..., ω2.No. There is no x e N such that 2*x >= omega. You have listed two
On 11.01.2025 10:41, joes wrote:Then don't claim that some sentence held for omega.
Am Sat, 11 Jan 2025 09:50:02 +0100 schrieb WM:Therefore it is irrelevant. No bijection from ℕ contains it.
On 10.01.2025 22:51, joes wrote:But not for omega, which is not a natural.
Am Fri, 10 Jan 2025 22:38:51 +0100 schrieb WM:The natural numbers are an infinite set. For all of them it is true,
On 10.01.2025 21:06, joes wrote:Good. It is not true for the infinite sets.
Am Fri, 10 Jan 2025 20:17:37 +0100 schrieb WM:Of course. Otherwise you would have to find a counterexample.
On 10.01.2025 19:28, joes wrote:But it is true for every natural
Am Fri, 10 Jan 2025 18:21:43 +0100 schrieb WM:No, I expect it is true for all natural numbers, none of which is >>>>>>> infinite.
I have no expectations about cardinality. I know that for every >>>>>>>>> finite initial segment the even numbers are about half of the >>>>>>>>> natural numbers.You wrongly expect this to hold in the infinite.
This does not change anywhere. It is true up to every natural >>>>>>>>> number.
Crawl back into your cave and marvel about infinity.There mathematics has flourished. Now mainly nonsense is produced.Informal reasoning gets you nowhere, see the centuries before that.No, that is only a habit of the last century.Mathematics is all about formalising.(if you formalise it correctly)!Irrelevant.
See:And which are not?Not what I said. Every natural is finite, and so are the startingFind an element of N or E that is not covered by the equation.∀n ∈ ℕ: |{1, 2, 3, ..., 2n}|/|{2, 4, 6, ..., 2n}| = 2.Those are not N and E.
segments of N and E.
Your claim does not hold for the sets.The whole sets (which can be seen as the limits) are not finite.My claim holds for all numbers only. That is mathematics. [?]
That's what I'm saying.Irrelevant. My claim holds for all natnumbers only.The *set* of all of them isn't.Every n is finite.No. For n->oo,That doesn't make it true for N and G.I am not interested in these letters but only in all natural
numbers.
All natural numbers are twice as many as all even natural numbers.
If your N and G denote all natural numbers and all even numbers,
then 2 is true also for them.
No numbers are "created" (I guess you mean the image is a subset of the domain?).All natnumbers which Cantor uses in bijections: "such that every elementAll what?E is both the set {2, 4, ..., 2n} and {2, 4, ..., 42n};And all of them can be denoted by n.
indeed, {2, 4, ..., 2kn} for every k e N.
of the set stands at a definite position of this sequence". If this has
been accomplished, and then more numbers are created, the bijection
fails. This must not happen.
Cantor bijectively maps the naturals to the algebraic numbers. You canCantor maps all natural numbers to a set. Afterwards these natural"thus we get the epitome (ω) of all real algebraic numbers [...] andThere is no "after" an infinity.
with respect to this order we can talk about the th algebraic number
where not a single one of this epitome () has been forgotten." [E.
Zermelo: "Georg Cantor – Gesammelte Abhandlungen mathematischen und
philosophischen Inhalts", Springer, Berlin (1932) p. 116]
Afterwards no extension by 42 is allowed.
numbers can be multiplied by 2. Not all remain those which Cantor has applied.
On 11.01.2025 01:28, Richard Damon wrote:
On 1/10/25 4:48 PM, WM wrote:
On 10.01.2025 21:08, Jim Burns wrote:But the set doesn't grow.
Where OUR infinityⁿᵒᵗᐧᵂᴹ "doesn't work",
it's you who's saying it doesn't work,
You are inconsistent. You claim that all natural numbers are an
invariable set. But when all elements are doubled then your set
grows, showing it is not invariable. That is nonsense.
Which element is in the doubled set that wasn't there in the first place?
The number of elements remains constant. All odd numbers of ℕ are
deleted. That implies that new even numbers are added.
"The infinite sequence thus defined has the peculiar property to contain
the positive rational numbers completely, and each of them only once at
a determined place." [G. Cantor, letter to R. Lipschitz (19 Nov 1883)]
If Cantor has constructed a sequence containing all even numbers of the original set ℕ, then the doubled even numbers are missing.
Regards, WM
On 10.01.2025 19:24, joes wrote:
The limit of the ratio is different from the ratio of the limit.
Let f(x) = 2x/x. The limit of the ratio is 2. The ratio of the limit is undefined. In mathematics we use the limit of the ratio.
Regards, WM
joes laid this down on his screen :If that was a joke, I didn't get it. Please enlighten me?
Am Sat, 11 Jan 2025 11:04:56 +0100 schrieb WM:All of the smallest inductive countably infinite sets. :)
On 11.01.2025 01:28, Richard Damon wrote:Nope. This is all just your conception of Aleph_0 as finite. It does
On 1/10/25 4:48 PM, WM wrote:The number of elements remains constant. All odd numbers of ℕ are
On 10.01.2025 21:08, Jim Burns wrote:But the set doesn't grow.
Where OUR infinityⁿᵒᵗᐧᵂᴹ "doesn't work",You are inconsistent. You claim that all natural numbers are an
it's you who's saying it doesn't work,
invariable set. But when all elements are doubled then your set
grows,
showing it is not invariable. That is nonsense.
Which element is in the doubled set that wasn't there in the first
place?
deleted. That implies that new even numbers are added.
not behave like that. All countable sets are bijective to each other.
On 1/11/2025 12:55 AM, WM wrote:
Every union of FISONs which stay below a certain threshold stays below
that threshold.
1, 2, 3, 4, 5, ..., ω becomes 2, 4, 6, 8, 10, ..., ω, ω+2, ω+4, ..., ω2.No. There is no x e N such that 2*x >= omega. You have listed two
consecutive infinities on the right.
On 1/11/2025 12:55 AM, WM wrote:
Every union of FISONs which stay below a certain threshold stays
below that threshold.
Nope. All FISONs are "subset smaller" than IN, but the union of all
FISONs equals IN.
Am Sat, 11 Jan 2025 12:44:59 +0100 schrieb WM:
∀n ∈ ℕ: |{1, 2, 3, ..., 2n}|/|{2, 4, 6, ..., 2n}| = 2.
Your claim does not hold for the sets.
My claim holds for all natnumbers only.That's what I'm saying.
No numbers are "created"
There is no "after" an infinity.
joes laid this down on his screen :
Am Sat, 11 Jan 2025 11:04:56 +0100 schrieb WM:
If Cantor has constructed a sequence containing all even numbers of theWhat? Doubled even numbers are also even numbers.
original set ℕ, then the doubled even numbers are missing.
He's a hopeless case.
Why is he a teacher? I still don't know.
On 1/11/2025 12:55 AM, WM wrote:
Every union of FISONs which stay below a certain threshold stays
below that threshold.
Nope. All FISONs are "subset smaller" than IN, but the union of all
FISONs equals IN.
On 1/10/25 4:48 PM, WM wrote:
You are inconsistent. You claim that all natural numbers are an
invariable set. But when all elements are doubled then your set grows,
showing it is not inc´variable. That is nonsense.
No, the set DOESN'T grow,
WM submitted this idea :
On 11.01.2025 15:09, FromTheRafters wrote:
joes laid this down on his screen :
Am Sat, 11 Jan 2025 11:04:56 +0100 schrieb WM:
If Cantor has constructed a sequence containing all even numbers ofWhat? Doubled even numbers are also even numbers.
the
original set ℕ, then the doubled even numbers are missing.
He's a hopeless case.
Yes, you cannot hope ever to understand the difference between
potential and actual infinity.
I've yet to see any useful application of the notion of potential
infinity.
IMO it was created to appease the philosophers and later
rejected as useless by modern mathematicians.
In actual infinity all numbers are present. No one is missing,
according to Cantor. None can be added.
If you multiply every number by 2, then larger even numbers than all
hitherto present even numbers are created because the number of
numbers remains constant but the odd numbers disappear.
Wrong, the numbers are not 'created' as they already existed.
On 12.01.2025 12:59, FromTheRafters wrote:*their
WM submitted this idea :
On 11.01.2025 15:09, FromTheRafters wrote:
joes laid this down on his screen :
It does not create new numbers.Am Sat, 11 Jan 2025 11:04:56 +0100 schrieb WM:
All existing numbers have been doubled. That creates a new set.In actual infinity all numbers are present. No one is missing,Wrong, the numbers are not 'created' as they already existed.
according to Cantor. None can be added.
If you multiply every number by 2, then larger even numbers than all
hitherto present even numbers are created because the number of
numbers remains constant but the odd numbers disappear.
On 11.01.2025 14:58, joes wrote:I am talking about the sets N and E. They are infinite, so you are
Am Sat, 11 Jan 2025 12:44:59 +0100 schrieb WM:
That may be if sets are considered as more than their elements. MyYour claim does not hold for the sets.∀n ∈ ℕ: |{1, 2, 3, ..., 2n}|/|{2, 4, 6, ..., 2n}| = 2.
claims hold for all their elements.
And not for infinity.My claim holds for all natnumbers only.
Turn it around and there is nothing before.No numbers are "created"
Wrong. 1 is after an infinity of (unit) fractions.There is no "after" an infinity.
On 11.01.2025 15:09, FromTheRafters wrote:Must be unexplainable...
joes laid this down on his screen :
Am Sat, 11 Jan 2025 11:04:56 +0100 schrieb WM:
Yes, you cannot hope ever to understand the difference between potentialHe's a hopeless case.If Cantor has constructed a sequence containing all even numbers ofWhat? Doubled even numbers are also even numbers.
the original set ℕ, then the doubled even numbers are missing.
and actual infinity.
In actual infinity all numbers are present. No one is missing, accordingWhy should you.
to Cantor. None can be added.
If you multiply every number by 2, then larger even numbers than allNo, not if you really multiply *every* number and not only finitely many.
hitherto present even numbers are created because the number of numbers remains constant but the odd numbers disappear.
On 10.01.2025 21:08, Jim Burns wrote:
</WM<JB>>Do you (WM) disagree with
'finite' meaning
'smaller.than fuller.by.one sets'?
That is also true for infinite sets.
Where OUR infinityⁿᵒᵗᐧᵂᴹ "doesn't work",
it's you who's saying it doesn't work,
You are inconsistent.
You claim that
all natural numbers are an invariable set.
(Losing all numbers but</WM>[1]
keeping infinitely many
can only be possible if
new numbers are acquired.)
Where OUR infinityⁿᵒᵗᐧᵂᴹ "doesn't work",
it's you who's saying it doesn't work,
You are inconsistent.
You claim that
all natural numbers are an invariable set.
But when all elements are doubled
then your set grows,
showing it is not invariable.
That is nonsense.
On 1/10/2025 4:28 PM, Richard Damon wrote:
On 1/10/25 10:32 AM, WM wrote:
On 10.01.2025 13:41, Richard Damon wrote:
On 1/9/25 11:48 AM, WM wrote:
It is true that {1, 2, 3, ...} is a set and {0, 1, 2, 3, ...} is a
greater set.
No, one may be the proper subset of the other, but it turns out that
due to the way that infinity works, they are both are the same size.
This has nothing to do with "how infinity works". It simply is a
result of an insufficient method to measure infinite sets.
Regards, WM
No, it DOES have a baring on how infinity "works", but that seems to be
beyond your comprehesion due to your stupidity.
Removing a finite part from and infinite thing does not make that
infinite thing "smaller", becausse the "finite" thing is not measurable
AT ALL compared to the infinite, as it is an infintesimal part of it.
Your logic is based on the concept that infihity isn't actually
infinite, just "inconceivably" huge, but that means the finite thing
*IS* a measurable part of the whole, just an inconceivably small portion.
IT is YOUR method that is insufficient to handle that actual nature of
infinity, because it is bigger than your system can handle.
The two sets do not have a measurable difference, because they have the
same number of elements. as can be shown with the proper pairing between
them. *YOU* only *THINK* they are different, because your logic can't do
that pairing, becuase for your logic, it take an infinite about of work,
and it can't do that.
Sorry, your logic is just blowing your brain up into smithereens by its
inconsistancies, resulting in your darkness out of that super nova of
error.
WM is a teacher! Holy MOLY!
Am Sun, 12 Jan 2025 12:19:15 +0100 schrieb WM:
On 11.01.2025 14:58, joes wrote:I am talking about the sets N and E. They are infinite, so you are
Am Sat, 11 Jan 2025 12:44:59 +0100 schrieb WM:That may be if sets are considered as more than their elements. My
Your claim does not hold for the sets.∀n ∈ ℕ: |{1, 2, 3, ..., 2n}|/|{2, 4, 6, ..., 2n}| = 2.
claims hold for all their elements.
saying nothing about them, only about finite subsets.
Am Sun, 12 Jan 2025 13:41:27 +0100 schrieb WM:
On 12.01.2025 12:59, FromTheRafters wrote:*their
WM submitted this idea :
On 11.01.2025 15:09, FromTheRafters wrote:
joes laid this down on his screen :
It does not create new numbers.
Am Sun, 12 Jan 2025 12:48:50 +0100 schrieb WM:
If you multiply every number by 2, then larger even numbers than allNo, not if you really multiply *every* number and not only finitely many.
hitherto present even numbers are created because the number of numbers
remains constant but the odd numbers disappear.
The *cardinality* is the same.
when this scandal has been pointed out, repeatedly, to the school
board of directors they invoked "academic freedom"!!!
On 1/10/2025 4:48 PM, WM wrote:
Elsethread:
(Losing all numbers but</WM>[1]
keeping infinitely many
can only be possible if
new numbers are acquired.)
It sounds as though
the only explanation which you (WM) accept
for the constancy of end.segment.size is
that elements are inserted (at the darkᵂᴹ end?)
as other elements are deleted (at the visibleᵂᴹ end?)
(Somehow this happens. Perhaps ℕ has homeostasis.)
Infiniteⁿᵒᵗᐧᵂᴹ sets do not change.
You are inconsistent.
You claim that
all natural numbers are an invariable set.
But when all elements are doubled
then your set grows, showing it is not invariable.
That is nonsense.
Perhaps this argument won't look like nonsense.
It features an utterly.familiar property,
being.finite,
⎜ ℕ is the set of finite ordinals.
Bob cannot become absent from a finite set
After serious thinking joes wrote :
Am Sun, 12 Jan 2025 13:41:27 +0100 schrieb WM:
On 12.01.2025 12:59, FromTheRafters wrote:*their
WM submitted this idea :
On 11.01.2025 15:09, FromTheRafters wrote:
joes laid this down on his screen :
Indeed! but this tagline is automatically generated by my newsreader. I
guess it needs a woke upgrade.
On 12.01.2025 15:26, Python wrote:
when this scandal has been pointed out, repeatedly, to the school
board of directors they invoked "academic freedom"!!!
The real scandal is the denial of this simple truth by stultified matheologians:
[whatever] which stay below a certain threshold stays below that threshold.
[snip more nonsense]
On 12.01.2025 16:48, FromTheRafters wrote:
After serious thinking joes wrote :
Am Sun, 12 Jan 2025 13:41:27 +0100 schrieb WM:
On 12.01.2025 12:59, FromTheRafters wrote:*their
WM submitted this idea :
On 11.01.2025 15:09, FromTheRafters wrote:
joes laid this down on his screen :
Indeed! but this tagline is automatically generated by my newsreader. I
guess it needs a woke upgrade.
Go woke, go broke.
Every* union of FISONs {1, 2, 3, ..., n} which stay below a certain*finite
threshold stays below that threshold. There is nothing that could
increase the set of definable natural numbers beyond it.
All FISONs are infinitesimal compared to |ℕ|, that means they all stay below every definable fraction 1/n 0f |ℕ|.As the name says.
On 12.01.2025 15:07, joes wrote:Not for x = omega.
Am Sun, 12 Jan 2025 12:48:50 +0100 schrieb WM:
When their number remains and half disappear, then half are added.If you multiply every number by 2, then larger even numbers than allNo, not if you really multiply *every* number and not only finitely
hitherto present even numbers are created because the number of
numbers remains constant but the odd numbers disappear.
many.
The *cardinality* is the same.Cardinality is irrelevant. x = x/2 + y.
On 12.01.2025 14:29, joes wrote:He uses *infinitely* many elements, not finite subsets.
Am Sun, 12 Jan 2025 12:19:15 +0100 schrieb WM:Wrong. I am talking about all numbers which Cantor uses for his
On 11.01.2025 14:58, joes wrote:I am talking about the sets N and E. They are infinite, so you are
Am Sat, 11 Jan 2025 12:44:59 +0100 schrieb WM:That may be if sets are considered as more than their elements. My
Your claim does not hold for the sets.∀n ∈ ℕ: |{1, 2, 3, ..., 2n}|/|{2, 4, 6, ..., 2n}| = 2.
claims hold for all their elements.
saying nothing about them, only about finite subsets.
bijections. He uses only elements, not sets.
"The infinite sequence thus defined has the peculiar property to containFor all numbers n e N, {1, 2, ..., n} != N.
the positive rational numbers completely"
"thus we get the epitome (ω) of all real algebraic numbers."
For all these numbers my equations are true.
On 12.01.2025 16:48, FromTheRafters wrote:Can you correct it when you quote me?
After serious thinking joes wrote :
Am Sun, 12 Jan 2025 13:41:27 +0100 schrieb WM:Indeed! but this tagline is automatically generated by my newsreader. I
On 12.01.2025 12:59, FromTheRafters wrote:*their
WM submitted this idea :
On 11.01.2025 15:09, FromTheRafters wrote:
joes laid this down on his screen :
guess it needs a woke upgrade.
Go woke, go broke.Fuck you.
Am Sun, 12 Jan 2025 16:56:03 +0100 schrieb WM:
On 12.01.2025 16:48, FromTheRafters wrote:Can you correct it when you quote me?
After serious thinking joes wrote :
Am Sun, 12 Jan 2025 13:41:27 +0100 schrieb WM:Indeed! but this tagline is automatically generated by my newsreader. I
On 12.01.2025 12:59, FromTheRafters wrote:*their
WM submitted this idea :
On 11.01.2025 15:09, FromTheRafters wrote:
joes laid this down on his screen :
guess it needs a woke upgrade.
It's not "woke" not to assume everybody is a male.
Go woke, go broke.Fuck you.
Am Sun, 12 Jan 2025 16:25:15 +0100 schrieb WM:
I am talking about all numbers which Cantor uses for hisHe uses *infinitely* many elements, not finite subsets.
bijections. He uses only elements, not sets.
"The infinite sequence thus defined has the peculiar property to containFor all numbers n e N
the positive rational numbers completely"
"thus we get the epitome (ω) of all real algebraic numbers."
For all these numbers my equations are true.
On 12.01.2025 17:16, joes wrote:No, you use FISONs {1, 2, ..., n}. N is not such one.
Am Sun, 12 Jan 2025 16:25:15 +0100 schrieb WM:So do I.
I am talking about all numbers which Cantor uses for his bijections.He uses *infinitely* many elements, not finite subsets.
He uses only elements, not sets.
They are not true for the infinite sets N and E.Yes. Otherwise you should find a number that is missing."The infinite sequence thus defined has the peculiar property toFor all numbers n e N
contain the positive rational numbers completely"
"thus we get the epitome (ω) of all real algebraic numbers."
For all these numbers my equations are true.
On 1/12/2025 10:54 AM, WM wrote:
No, it depends on completeness.
It is completely true
that each natural number is a natural number and
that only natural numbers are natural numbers.
ℕ is the set of finite ordinals.
On 12.01.2025 15:40, Jim Burns wrote:
On 1/10/2025 4:48 PM, WM wrote:
You are inconsistent.
You claim that
all natural numbers are an invariable set.
But when all elements are doubled
then your set grows, showing it is not invariable.
That is nonsense.
Perhaps this argument won't look like nonsense.
It features an utterly.familiar property,
being.finite,
No, it depends on completeness.
If all natural numbers are there
such than none can be added,
If all natural numbers are there
such than none can be added,
then doubling all of them
deletes odd numbers and
must create new even numbers which
cannot be natural numbers.
⎜ ℕ is the set of finite ordinals.
Of all.
None can be added.
If all are doubled,
then 50 % odd nubers are deleted,
50 % even numbers are added.
Because the total sum remains constant.
x = x/2 + y.
On 12.01.2025 03:09, Chris M. Thomasson wrote:
Why is he a teacher? I still don't know.
In order to correct the nonsense taught in set theory, like this: All
FISONs are less than |ℕ|/n for every definable n, but their union is | ℕ|. Only very stultified brains can claim that nonsense.
Regards, WM
On 12.01.2025 20:33, Jim Burns wrote:
On 1/12/2025 10:54 AM, WM wrote:
No, it depends on completeness.
It is completely true
that each natural number is a natural number and
that only natural numbers are natural numbers.
and that nothing fits between them and ω.
ℕ is the set of finite ordinals.
such than none can be added.
Regular distances in (0, ω) multiplied by 2 remain regular distances in
(0, 2ω).
Regards, WM
On 1/12/25 7:47 AM, WM wrote:
All FISONs are less than |ℕ|/n for every n, but their union is |ℕ|.
Es
gilt dann also |ℕ|/n = |ℕ|.
Was Du vermutlich meinst/ausdrücken willst, ist, dass für alle k e IN:
An e IN: |F(k)|*n < |IN|
gilt. D. h. dass jeder Anfangsabschnitt im Vergleich zu IN geradezu "verschwindend klein" ist.
On 1/12/25 5:58 AM, WM wrote:
On 11.01.2025 14:34, joes wrote:
1, 2, 3, 4, 5, ..., ω becomes 2, 4, 6, 8, 10, ..., ω, ω+2, ω+4, ..., >>>> ω2.No. There is no x e N such that 2*x >= omega. You have listed two
consecutive infinities on the right.
There is a basic law: When a sequence of regular distances is
multiplied by 2, then a sequence of regular distances results.
Which isn't applicable, since their isn't such a "regular distance"
your starting premise is just
The interval (0, ω)*2 becomes (0, ω*2) with ω in the middle. Below ω
the newly created even numbers cannot be inserted, because more than
all even natural numbers do not exist in actual infinity.
Every contrary opinion is based on potential infinity.
Nope, your "opinion" is just based on your
On 12.01.2025 20:33, Jim Burns wrote:
On 1/12/2025 10:54 AM, WM wrote:
No, it depends on completeness.
It is completely true
that each natural number is a natural number and
that only natural numbers are natural numbers.
and that nothing fits between them and ω.
ℕ is the set of finite ordinals.
such than none can be added.
Regular distances in (0, ω) multiplied by 2
remain regular distances in (0, 2ω).
On 13.01.2025 05:51, Moebius wrote:
Es gilt dann also |ℕ|/n = |ℕ|.
Wrong. |ℕ| is a fixed number.
On 1/12/2025 2:39 PM, WM wrote:
On 12.01.2025 20:33, Jim Burns wrote:
On 1/12/2025 10:54 AM, WM wrote:
No, it depends on completeness.
It is completely true
that each natural number is a natural number and
that only natural numbers are natural numbers.
and that nothing fits between them and ω.
Yes.
ω is first infiniteᵒʳᵈ.
No infiniteᵒʳᵈ is before ω
No finiteᵒʳᵈ is after ω
Regular distances in ⦅0,ω⦆ multiplied by 2
remain regular distances in ⦅0,ω⦆, not.in ⟦ω,2ω⦆
On 1/13/2025 7:48 AM, WM wrote:
On 13.01.2025 05:51, Moebius wrote:
Es gilt dann also |ℕ|/n = |ℕ|.
Wrong. |ℕ| is a fixed number.
By 'fixed', you mean that #ℕ > #(ℕ\{0})
However,
ℕ is the set of all 'fixed' (finite) ordinals.
⎛ Assume ℕ is 'fixed'.
⎜
⎜ A 'fixed' ordinal ⟦0,𝔑⦆ the size of ℕ exists.
⎜ #⟦0,𝔑⦆ = #ℕ
⎜
⎜ ⟦0,𝔑+1⦆ is 'fixed', too.
⎜ ⟦0,𝔑+1⦆ ⊆ ℕ
Therefore,
ℕ is not 'fixed'.
On 13.01.2025 17:33, Jim Burns wrote:
On 1/12/2025 2:39 PM, WM wrote:
On 12.01.2025 20:33, Jim Burns wrote:
On 1/12/2025 10:54 AM, WM wrote:
No, it depends on completeness.
It is completely true
that each natural number is a natural number and
that only natural numbers are natural numbers.
and that nothing fits between them and ω.
Yes.
Therefore
doubling of all natural numbers
creates numbers larger than ω.
ω is first infiniteᵒʳᵈ.
No infiniteᵒʳᵈ is before ω
No finiteᵒʳᵈ is after ω
Right.
Regular distances in ⦅0,ω⦆ multiplied by 2
remain regular distances in ⦅0,ω⦆, not.in ⟦ω,2ω⦆
Doubling of all n
deletes the odd numbers
but cannot change the number of numbers,
therefore creates even numbers.
They do not fit below ω.
Remember:
nothing fits between them and ω.
On 13.01.2025 18:06, Jim Burns wrote:
On 1/13/2025 7:48 AM, WM wrote:
On 13.01.2025 05:51, Moebius wrote:
Es gilt dann also |ℕ|/n = |ℕ|.
Wrong. |ℕ| is a fixed number.
By 'fixed', you mean that #ℕ > #(ℕ\{0})
However,
ℕ is the set of all 'fixed' (finite) ordinals.
Which is ivariable.
⎛ Assume ℕ is 'fixed'.
⎜
⎜ A 'fixed' ordinal ⟦0,𝔑⦆ the size of ℕ exists.
|ℕ| = ω-1.
⎜ #⟦0,𝔑⦆ = #ℕ
⎜
⎜ ⟦0,𝔑+1⦆ is 'fixed', too.
ω
⎜ ⟦0,𝔑+1⦆ ⊆ ℕ
No.
On 13.01.2025 03:54, Richard Damon wrote:
On 1/12/25 5:58 AM, WM wrote:
On 11.01.2025 14:34, joes wrote:
1, 2, 3, 4, 5, ..., ω becomes 2, 4, 6, 8, 10, ..., ω, ω+2,No. There is no x e N such that 2*x >= omega. You have listed two
ω+4, ..., ω2.
consecutive infinities on the right.
There is a basic law: When a sequence of regular distances is
multiplied by 2, then a sequence of regular distances results.
Which isn't applicable, since their isn't such a "regular distance"
Between all natural numbers, there is a regular distance. When doubled,
the new ones do not fit between the old ones and ω because nothing fits between ℕ and ω.
your starting premise is just
to double all numbers which fit between 1 and ω.
The interval (0, ω)*2 becomes (0, ω*2) with ω in the middle. Below ω >>> the newly created even numbers cannot be inserted, because more than
all even natural numbers do not exist in actual infinity.
Every contrary opinion is based on potential infinity.
Nope, your "opinion" is just based on your
correct understanding that all between 1 and ω is available for doubling
and nothing can be inserted.
Regards, WM
On 1/13/2025 9:17 AM, WM wrote:
doubling of all natural numbers creates numbers larger than ω.
No. Double all of the natural numbers:
{1*2, 2*2, 3*2, 4*2, ...} = {2, 4, 6, 8, ...} .
All of those results are natural numbers. They were already there...
ω is first infiniteᵒʳᵈ.
No infiniteᵒʳᵈ is before ω
No finiteᵒʳᵈ is after ω
distances in ⦅0,ω⦆ multiplied by 2 remain regular distances in ⦅0,ω⦆, not.in ⟦ω,2ω⦆
Doubling of all n deletes the odd numbers
but cannot change the number of numbers,
therefore creates even numbers.
They do not fit below ω
Remember: [no ordinal number] fits between them and ω.
On 1/13/25 7:42 AM, WM wrote:
On 13.01.2025 03:54, Richard Damon wrote:
On 1/12/25 5:58 AM, WM wrote:
On 11.01.2025 14:34, joes wrote:
1, 2, 3, 4, 5, ..., ω becomes 2, 4, 6, 8, 10, ..., ω, ω+2, ω+4, >>>>>> ..., ω2.No. There is no x e N such that 2*x >= omega. You have listed two
consecutive infinities on the right.
There is a basic law: When a sequence of regular distances is
multiplied by 2, then a sequence of regular distances results.
Which isn't applicable, since their isn't such a "regular distance"
Between all natural numbers, there is a regular distance. When
doubled, the new ones do not fit between the old ones and ω because
nothing fits between ℕ and ω.
But omega isn't a natural number, so the space between the ... and omega isn't the same as the space between two consecutive natural numbers.
And the distance between the ... and omega *IS* big enough to fit the doubling of all the numbers (without needing to make any new ones).
your starting premise is just
to double all numbers which fit between 1 and ω.
Which are still just the even numbers between 1 and omega.
correct understanding that all between 1 and ω is available for
doubling and nothing can be inserted.
And nothing NEEDS to be inserted.
Am 13.01.2025 um 21:55 schrieb Chris M. Thomasson:
On 1/13/2025 9:17 AM, WM wrote:
doubling of all natural numbers creates numbers larger than ω.
Idiotic nonsense.
All of those results are natural numbers. They were already there...
Indeed. (At least in the context of classical mathematics/set theory.)
Using symols:
An e IN: n*2 e IN .
On 1/13/2025 9:17 AM, WM wrote:
On 13.01.2025 17:33, Jim Burns wrote:
On 1/12/2025 2:39 PM, WM wrote:
On 12.01.2025 20:33, Jim Burns wrote:
On 1/12/2025 10:54 AM, WM wrote:
No, it depends on completeness.
It is completely true
that each natural number is a natural number and
that only natural numbers are natural numbers.
and that nothing fits between them and ω.
Yes.
Therefore doubling of all natural numbers creates numbers larger than ω.
No. double all of the natural numbers:
{ 1 * 2, 2 * 2, 3 * 2, 4 * 2, ... } = { 2, 4, 6, 8, ... }
All of those results are natural numbers. They were already there...
Doubling of all n deletes the odd numbers but cannot change the number
of numbers, therefore creates even numbers. They do not fit below ω.
Remember: nothing fits between them and ω.
On 1/13/2025 12:17 PM, WM wrote:
Doubling of all n
deletes the odd numbers
but cannot change the number of numbers,
ℕ is the set of finite ordinals.
There is no finite set larger than ℕ
thus ℕ is infinite.
There is no infinite set smaller than ℕ
𝔼 is the set of even finite ordinals.
There is no finite set larger than 𝔼
thus 𝔼 is infinite
𝔼 ⊆ ℕ
#𝔼 ≤ #ℕ
There is no infinite set smaller than ℕ
#𝔼 ≥ #ℕ
therefore creates even numbers.
They do not fit below ω.
No.
They fit below ω
A step is never from finite to infinite.
Therefore, a step never crosses ω
Therefore, a sum never crosses ω
Therefore, a product never crosses ω
Therefore, a power never crosses ω
On 1/13/2025 12:29 PM, WM wrote:
On 13.01.2025 18:06, Jim Burns wrote:
On 1/13/2025 7:48 AM, WM wrote:
ℕ is only invariable in the sense which we use.
However,
you (WM) are convinced that
a set (such as ℕ) larger than
any set with sets.different.in.size.by.one
changes (has elements inserted or deleted)
in order to not.change.in.size.by.one.
⎜ #⟦0,𝔑⦆ = #ℕ
⎜
⎜ ⟦0,𝔑+1⦆ is 'fixed', too.
ω
Is ω = ⟦0,𝔑+1⦆ your 'fixed' (our 'finite')?
What about ω+1 and ω+2?
⎜ ⟦0,𝔑+1⦆ ⊆ ℕ
No.
Yes.
∀𝔑 ∈ ℕ: #⟦0,𝔑⦆ < #⟦0,𝔑+1⦆ ≤ #ℕ
On 14.01.2025 00:47, Richard Damon wrote:
On 1/13/25 7:42 AM, WM wrote:
On 13.01.2025 03:54, Richard Damon wrote:
On 1/12/25 5:58 AM, WM wrote:
On 11.01.2025 14:34, joes wrote:
1, 2, 3, 4, 5, ..., ω becomes 2, 4, 6, 8, 10, ..., ω, ω+2,No. There is no x e N such that 2*x >= omega. You have listed two
ω+4, ..., ω2.
consecutive infinities on the right.
There is a basic law: When a sequence of regular distances is
multiplied by 2, then a sequence of regular distances results.
Which isn't applicable, since their isn't such a "regular distance"
Between all natural numbers, there is a regular distance. When
doubled, the new ones do not fit between the old ones and ω because
nothing fits between ℕ and ω.
But omega isn't a natural number, so the space between the ... and
omega isn't the same as the space between two consecutive natural
numbers.
Maybe. But the space between the natural numbers is doubled while the accessible space remains constant in actual infinity.
And the distance between the ... and omega *IS* big enough to fit the
doubling of all the numbers (without needing to make any new ones).
If there is space, then it is filled before doubling already. Why
shouldn't it?
your starting premise is just
to double all numbers which fit between 1 and ω.
Which are still just the even numbers between 1 and omega.
∀n ∈ ℕ: 2n > n. All numbers are doubled. Their number remains the same (not only the cardinality, but the reality). Half of all are deleted.
Half are new.
correct understanding that all between 1 and ω is available for
doubling and nothing can be inserted.
And nothing NEEDS to be inserted.
It cannot. Therefore half of the results is larger than ω.
Regards, WM
On 13.01.2025 20:31, Jim Burns wrote:
On 1/13/2025 12:17 PM, WM wrote:
[...]
A step is never from finite to infinite.
Therefore, a step never crosses ω
Therefore, a sum never crosses ω
Therefore, a product never crosses ω
Therefore, a power never crosses ω
All that is true in potential infinity,
however it is wrong in completed infinity.
Doubling of all n
deletes the odd numbers
but cannot change the number of numbers,
ℕ is the set of finite ordinals.
There is no finite set larger than ℕ
thus ℕ is infinite.
There is no infinite set smaller than ℕ
𝔼 is the set of even finite ordinals.
There is no finite set larger than 𝔼
thus 𝔼 is infinite
𝔼 ⊆ ℕ
#𝔼 ≤ #ℕ
There is no infinite set smaller than ℕ
#𝔼 ≥ #ℕ
That is obviously wrong.
On 13.01.2025 20:31, Jim Burns wrote:
On 1/13/2025 12:17 PM, WM wrote:
[...]
A step is never from finite to infinite.
Therefore, a step never crosses ω
Therefore, a sum never crosses ω
Therefore, a product never crosses ω
Therefore, a power never crosses ω
All that is true in potential infinity,
however it is wrong in completed infinity.
therefore creates even numbers.
They do not fit below ω.
No.
They fit below ω
In completed infinity
all available places are occupied.
Half are new.
On 1/14/25 3:47 AM, WM wrote:
Therefore half of the results is larger than ω.
YOu *THINK* it can not, because
EVERY Natural Number is "Definable",
How is division of ordinals defined?
And nothing more needs to fit inbetween, as double evey Natural Number
is an already existing Natural Number, so none were created.
On 1/14/2025 4:07 AM, WM wrote:
On 13.01.2025 20:31, Jim Burns wrote:
On 1/13/2025 12:17 PM, WM wrote:
[...]
A step is never from finite to infinite.
There is no infinite set smaller than ℕ
#𝔼 ≥ #ℕ
That is obviously wrong.
There is no infinite set smaller than ℕ.
In each of our sets,
each of its elements is in the set,
each available place is occupied.
A potentiallyᵂᴹ infiniteˢᵉᵗ set,
the same as any other set,
has all available places occupied
and is completeᵂᴹ.
Half are new.
A step is never from finite to infinite.
On 14.01.2025 13:37, Richard Damon wrote:Yes, N \ {1, 2, 3, ...} = {}.
EVERY Natural Number is "Definable",Then remove the set ℕ by application of only definable numbers:
ℕ \ {1} \ {1, 2} \ {1, 2, 3} \ ...
On 15.01.2025 16:16, Jim Burns wrote:
On 1/14/2025 4:07 AM, WM wrote:
Half are new.
A step is never from finite to infinite.
∀n ∈ ℕ: 2n =/= n.
A step is never from finite to infinite.
∀n ∈ ℕ: 2n =/= n.
On 14.01.2025 13:37, Richard Damon wrote:
EVERY Natural Number is "Definable",
Then remove the set ℕ by application of only definable numbers:
ℕ \ {1} \ {1, 2} \ {1, 2, 3} \ ...
Regards, WM
On 15.01.2025 16:16, Jim Burns wrote:
On 1/14/2025 4:07 AM, WM wrote:
On 13.01.2025 20:31, Jim Burns wrote:
On 1/13/2025 12:17 PM, WM wrote:
therefore creates even numbers.
They do not fit below ω.
No.
They fit below ω
In completed infinity
all available places are occupied.
In each of our sets,
each of its elements is in the set,
each available place is occupied.
Therefore new numbers are not accepted.
A potentiallyᵂᴹ infiniteˢᵉᵗ set,
the same as any other set,
has all available places occupied
and is completeᵂᴹ.
Potential infinity is growing.
"In analysis we have to deal only
with the infinitely small and
the infinitely large
as a limit-notion,
as something becoming, emerging, produced,
i.e., as we put it, with the potential infinite.
But this is not the proper infinite.
That we have
for instance
when we consider
the entirety of the numbers 1, 2, 3, 4, ... itself
as a completed unit, or
the points of a line as
an entirety of things which is completely available.
That sort of infinity is named actual infinite."
[D. Hilbert: "Über das Unendliche",
Mathematische Annalen 95 (1925) p. 167]
On 1/15/2025 1:17 PM, WM wrote:
On 15.01.2025 16:16, Jim Burns wrote:
On 1/14/2025 4:07 AM, WM wrote:
On 13.01.2025 20:31, Jim Burns wrote:
On 1/13/2025 12:17 PM, WM wrote:
therefore creates even numbers.
They do not fit below ω.
No.
They fit below ω
In completed infinity
all available places are occupied.
In each of our sets,
each of its elements is in the set,
each available place is occupied.
Therefore new numbers are not accepted.
And all even numbers fit below ω
None are created.
A potentiallyᵂᴹ infiniteˢᵉᵗ set,
the same as any other set,
has all available places occupied
and is completeᵂᴹ.
Potential infinity is growing.
In each of our sets,
each element has an available space, and
only its elements have available spaces.
A place in a set is occupied by virtue of
its element being in the set.
In each of our sets,
each of its elements is in the set,
each available place is occupied.
A potentiallyᵂᴹ infiniteˢᵉᵗ set
has all available places occupied,
the same as any other set,
which is to say,
it is (has been, will be) completeᵂᴹ.
"In analysis we have to deal only
with the infinitely small and
the infinitely large
as a limit-notion,
as something becoming, emerging, produced,
i.e., as we put it, with the potential infinite.
But this is not the proper infinite.
That we have for instance
when we consider
the entirety of the numbers 1, 2, 3, 4, ... itself
as a completed unit, or
the points of a line as
an entirety of things which is completely available.
That sort of infinity is named actual infinite."
[D. Hilbert: "Über das Unendliche", Mathematische Annalen 95 (1925) p.
167]
A finite set has
emptier.by.one sets which are smaller.
For each finite set,
a finite ordinal larger than that set
exists.
For the set ℕ of all finite ordinals,
a finite ordinal larger than ℕ
doesn't exist.
Therefore,
the set ℕ of all finite ordinals
isn't itself finite, and,
unlike a finite set, ℕ doesn't have
emptier.by.one sets which are smaller.
----
Finite people are able to reason about infinity
by describing an indefinite one of infinitely.many
and then supplementing the descriptive claims
with visibly not.first.false claims.
As finite people,
we have not and _cannot_ witness
the infinitely.many described.
What we can witness, instead, are
the finitely.many finite.length claims themselves,
and witness the correctness of the description of
that which we are currently discussing,
and witness the not.first.false.ness of
the other claims.
Upon witnessing all that,
we know the claims are true.
On 14.01.2025 19:41, Jim Burns wrote:
On 1/14/2025 4:07 AM, WM wrote:
On 13.01.2025 20:31, Jim Burns wrote:
On 1/13/2025 12:17 PM, WM wrote:
[...]
A step is never from finite to infinite.
The dark realm is appears infinite.
There is no infinite set smaller than ℕ
#𝔼 ≥ #ℕ
That is obviously wrong.
The rule of subset proves that every proper subset has fewer elements than its superset. So there are more natural numbers than prime numbers, || > ||, and more complex numbers than real numbers, || > ||. Even
finitely many exceptions from the subset-relation are admitted for
infinite subsets. Therefore there are more odd numbers than prime
numbers || > ||.
The rule of construction yields the number of integers || = 2|| + 1 and the number of fractions || = 2||2 + 1 (there are fewer rational numbers # ). Since all products of rational numbers with an irrational number are irrational, there are many more irrational numbers than
rational numbers || > |#|.
The rule of symmetry yields precisely the same number of real
geometric points in every interval (n, n+1] and with at most a small
error same number of odd numbers and of even numbers in every finite
interval and in the whole real line.
There is no infinite set smaller than ℕ.
{2,3, 4, ...} is smaller by one element.
Regards, WM
Am Wed, 15 Jan 2025 18:58:40 +0100 schrieb WM:
On 14.01.2025 13:37, Richard Damon wrote:Yes, N \ {1, 2, 3, ...} = {}.
EVERY Natural Number is "Definable",Then remove the set ℕ by application of only definable numbers:
ℕ \ {1} \ {1, 2} \ {1, 2, 3} \ ...
On 1/15/2025 1:17 PM, WM wrote:
On 15.01.2025 16:16, Jim Burns wrote:
On 1/14/2025 4:07 AM, WM wrote:
Half are new.
A step is never from finite to infinite.
∀n ∈ ℕ: 2n =/= n.
A step is never from finite to infinite.
A potentiallyᵂᴹ infiniteˢᵉᵗ set
has all available places occupied,
the same as any other set,
which is to say,
it is (has been, will be) completeᵂᴹ.
"In analysis we have to deal only
with the infinitely small and
the infinitely large
as a limit-notion,
as something becoming, emerging, produced,
i.e., as we put it, with the potential infinite.
But this is not the proper infinite.
That we have for instance
when we consider
the entirety of the numbers 1, 2, 3, 4, ... itself
as a completed unit, or
the points of a line as
an entirety of things which is completely available.
That sort of infinity is named actual infinite."
[D. Hilbert: "Über das Unendliche", Mathematische Annalen 95 (1925) p.
167]
For the set ℕ of all finite ordinals,
a finite ordinal larger than ℕ
doesn't exist.
The potential infinity
itself isn't growing, our KNOWLEDGE of it grows as we generate its members.
On 14.01.2025 19:41, Jim Burns wrote:
On 1/14/2025 4:07 AM, WM wrote:
On 13.01.2025 20:31, Jim Burns wrote:
A step is never from finite to infinite.
The dark realm is appears infinite.
There is no infinite set smaller than ℕ
#𝔼 ≥ #ℕ
That is obviously wrong.
The rule of subset proves that
every proper subset has
fewer elements than its superset.
On 15.01.2025 20:54, joes wrote:
Am Wed, 15 Jan 2025 18:58:40 +0100 schrieb WM:
On 14.01.2025 13:37, Richard Damon wrote:Yes, N \ {1, 2, 3, ...} = {}.
EVERY Natural Number is "Definable",Then remove the set ℕ by application of only definable numbers:
ℕ \ {1} \ {1, 2} \ {1, 2, 3} \ ...
Of course. But what FISON can be dropped?
Regards, WM
On 16.01.2025 02:23, Jim Burns wrote:
On 1/15/2025 1:17 PM, WM wrote:
On 15.01.2025 16:16, Jim Burns wrote:
On 1/14/2025 4:07 AM, WM wrote:
On 13.01.2025 20:31, Jim Burns wrote:
On 1/13/2025 12:17 PM, WM wrote:
therefore creates even numbers.
They do not fit below ω.
No.
They fit below ω
In completed infinity
all available places are occupied.
In each of our sets,
each of its elements is in the set,
each available place is occupied.
Therefore new numbers are not accepted.
And all even numbers fit below ω
None are created.
A potentiallyᵂᴹ infiniteˢᵉᵗ set
has all available places occupied,
But it can grow.
Multiplying all its elements by 2
creates new elements.
the same as any other set,
which is to say,
it is (has been, will be) completeᵂᴹ.
Then new numbers must be outside.
On 1/15/2025 1:13 PM, WM wrote:
On 14.01.2025 19:41, Jim Burns wrote:
On 1/14/2025 4:07 AM, WM wrote:
On 13.01.2025 20:31, Jim Burns wrote:
A step is never from finite to infinite.
The dark realm is appears infinite.
Nowhere,
among what appears and
among what doesn't appear,
is there a step from finite to infinite.
Nowhere,
among what appears and
among what doesn't appear,
is there finite ω-1 and infinite (ω-1)+1
Also, more generally,
there is no infiniteˢᵉᵗ smaller than ℕ
⎛ For each finiteᵒʳᵈ k
⎜ ∃fₖ one.to.one: ⟦0,k⦆ ⇉ 𝕌
None of those finitesᵒʳᵈ is the size of ℕ
On 1/16/25 10:19 AM, WM wrote:
On 16.01.2025 13:27, Richard Damon wrote:
The potential infinity itself isn't growing, our KNOWLEDGE of it
grows as we generate its members.
But the knowledge of actually infinite sets isn't growing?
Depends how good you can think.
On 1/16/25 8:18 AM, WM wrote:
On 15.01.2025 20:54, joes wrote:Of course no FINITE (the F of FISON) will be the match for the INFINITE
Am Wed, 15 Jan 2025 18:58:40 +0100 schrieb WM:
On 14.01.2025 13:37, Richard Damon wrote:Yes, N \ {1, 2, 3, ...} = {}.
EVERY Natural Number is "Definable",Then remove the set ℕ by application of only definable numbers:
ℕ \ {1} \ {1, 2} \ {1, 2, 3} \ ...
Of course. But what FISON can be dropped?
set of Natural Numbers.
That doesn't mean the whole set of them fail, but of course, you need to
be using logic that CAN use *ALL* of them
On 1/16/2025 8:32 AM, WM wrote:
A potentiallyᵂᴹ infiniteˢᵉᵗ set
has all available places occupied,
But it can grow.
No, it can't.
Multiplying all its elements by 2
creates new elements.
No. it doesn't.
the same as any other set,
which is to say,
it is (has been, will be) completeᵂᴹ.
Then new numbers must be outside.
There are no new numbers.
A step is never from finite to infinite.
Therefore, a step never crosses ω
The fact that you think more can be created, just means you never
understood how to have all of them in the first place.
On 17.01.2025 01:37, Richard Damon wrote:"Logic" (by which you mean common sense) does not say you can drop all segments, only finitely many.
On 1/16/25 8:18 AM, WM wrote:
On 15.01.2025 20:54, joes wrote:Of course no FINITE (the F of FISON) will be the match for the INFINITE
Am Wed, 15 Jan 2025 18:58:40 +0100 schrieb WM:
On 14.01.2025 13:37, Richard Damon wrote:Yes, N \ {1, 2, 3, ...} = {}.
EVERY Natural Number is "Definable",Then remove the set ℕ by application of only definable numbers:
ℕ \ {1} \ {1, 2} \ {1, 2, 3} \ ...
Of course. But what FISON can be dropped?
set of Natural Numbers.
That doesn't mean the whole set of them fail, but of course, you need
to be using logic that CAN use *ALL* of them
Logic says that every FISON which is smaller than another FISON can be dropped. Logic says even more: Every FISON which is smaller than ℕ can
be dropped. Therefore all can be dropped.
Do you know a FISON that is smaller than ℕ but cannot be dropped?
On 17.01.2025 01:37, Richard Damon wrote:On the contrary. Only if you have finitely many, does the "doubling"
The fact that you think more can be created, just means you neverOnly if we have all of them and can double all of them then by 2n > n
understood how to have all of them in the first place.
greater numbers than all of them are created.
Am Fri, 17 Jan 2025 11:39:55 +0100 schrieb WM:
Logic says that every FISON which is smaller than another FISON can be"Logic" (by which you mean common sense) does not say you can drop all segments, only finitely many.
dropped. Logic says even more: Every FISON which is smaller than ℕ can
be dropped. Therefore all can be dropped.
Do you know a FISON that is smaller than ℕ but cannot be dropped?
Am Fri, 17 Jan 2025 11:42:02 +0100 schrieb WM:
On 17.01.2025 01:37, Richard Damon wrote:On the contrary. Only if you have finitely many, does the "doubling"
The fact that you think more can be created, just means you neverOnly if we have all of them and can double all of them then by 2n > n
understood how to have all of them in the first place.
greater numbers than all of them are created.
of the elements (not the set) result in larger numbers.
The results
might also not be larger than *every* input number.
On 17.01.2025 12:18, joes wrote:
Am Fri, 17 Jan 2025 11:39:55 +0100 schrieb WM:
Logic says that every FISON which is smaller than another FISON can be"Logic" (by which you mean common sense) does not say you can drop all
dropped. Logic says even more: Every FISON which is smaller than ℕ can >>> be dropped. Therefore all can be dropped.
Do you know a FISON that is smaller than ℕ but cannot be dropped?
segments, only finitely many.
By logic I mean logic. Since the small numbers are always covered by the greater FIS=ONs, the smaller FISONs can be dropped. Either the complete
set ℕ is produced by one FISON, which then is not finite and therefore
is not a FISON, or it is not produced by FISONs.
This can also be proved by induction: Every FISON is missing some
natural numbers and is therefore irrelevant for the union.
Regards, WM
On 16.01.2025 23:22, Jim Burns wrote:
Nowhere,
among what appears and
among what doesn't appear,
is there finite ω-1 and infinite (ω-1)+1
So it appears because ω and ω-1 are dark.
So it appears because ω and ω-1 are dark.
But if ω is assumed to exist,
then there is a set cotaining ω elements.
From this set one element can
be subtracted.
Also, more generally,
there is no infiniteˢᵉᵗ smaller than ℕ
Therefore ℕ \ {1} is finite.
But it appears infinite like all sets which
cannot be counted by FISONs.
On 1/17/25 5:39 AM, WM wrote:
On 17.01.2025 01:37, Richard Damon wrote:
On 1/16/25 8:18 AM, WM wrote:
On 15.01.2025 20:54, joes wrote:
Am Wed, 15 Jan 2025 18:58:40 +0100 schrieb WM:
Logic says that every FISON which is smaller than another FISON can be
dropped. Logic says even more: Every FISON which is smaller than ℕ can
be dropped. Therefore all can be dropped.
ANY can be dropped, but not ALL.
Do you know a FISON that is smaller than ℕ but cannot be dropped?n
No, and all that shows is that there isn't a "largest" FISON that has
all the Natural Numbers.
On 1/17/25 8:00 AM, WM wrote:All which are smaller than the greatest are useless. That should even be understood by a very limited brain. There is no greatest. Therefore all
By logic I mean logic. Since the small numbers are always covered by
the greater FIS=ONs, the smaller FISONs can be dropped. Either the
complete set ℕ is produced by one FISON, which then is not finite and
therefore is not a FISON, or it is not produced by FISONs.
It is produced by the INFINITE set of FISONs,
On 1/17/2025 4:08 AM, WM wrote:
On 16.01.2025 23:22, Jim Burns wrote:
Nowhere,
among what appears and
among what doesn't appear,
is there finite ω-1 and infinite (ω-1)+1
So it appears because ω and ω-1 are dark.
We never see ω and ω-1
We see descriptions of ω and ω-1
That is sufficient for knowledge of ω and ω-1
You (WM) introduce
negative cardinality (darkᵂᴹ numbers)
in an attempt to fit these claims together.
Matheologians don't use anything so fancy,
But it appears infinite like all sets which cannot be counted by FISONs.
"cannot be counted by FISONs"
and
"larger.than.any.finite"
sound similar.
However, matheologians don't have darkᵂᴹ sets
allegedly capable of backing a larger.than.any.finite set
down to a not.larger.than.any.finite set,
by _inserting_ elements.
That is why you (WM) think we're crazy or stupid,
negative cardinality.
On 1/17/25 5:50 AM, WM wrote:
On 17.01.2025 01:37, Richard Damon wrote:That "definition" violates to definition that set don't change.
On 1/16/25 10:19 AM, WM wrote:
On 16.01.2025 13:27, Richard Damon wrote:
The potential infinity itself isn't growing, our KNOWLEDGE of it
grows as we generate its members.
But the knowledge of actually infinite sets isn't growing?
Depends how good you can think.
No, it has nothing to do with your missing knowledge or your lack of
thinking capability. "Potential infinity refers to a procedure that
gets closer and closer to, but never quite reaches, an infinite end.
[...] Completed infinity, or actual infinity, is an infinity that one
actually reaches; the process is already done."[E. Schechter:
"Potential versus completed infinity: Its history and controversy" (5
Dec 2009)]
Some may talk of a growing set, but then you can't use any logic based
on "fixed" sets.
Am 17.01.2025 um 17:53 schrieb Jim Burns:
On 1/17/2025 4:08 AM, WM wrote:
On 16.01.2025 23:22, Jim Burns wrote:
Nowhere,
among what appears and
among what doesn't appear,
is there finite ω-1 and infinite (ω-1)+1
So it appears because ω and ω-1 are dark.
We never see ω and ω-1
We see descriptions of ω and ω-1
That is sufficient for knowledge of ω and ω-1
Dark numbers cannot be seen,
if you understand by that phrase
be put in a FISON.
You (WM) introduce
negative cardinality (darkᵂᴹ numbers)
in an attempt to fit these claims together.
No, I don't.
On 1/17/2025 2:40 PM, WM wrote:
Am 17.01.2025 um 17:53 schrieb Jim Burns:
On 1/17/2025 4:08 AM, WM wrote:
On 16.01.2025 23:22, Jim Burns wrote:
Nowhere,
among what appears and
among what doesn't appear,
is there finite ω-1 and infinite (ω-1)+1
So it appears because ω and ω-1 are dark.
We never see ω and ω-1
We see descriptions of ω and ω-1
That is sufficient for knowledge of ω and ω-1
Dark numbers cannot be seen,
if you understand by that phrase
be put in a FISON.
Definitions can be seen.
Finite sequences of claims, each claim of which
is true.or.not.first.false
can be seen.
----
The finite extends
much further than you (WM) think it does.
Infinitely further than you think it does.
No finite ordinal has
an infinite immediate successor.
You (WM) introduce
negative cardinality (darkᵂᴹ numbers)
in an attempt to fit these claims together.
No, I don't.
I'm willing to believe that
you didn't intend to introduce negative cardinality.
Nonetheless, you did.
A potentiallyᵂᴹ infinite set is larger.than.any.finite.
An actuallyᵂᴹ infinite set A isn't potentially infinite.
It isn't larger.than.any.finite.
There is a larger finite set F.
Actuallyᵂᴹ infinite A has
a potentiallyᵂᴹ infinite subset P
On 17.01.2025 15:52, Richard Damon wrote:And that is why set theory doesn't talk about "potential infinity".
On 1/17/25 5:50 AM, WM wrote:So it is. But if infinity is potential, then we cannot change this in
On 17.01.2025 01:37, Richard Damon wrote:That "definition" violates to definition that set don't change.
On 1/16/25 10:19 AM, WM wrote:No, it has nothing to do with your missing knowledge or your lack of
On 16.01.2025 13:27, Richard Damon wrote:Depends how good you can think.
The potential infinity itself isn't growing, our KNOWLEDGE of itBut the knowledge of actually infinite sets isn't growing?
grows as we generate its members.
thinking capability. "Potential infinity refers to a procedure that
gets closer and closer to, but never quite reaches, an infinite end.
[...] Completed infinity, or actual infinity, is an infinity that one
actually reaches; the process is already done."[E. Schechter:
"Potential versus completed infinity: Its history and controversy" (5
Dec 2009)]
order to keep set theory, but then set theory is wrong.
Some may talk of a growing set, but then you can't use any logic basedCorrect. If infinity is potential. set theory is wrong.
on "fixed" sets.
On 18.01.2025 00:08, Jim Burns wrote:
On 1/17/2025 2:40 PM, WM wrote:
Am 17.01.2025 um 17:53 schrieb Jim Burns:
On 1/17/2025 4:08 AM, WM wrote:
On 16.01.2025 23:22, Jim Burns wrote:
Nowhere,
among what appears and
among what doesn't appear,
is there finite ω-1 and infinite (ω-1)+1
So it appears because ω and ω-1 are dark.
We never see ω and ω-1
We see descriptions of ω and ω-1
That is sufficient for knowledge of ω and ω-1
Dark numbers cannot be seen,
if you understand by that phrase
be put in a FISON.
Definitions can be seen.
Yes,
dark numbers however can be handled only collectively.
That distinguishes them from visible numbers.
Finite sequences of claims, each claim of which
is true.or.not.first.false
can be seen.
Like the visible numbers.
Am 17.01.2025 um 15:52 schrieb Richard Damon:
On 1/17/25 8:00 AM, WM wrote:
All which are smaller than the greatest are useless. That should even be understood by a very limited brain. There is no greatest. Therefore allBy logic I mean logic. Since the small numbers are always covered by
the greater FIS=ONs, the smaller FISONs can be dropped. Either the
complete set ℕ is produced by one FISON, which then is not finite and
therefore is not a FISON, or it is not produced by FISONs.
It is produced by the INFINITE set of FISONs,
are useless.
Regards, WM
On 18.01.2025 00:08, Jim Burns wrote:
The finite extends
much further than you (WM) think it does.
Infinitely further than you think it does.
No.
As long as
you deny Bob's existence and violate logic
you are not a reliable source.
No finite ordinal has
an infinite immediate successor.
Maybe. But then there is no infinite ordinal.
On 1/18/2025 3:41 AM, WM wrote:
Definitions can be seen.
Yes,
dark numbers however can be handled only collectively.
That distinguishes them from visible numbers.
Definitions of visibleᵂᴹ.or.darkᵂᴹ numbers
are true claims about visibleᵂᴹ.or.darkᵂᴹ numbers
and do not distinguish them.
Visible.
On paper. In chalk. Or in glowing dots.
Carved into marble. Or into clay tablets.
⎛
⎜ Visibleᵂᴹ.or.darkᵂᴹ ordinals are
⎜ well.ordered.
⎜ Visibleᵂᴹ.or.darkᵂᴹ ordinals are
⎜ linearly ordered.
⎜
⎜ Darkᵂᴹ ordinals are linearly ordered
⎜ with respect to visibleᵂᴹ ordinals and
⎝ with respect to other darkᵂᴹ ordinals.
Somewhere Out There, there might be
a person thinking that
"This is darkᵂᴹ"
implies
"This is something we must remain ignorant of".
A person thinking that would be wrong.
On 1/17/25 2:34 PM, WM wrote:
Am 17.01.2025 um 15:52 schrieb Richard Damon:
On 1/17/25 8:00 AM, WM wrote:All which are smaller than the greatest are useless. That should even
By logic I mean logic. Since the small numbers are always covered by
the greater FIS=ONs, the smaller FISONs can be dropped. Either the
complete set ℕ is produced by one FISON, which then is not finite
and therefore is not a FISON, or it is not produced by FISONs.
It is produced by the INFINITE set of FISONs,
be understood by a very limited brain. There is no greatest. Therefore
all are useless.
So, you agree that your logic makes everything worthless.
THus, your false premise that without a biggest, the smaller are
worthless is just incorrect, and thus your conclusion is incorrect.
It may be that without being able to have a biggest Natural Number you
think that it makes that set worthless, but that is just your error.
Am Fri, 17 Jan 2025 22:56:13 +0100 schrieb WM:
Correct. If infinity is potential. set theory is wrong.And that is why set theory doesn't talk about "potential infinity".
WM presented the following explanation :
On 18.01.2025 12:03, joes wrote:
Am Fri, 17 Jan 2025 22:56:13 +0100 schrieb WM:
Correct. If infinity is potential. set theory is wrong.And that is why set theory doesn't talk about "potential infinity".
Nevertheless it uses potential infinity.
No, it doesn't.
All "bijections" yield the same cardinality because only the
potentially infinite parts of the sets are applied.
No, it is because these bijections show that some infinite sets' sizes
can be shown to be equal even if no completed count exists.
On 1/17/25 4:56 PM, WM wrote:
That "definition" violates to definition that set don't change.
So it is. But if infinity is potential, then we cannot change this in
order to keep set theory, but then set theory is wrong.
So, you are just agreeing that your logic is based on contradictory
premsises and thus is itself contradictory and worthless.
On 1/18/2025 3:41 AM, WM wrote:
On 18.01.2025 00:08, Jim Burns wrote:
The finite extends
much further than you (WM) think it does.
Infinitely further than you think it does.
No.
As long as
you deny Bob's existence and violate logic
you are not a reliable source.
https://en.wikipedia.org/wiki/Finite_set
⎛
⎜ Informally, a finite set is a set which
⎜ one could in principle count and finish counting.
There is no step from finite to infinite.
On 18.01.2025 14:46, Richard Damon wrote:
On 1/17/25 4:56 PM, WM wrote:
That "definition" violates to definition that set don't change.
So it is. But if infinity is potential, then we cannot change this in
order to keep set theory, but then set theory is wrong.
So, you are just agreeing that your logic is based on contradictory
premsises and thus is itself contradictory and worthless.
No, set theory claims actual infinity but in fact useses potential
infinity with its "bijections". They contain only natnumbers which have ℵ₀ successors. If all natural numbers were applied, there would not be successors:
ℕ \ {1, 2, 3, ...} = { }.
Regards, WM
On 19.01.2025 11:42, FromTheRafters wrote:Set theory doesn't use "potential infinity".
WM presented the following explanation :Use all natnumbers individually such that none remains. Fail.
On 18.01.2025 12:03, joes wrote:No, it doesn't.
Am Fri, 17 Jan 2025 22:56:13 +0100 schrieb WM:Nevertheless it uses potential infinity.
Correct. If infinity is potential. set theory is wrong.And that is why set theory doesn't talk about "potential infinity".
Quite the opposite.All "bijections" yield the same cardinality because only the
potentially infinite parts of the sets are applied.
The "complete count" is infinite.No, it is because these bijections show that some infinite sets' sizesThey appear equal because no completed count exists.
can be shown to be equal even if no completed count exists.
All natnumbers in bijections have ℵ₀ not applied successors.A bijection is not meant to be thought about sequentially?
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo Only potential infinity is applied.This is not infinity.
In actual infinity all natnumbers would be applied:It absolutely is. Just give a rule for every natural.
ℕ \ {1, 2, 3, ...} = { }
But that is not possible in bijections.
On 18.01.2025 19:38, Jim Burns wrote:That's not what that means. Some infinite sets are countable, even though
On 1/18/2025 3:41 AM, WM wrote:
On 18.01.2025 00:08, Jim Burns wrote:
The finite extends much further than you (WM) think it does.No.
Infinitely further than you think it does.
As long as you deny Bob's existence and violate logic you are not a
reliable source.
https://en.wikipedia.org/wiki/Finite_set ⎛
⎜ Informally, a finite set is a set which ⎜ one could in principle
count and finish counting.
Cantor claims this also for infinite sets: "The infinite sequence thus defined has the peculiar property to contain the positive rational
numbers completely, and each of them only once at a determined place."
[G. Cantor, letter to R. Lipschitz (19 Nov 1883)]
There is only a limit, which does have different properties.There is no step from finite to infinite.Not in the visible domain. But there is no loss in lossless exchange -
even in the dark domain. There lies your fault.
On 18.01.2025 14:46, Richard Damon wrote:Yes, with mutable sets.
On 1/17/25 4:56 PM, WM wrote:
So, you are just agreeing that your logic is based on contradictoryThat "definition" violates to definition that set don't change.So it is. But if infinity is potential, then we cannot change this in
order to keep set theory, but then set theory is wrong.
premsises and thus is itself contradictory and worthless.
No, set theory claims actual infinity but in fact useses potential"Set theory" uses neither.
infinity with its "bijections".
They contain only natnumbers which have ℵ₀ successors.There are no naturals with a finite number of successors, otherwise
If all natural numbers were applied, there would not beThere are no successors if only you would actually "apply" the infinitely
successors: ℕ \ {1, 2, 3, ...} = { }.
On 1/19/25 5:43 AM, WM wrote:
So, you admit that your FISONs are worthless?
Name any FISON that is required to produce ℕ by a union of FISONs.
There is non, because you need almost ALL of the FISONS to do that,
The set of required FISONs is well defined because for every FISON we
can decide whether it is required in the union.
But YOU can only do that for a finite number of them, not almost all of
them.
Further, according to set theory, every well defined set of ordinal
numbers has a first element. The FISONs F(n) must obey this theorem
because they can be ordered by their greates ordinals n.
You simply violate this fact.
Where?
The error is your handwaving claim that infinitely many FISONs are
required. Infinitely many failures will not yield a success.
And YOUR "handwaving" just shows your stupidity, and inability to
understand your stupidity.
I don't "Handwave" the claim of needing infinitely many FISONs,
Note, even in actual infinity, every Natural Number has Aleph_0
successors
Am Sun, 19 Jan 2025 11:52:33 +0100 schrieb WM:
Cantor claims this also for infinite sets: "The infinite sequence thusThat's not what that means. Some infinite sets are countable, even though
defined has the peculiar property to contain the positive rational
numbers completely, and each of them only once at a determined place."
[G. Cantor, letter to R. Lipschitz (19 Nov 1883)]
you don't "finish" them. The quote refers to a bijection.
There is only a limit, which does have different properties.There is no step from finite to infinite.Not in the visible domain. But there is no loss in lossless exchange -
even in the dark domain. There lies your fault.
WM formulated the question :
On 19.01.2025 11:42, FromTheRafters wrote:
WM presented the following explanation :
On 18.01.2025 12:03, joes wrote:
Am Fri, 17 Jan 2025 22:56:13 +0100 schrieb WM:
Correct. If infinity is potential. set theory is wrong.And that is why set theory doesn't talk about "potential infinity".
Nevertheless it uses potential infinity.
No, it doesn't.
Use all natnumbers individually such that none remains. Fail.
This makes no sense.
All "bijections" yield the same cardinality because only the
potentially infinite parts of the sets are applied.
No, it is because these bijections show that some infinite sets'
sizes can be shown to be equal even if no completed count exists.
They appear equal because no completed count exists.
No, they are the same size when it is shown there is at least one
bijection.
All natnumbers in bijections have ℵ₀ not applied successors.
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
Only potential infinity is applied.
You mean that only finite sets are involved.
Am Sun, 19 Jan 2025 11:47:50 +0100 schrieb WM:
They contain only natnumbers which have ℵ₀ successors.There are no naturals with a finite number of successors, otherwise
you could count backwards from the end.
If all natural numbers were applied, there would not beThere are no successors if only you would actually "apply" the infinitely many naturals.
successors: ℕ \ {1, 2, 3, ...} = { }.
On 19.01.2025 13:32, Richard Damon wrote:Wrong:
Note, even in actual infinity, every Natural Number has Aleph_0Then not all could be subtracted from ℕ.
successors
But that is possible ℕ \ {1, 2, 3, ...} = { }.
On 18.01.2025 19:38, Jim Burns wrote:
On 1/18/2025 3:41 AM, WM wrote:
On 18.01.2025 00:08, Jim Burns wrote:
The finite extends
much further than you (WM) think it does.
Infinitely further than you think it does.
No.
As long as
you deny Bob's existence and violate logic
you are not a reliable source.
https://en.wikipedia.org/wiki/Finite_set
⎛
⎜ Informally, a finite set is a set which
⎜ one could in principle count and finish counting.
Cantor claims this also for infinite sets:
"The infinite sequence thus defined has
the peculiar property to contain
the positive rational numbers completely, and
each of them only once
at a determined place."
[G. Cantor, letter to R. Lipschcitz c(19 Nov 1883)]
There is no step from finite to infinite.
Not in the visible domain.
But there is no loss in lossless exchange -
even in the dark domain.
There lies your fault.
WM explained :
On 19.01.2025 13:32, Richard Damon wrote:
Note, even in actual infinity, every Natural Number has Aleph_0
successors
Then not all could be subtracted from ℕ. But that is possible ℕ \ {1, >> 2, 3, ...} = { }.
All that really says is that
the difference set between the set of natural numbers and the set of
natural numbers is empty.
It happens that WM formulated :
On 19.01.2025 16:10, joes wrote:
Am Sun, 19 Jan 2025 11:47:50 +0100 schrieb WM:
They contain only natnumbers which have ℵ₀ successors.There are no naturals with a finite number of successors, otherwise
you could count backwards from the end.
That is possible: ω-1, ω-2, ω-3, ...
Wrong, omega has no immediate predecessor.
On 1/19/2025 5:52 AM, WM wrote:
⎜ Informally, a finite set is a set which
⎜ one could in principle count and finish counting.
Cantor claims this also for infinite sets:
The finite extends
much further than you (WM) think it does.
Infinitely further than you think it does.
"The infinite sequence thus defined has
the peculiar property to contain
the positive rational numbers completely, and
each of them only once
at a determined place."
[G. Cantor, letter to R. Lipschcitz c(19 Nov 1883)]
There is no step from finite to infinite.
Not in the visible domain.
How do we know "not in visible domain"?
Yes,
there is no step from finite to infinite
in the visibleᵂᴹ domain.
As well,
there is no step from finite to infinite
in the visibleᵂᴹ.or.darkᵂᴹ domain.
There is no step from finite to infinite
anywhere.
But there is no loss in lossless exchange - even in the dark domain.
In the visibleᵂᴹ.or.darkᵂᴹ domain:
⎛
⎜ There are no sets A ≠ Aᵃ and B ≠ Bᵇ such that
⎜⎛ A is smaller than B but
⎜⎝ Aᵃ is not.smaller than Bᵇ
On 19.01.2025 14:29, FromTheRafters wrote:
WM formulated the question :
On 19.01.2025 11:42, FromTheRafters wrote:
WM presented the following explanation :
On 18.01.2025 12:03, joes wrote:
Am Fri, 17 Jan 2025 22:56:13 +0100 schrieb WM:Nevertheless it uses potential infinity.
Correct. If infinity is potential. set theory is wrong.And that is why set theory doesn't talk about "potential infinity". >>>>>
No, it doesn't.
Use all natnumbers individually such that none remains. Fail.
This makes no sense.
It is impossible.
All "bijections" yield the same cardinality because only the
potentially infinite parts of the sets are applied.
No, it is because these bijections show that some infinite sets'
sizes can be shown to be equal even if no completed count exists.
They appear equal because no completed count exists.
No, they are the same size when it is shown there is at least one
bijection.
Every element of the bijection has almost all elements as successors. Therefore the bijection is none.
All natnumbers in bijections have ℵ₀ not applied successors.
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
Only potential infinity is applied.
You mean that only finite sets are involved.
Of course Infinitely many successors prevent that their predecessors are infinitely many.
Regards, WM
On 20.01.2025 15:31, FromTheRafters wrote:This is not theology.
It happens that WM formulated :You cannot see the whole sequence from 1 to ω-1. But there are more
On 19.01.2025 16:10, joes wrote:Wrong, omega has no immediate predecessor.
Am Sun, 19 Jan 2025 11:47:50 +0100 schrieb WM:That is possible: ω-1, ω-2, ω-3, ...
They contain only natnumbers which have ℵ₀ successors.There are no naturals with a finite number of successors, otherwise
you could count backwards from the end.
things in heaven and earth than you can see.
On 20.01.2025 15:22, FromTheRafters wrote:It does not mention nonexistent numbers.
WM explained :It says that all natural numbers without ℵ₀ successors can be handled.
On 19.01.2025 13:32, Richard Damon wrote:All that really says is that the difference set between the set of
Note, even in actual infinity, every Natural Number has Aleph_0Then not all could be subtracted from ℕ. But that is possible ℕ \ {1, >>> 2, 3, ...} = { }.
successors
natural numbers and the set of natural numbers is empty.
This is different for definable natural numbers because there not allYes they can. The number of naturals is not a natural. Why do you
natural numbers can be handled:
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
On 1/20/25 7:33 AM, WM wrote:
On 19.01.2025 14:29, FromTheRafters wrote:
WM formulated the question :
On 19.01.2025 11:42, FromTheRafters wrote:
WM presented the following explanation :
On 18.01.2025 12:03, joes wrote:
Am Fri, 17 Jan 2025 22:56:13 +0100 schrieb WM:Nevertheless it uses potential infinity.
Correct. If infinity is potential. set theory is wrong.And that is why set theory doesn't talk about "potential infinity". >>>>>>
No, it doesn't.
Use all natnumbers individually such that none remains. Fail.
This makes no sense.
It is impossible.
Because logic that insists on dealing with an INFINITE set one by one is illogical
Every element of the bijection has almost all elements as successors.
Therefore the bijection is none.
Nope, the logic that can't see the completion at infinity is broken.
On 20.01.2025 19:07, Richard Damon wrote:
On 1/20/25 7:33 AM, WM wrote:
On 19.01.2025 14:29, FromTheRafters wrote:
WM formulated the question :
On 19.01.2025 11:42, FromTheRafters wrote:
WM presented the following explanation :
On 18.01.2025 12:03, joes wrote:
Am Fri, 17 Jan 2025 22:56:13 +0100 schrieb WM:Nevertheless it uses potential infinity.
Correct. If infinity is potential. set theory is wrong.And that is why set theory doesn't talk about "potential infinity". >>>>>>>
No, it doesn't.
Use all natnumbers individually such that none remains. Fail.
This makes no sense.
It is impossible.
Because logic that insists on dealing with an INFINITE set one by one
is illogical
Yes. Therefore only the elements of a (potentially in-) finite set can
be dealt with individually, i.e., one by one.
Every element of the bijection has almost all elements as successors.
Therefore the bijection is none.
Nope, the logic that can't see the completion at infinity is broken.
You contradict yourself. Bijections need individual elements.
Regards, WM
the matching happens as a
collective whole. It can't complete if you don't move from the
individual step to the collective step.
On 21.01.2025 13:17, Richard Damon wrote:
the matching happens as a collective whole. It can't complete if you
don't move from the individual step to the collective step.
So it is.
Regards, WM
On 1/21/25 7:48 AM, WM wrote:
On 21.01.2025 13:17, Richard Damon wrote:And thus,
the matching happens as a collective whole. It can't complete if you
don't move from the individual step to the collective step.
So it is.
On 22.01.2025 00:41, Richard Damon wrote:
On 1/21/25 7:48 AM, WM wrote:
On 21.01.2025 13:17, Richard Damon wrote:And thus,
the matching happens as a collective whole. It can't complete if you
don't move from the individual step to the collective step.
So it is.
there are visible and dark numbers. As you said: We have to move from
the individual step to the collective step.
Regards, WM
On 1/22/25 5:36 AM, WM wrote:
On 22.01.2025 00:41, Richard Damon wrote:But the numbers being used didn't change, only the logic.
On 1/21/25 7:48 AM, WM wrote:
On 21.01.2025 13:17, Richard Damon wrote:And thus,
the matching happens as a collective whole. It can't complete if
you don't move from the individual step to the collective step.
So it is.
there are visible and dark numbers. As you said: We have to move from
the individual step to the collective step.
I guess when we talk about the set { 0, 1, 2, 3, ... } you must be
defining "4" as a "dark number" as that was the point we shifted from
listing them individually to collectively.
The basic problem with your "dark numbers" is that no number itself is "dark", the collective nature is just how we chose to use that number.
On 22.01.2025 13:10, Richard Damon wrote:
On 1/22/25 5:36 AM, WM wrote:
On 22.01.2025 00:41, Richard Damon wrote:But the numbers being used didn't change, only the logic.
On 1/21/25 7:48 AM, WM wrote:
On 21.01.2025 13:17, Richard Damon wrote:And thus,
the matching happens as a collective whole. It can't complete if
you don't move from the individual step to the collective step.
So it is.
there are visible and dark numbers. As you said: We have to move from
the individual step to the collective step.
Numbers do not change. Only their state of being known.
I guess when we talk about the set { 0, 1, 2, 3, ... } you must be
defining "4" as a "dark number" as that was the point we shifted from
listing them individually to collectively.
No. Every number that is defined in a system by its FISON is visible.
Many numbers smaller than 10^99 are defined on the pocket calculator. No greater number can be defined in that system.
The basic problem with your "dark numbers" is that no number itself is
"dark", the collective nature is just how we chose to use that number.
It is how we *can* chose them.
Regards, WM
On 1/23/25 3:32 AM, WM wrote:
Numbers do not change. Only their state of being known.
So, why do you claim they changed?
You seem to think that you can't use "visible" numbers collectively,
they become dark when you do, nor can your "dark" numbers be used individually,
No. Every number that is defined in a system by its FISON is visible.
Many numbers smaller than 10^99 are defined on the pocket calculator.
No greater number can be defined in that system.
Numbers are not definied by its "FISON", its FISON is defined by the
number.
ALL Natural Numbers are defined, and thus visible, and not "dark"
There are an infinite set of FISONs, one for every Natural Number.
On 23.01.2025 13:01, Richard Damon wrote:
On 1/23/25 3:32 AM, WM wrote:
Numbers do not change. Only their state of being known.
So, why do you claim they changed?
I did not.
You seem to think that you can't use "visible" numbers collectively,
they become dark when you do, nor can your "dark" numbers be used
individually,
You have not understood anything. All numbers can be used collectively
but visible numbers can be used as individuals.
No. Every number that is defined in a system by its FISON is visible.
Many numbers smaller than 10^99 are defined on the pocket calculator.
No greater number can be defined in that system.
Numbers are not definied by its "FISON", its FISON is defined by the
number.
Both is correct for visible numbers.
ALL Natural Numbers are defined, and thus visible, and not "dark"
Up to every FISON |ℕ \ {1, 2, 3, ..., n}| = ℵo. Since every FISON is the union of all its predecessors we get
F(n): |ℕ \ UF(n)| = ℵo.
If you don't believe in the union of all F(n), find the first exception.
There are an infinite set of FISONs, one for every Natural Number.
The union of all FISONs does not cover ℕ. Otherwise Cantor's theorem
would require the existence of a first necessary FISON.
Regards, WM
On 18.01.2025 14:46, Richard Damon wrote:
On 1/17/25 4:56 PM, WM wrote:
That "definition" violates to definition that set don't change.
So it is. But if infinity is potential, then we cannot change this in
order to keep set theory, but then set theory is wrong.
So, you are just agreeing that your logic is based on contradictory
premsises and thus is itself contradictory and worthless.
No, set theory claims actual infinity but in fact useses potential
infinity with its "bijections". They contain only natnumbers which have ℵ₀ successors. If all natural numbers were applied, there would not be successors:
ℕ \ {1, 2, 3, ...} = { }.
Regards, WM
no
finite set of numbers is teh full set of the Natural Numbers.
The union of all FISONs does not cover ℕ. Otherwise Cantor's theorem
would require the existence of a first necessary FISON.
Sure it does, but you need to use an INFINITE set of them,
On 1/19/25 5:47 AM, WM wrote:
On 18.01.2025 14:46, Richard Damon wrote:
On 1/17/25 4:56 PM, WM wrote:
That "definition" violates to definition that set don't change.
So it is. But if infinity is potential, then we cannot change this
in order to keep set theory, but then set theory is wrong.
So, you are just agreeing that your logic is based on contradictory
premsises and thus is itself contradictory and worthless.
No, set theory claims actual infinity but in fact useses potential
infinity with its "bijections". They contain only natnumbers which
have ℵ₀ successors. If all natural numbers were applied, there would
not be successors:
ℕ \ {1, 2, 3, ...} = { }.
No, set theory claims that the set is infinite.
you are trying to use a non-set compatible distinction between actual
and potential infinity
On 24.01.2025 13:29, Richard Damon wrote:No, the successors MAKE it infinite.
On 1/19/25 5:47 AM, WM wrote:
On 18.01.2025 14:46, Richard Damon wrote:
On 1/17/25 4:56 PM, WM wrote:
That "definition" violates to definition that set don't change.
So it is. But if infinity is potential, then we cannot change this
in order to keep set theory, but then set theory is wrong.
So, you are just agreeing that your logic is based on contradictory
premsises and thus is itself contradictory and worthless.
No, set theory claims actual infinity but in fact useses potential
infinity with its "bijections". They contain only natnumbers which
have ℵ₀ successors. If all natural numbers were applied, there would >>> not be successors:
ℕ \ {1, 2, 3, ...} = { }.
No, set theory claims that the set is infinite.
But it is only potentially infinite. ℵo successors prevent actual
infinity.
Why don't you use them?you are trying to use a non-set compatible distinction between actualI prove it. Only finite numbers can be chosen individually.
and potential infinity
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
To have infinitely many would require to use also the ℵo successors.
On 24.01.2025 13:29, Richard Damon wrote:
On 1/19/25 5:47 AM, WM wrote:
On 18.01.2025 14:46, Richard Damon wrote:
On 1/17/25 4:56 PM, WM wrote:
That "definition" violates to definition that set don't change.
So it is. But if infinity is potential, then we cannot change this
in order to keep set theory, but then set theory is wrong.
So, you are just agreeing that your logic is based on contradictory
premsises and thus is itself contradictory and worthless.
No, set theory claims actual infinity but in fact useses potential
infinity with its "bijections". They contain only natnumbers which
have ℵ₀ successors. If all natural numbers were applied, there would >>> not be successors:
ℕ \ {1, 2, 3, ...} = { }.
No, set theory claims that the set is infinite.
But it is only potentially infinite. ℵo successors prevent actual infinity.
you are trying to use a non-set compatible distinction between actual
and potential infinity
I prove it. Only finite numbers can be chosen individually.
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
To have infinitely many would require to use also the ℵo successors.
Regards, WM
Am Sat, 25 Jan 2025 12:14:18 +0100 schrieb WM:
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.Why don't you use them?
To have infinitely many would require to use also the ℵo successors.
Give the first necessary FISON which according to Cantor's theorem
exists in case that their union is ℕ.
The question is based on a false premise.
On 25.01.2025 15:16, Richard Damon wrote:
Give the first necessary FISON which according to Cantor's theorem
exists in case that their union is ℕ.
The question is based on a false premise.
Go through the sequence of all FISONs. Every not necessary FISON can be discarded, one after the other. Otherwise it would be necessary. What remains? Nothing. In the same way every not sufficient FISON can be wasted.
Regards, WM
On 25.01.2025 14:09, joes wrote:Mais non, there are Aleph_0 FISONs!
Am Sat, 25 Jan 2025 12:14:18 +0100 schrieb WM:
They are outside of FISONs F(n) = {1, 2, 3, ..., n}.∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.Why don't you use them?
To have infinitely many would require to use also the ℵo successors.
Note the universal quantifier in ∀n ∈ UF(n): |ℕ \ {1, 2, 3, ..., n}| = ℵoYeah, what about it? It’s not inside the set to be subtracted.
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