WM <invalid@no.org> wrote:
What I think you're trying to say is that the sum of all terms
containing both 8 and 9 diverges. It is far from clear that this is the case.
We can continue and remove all terms containing 1, 2, 3, 4, 5, 6, 7, 8,
9, 0 in the denominator without changing this.
What do you mean by "this"? What does it refer to?
That means that only the terms containing all these digits together
constitute the diverging series. (*)
There are many diverging sub-series possible. I think you mean "a"
diverging series.
But that's not the end! We can remove any number, like 2025, ....
From what? 2025 isn't a digit.
.... and the remaining series will converge. For proof use base 2026.
This extends to every definable number.
Meaningless. "Definable number" is itself undefined.
The harmonic series diverges. Kempner has shown in 1914 that when all
terms containing the digit 9 are removed, the series converges.
That means that the terms containing 9 diverge. Same is true when all
terms containing 8 are removed. That means all terms containing 8 and 9 simultaneously diverge.
We can continue and remove all terms containing 1, 2, 3, 4, 5, 6, 7, 8,
9, 0 in the denominator without changing this.
That means that only the terms containing all these digits together constitute the diverging series. (*)
But that's not the end! We can remove any number, like 2025, ....
.... and the remaining series will converge. For proof use base 2026.
This extends to every definable number.
Therefore the diverging part of the harmonic series is constituted
only by terms containing a digit sequence of all definable numbers.
Note that here not only the first terms are cut off but that many
following terms are excluded from the diverging remainder.
This is a proof of the huge set of undefinable or dark numbers.
(*) At this point the diverging series starts with the smallest term 1023456789 and contains further terms like 1203456789 or 1234567891010
or 123456789111 or 1234567891011. Only those containing the digit
sequence 10 will survive the next step, and only those containing the
digit sequence 1234567891011 (where the order of the first nine digits
is irrelevant) will survive the next step.
Regards, WM
On 12.03.2025 11:22, Alan Mackenzie wrote:
Meaningless. "Definable number" is itself undefined.
Definition: A natural number is "named" or "addressed" or "identified"
or "(individually) defined" or "instantiated" if it can be communicated, necessarily by a finite amount of information, in the sense of
Poincaré[1], such that sender and receiver understand the same and can
link it by a finite initial segment (1, 2, 3, ..., n) of natural numbers
to the origin 0. All other natural numbers are called dark natural numbers.
Communication can occur
- by direct description in the unary system like ||||||| or as many
beeps, raps, or flashes,
- by a finite initial segment of natural numbers (1, 2, 3, 4, 5, 6, 7),
- as n-ary representation, for instance binary 111 or decimal 7,
- by indirect description like "the number of colours of the rainbow",
- by other words known to sender and receiver like "seven".
[1] "In my opinion a subject is only conceivable if it can be defined
by a finite number of words."
Regards, WM
WM <wolfgang.mueckenheim@tha.de> wrote:
On 12.03.2025 11:22, Alan Mackenzie wrote:
Meaningless. "Definable number" is itself undefined.
Definition: A natural number is "named" or "addressed" or "identified"
or "(individually) defined" or "instantiated" if it can be communicated,
necessarily by a finite amount of information, in the sense of
Poincaré[1], such that sender and receiver understand the same and can
link it by a finite initial segment (1, 2, 3, ..., n) of natural numbers
to the origin 0. All other natural numbers are called dark natural numbers.
This is bullshit.
Communication can occur
- by direct description in the unary system like ||||||| or as many
beeps, raps, or flashes,
- by a finite initial segment of natural numbers (1, 2, 3, 4, 5, 6, 7),
- as n-ary representation, for instance binary 111 or decimal 7,
- by indirect description like "the number of colours of the rainbow",
- by other words known to sender and receiver like "seven".
Your "dark numbers" have no part in mathematics, don't exist, and can't exist. A proof, which I've given to you before, is as follows:
1. Assume that "dark numbers" exist.
2. Every non-empty set of natural numbers contains a least element.
3. The least element of the set of dark numbers, by its very
definition, has been "named", "addressed", "defined", and
"instantiated".
Jim has supplied at least one other proof.
On 12.03.2025 13:12, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 12.03.2025 11:22, Alan Mackenzie wrote:
Meaningless. "Definable number" is itself undefined.
Definition: A natural number is "named" or "addressed" or
"identified" or "(individually) defined" or "instantiated" if it can
be communicated, necessarily by a finite amount of information, in
the sense of Poincaré[1], such that sender and receiver understand
the same and can link it by a finite initial segment (1, 2, 3, ...,
n) of natural numbers to the origin 0. All other natural numbers are
called dark natural numbers.
This is bullshit.
Perhaps in your head.
Communication can occur
- by direct description in the unary system like ||||||| or as many
beeps, raps, or flashes,
- by a finite initial segment of natural numbers (1, 2, 3, 4, 5, 6, 7), >>> - as n-ary representation, for instance binary 111 or decimal 7,
- by indirect description like "the number of colours of the rainbow", >>> - by other words known to sender and receiver like "seven".
Your "dark numbers" have no part in mathematics, don't exist, and can't
exist. A proof, which I've given to you before, is as follows:
1. Assume that "dark numbers" exist.
Wrong.
2. Every non-empty set of natural numbers contains a least element.
If the numbers are definable.
Learn what potential infinity is.
3. The least element of the set of dark numbers, by its very
definition, has been "named", "addressed", "defined", and
"instantiated".
Try to remove all numbers individually from the harmonic series such
that none remains. If you can't, find the first one which resists.
Jim has supplied at least one other proof.
He claims that lossless exchange can produce losses. He is in
contradiction with logic.
Regards, WM
Alan Mackenzie brought next idea :
WM <wolfgang.mueckenheim@tha.de> wrote:
Jim has supplied at least one other proof.
Haha, at least! :D
WM <wolfgang.mueckenheim@tha.de> wrote:
Jim has supplied at least one other proof.
WM <wolfgang.mueckenheim@tha.de> wrote:
If the numbers are definable.
Meaningless. Or are you admitting that your "dark numbers" aren't
natural numbers after all?
Learn what potential infinity is.
I know what it is. It's an outmoded notion of infinity, popular in the 1880s, but which is entirely unneeded in modern mathematics.
3. The least element of the set of dark numbers, by its very
definition, has been "named", "addressed", "defined", and
"instantiated".
So you counter my proof by silently snipping elements 4, 5 and 6 of it? That's not a nice thing to do.
Try to remove all numbers individually from the harmonic series such
that none remains. If you can't, find the first one which resists.
Why should I want to do that?
Jim has supplied at least one other proof.
He claims that lossless exchange can produce losses. He is in
contradiction with logic.
Irrelevant to the current discussion. He has supplied at least one other proof of the non-existence of "dark numbers".
On 12.03.2025 13:12, Alan Mackenzie wrote:
Jim has supplied at least one other proof.
He claims that lossless exchange can produce losses.
On 3/12/2025 9:21 AM, WM wrote:
He claims that lossless exchange can produce losses.
I claim that the set of all and only finite ordinals
holds all and only finite ordinals.
WM <wolfgang.mueckenheim@tha.de> wrote:
If you were able to learn, then you would have the chance here:
ℕ \ {1} = ℵo
Where do you get that from? You're trying to say a subset of N is
identical to the first transfinite cardinal. That cannot be true.
and if ℕ \ {1, 2, 3, ..., n} = ℵo>> then ℕ \ {1, 2, 3, ..., n+1} = ℵo.
Where do you get that from? It is clearly false - the first of these
sets contains an element, n+1, that the second one doesn't. Therefore
they are distinct sets.
Induction cannot cover all natural numbers but only less than remain
uncovered.
The second part of that sentence is gibberish. Nobody has been talking
about "uncovering" numbers, whatever that might mean. [...]
On 12.03.2025 18:42, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
If the numbers are definable.
Meaningless. Or are you admitting that your "dark numbers" aren't
natural numbers after all?
They
Learn what potential infinity is.
I know what it is. It's an outmoded notion of infinity, popular in the
1880s, but which is entirely unneeded in modern mathematics.
That makes "modern mathematics" worthless.
3. The least element of the set of dark numbers, by its very
definition, has been "named", "addressed", "defined", and
"instantiated".
So you counter my proof by silently snipping elements 4, 5 and 6 of it?
That's not a nice thing to do.
They were based on the mistaken 3 and therefore useless.
Try to remove all numbers individually from the harmonic series such
that none remains. If you can't, find the first one which resists.
Why should I want to do that?
In order to experience that dark numbers exist and can't be manipulated.
Jim has supplied at least one other proof.
He claims that lossless exchange can produce losses. He is in
contradiction with logic.
Irrelevant to the current discussion. He has supplied at least one other
proof of the non-existence of "dark numbers".
As invalid as yours.
If you were able to learn, then you would have the chance here:
ℕ \ {1} = ℵo
and if ℕ \ {1, 2, 3, ..., n} = ℵo
then ℕ \ {1, 2, 3, ..., n+1} = ℵo.
Induction cannot cover all natural numbers but only less than remain uncovered.
Regards, WM
On 12.03.2025 20:20, Jim Burns wrote:
On 3/12/2025 9:21 AM, WM wrote:
He claims that lossless exchange can produce losses.
I claim that the set of all and only finite ordinals
holds all and only finite ordinals.
You claim that
the lossless exchange of X and O could delete an O.
That is counter logic.
On 12.03.2025 20:20, Jim Burns wrote:It actually isn't. We are doing an infinite number of exchanges.
On 3/12/2025 9:21 AM, WM wrote:
He claims that lossless exchange can produce losses.
I claim that the set of all and only finite ordinals holds all and only
finite ordinals.
You claim that the lossless exchange of X and O could delete an O. That
is counter logic.
Am 12.03.2025 um 22:31 schrieb Alan Mackenzie:
WM <wolfgang.mueckenheim@tha.de> wrote:
If you were able to learn, then you would have the chance here:
ℕ \ {1} = ℵo
Where do you get that from? You're trying to say a subset of N is
identical to the first transfinite cardinal. That cannot be true.
Should read: |ℕ \ {1}| = ℵo
and if ℕ \ {1, 2, 3, ..., n} = ℵo>> then ℕ \ {1, 2, 3, ..., n+1} = ℵo.
Where do you get that from? It is clearly false - the first of these
sets contains an element, n+1, that the second one doesn't. Therefore
they are distinct sets.
Should read: and if |ℕ \ {1, 2, 3, ..., n}| = ℵo then |ℕ \ {1, 2, 3, ..., n+1}| = ℵo.
Induction cannot cover all natural numbers but only less than remain
uncovered.
WM <wolfgang.mueckenheim@tha.de> wrote:
On 12.03.2025 18:42, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
If the numbers are definable.
Meaningless. Or are you admitting that your "dark numbers" aren't
natural numbers after all?
They
They?
Learn what potential infinity is.
I know what it is. It's an outmoded notion of infinity, popular in the
1880s, but which is entirely unneeded in modern mathematics.
That makes "modern mathematics" worthless.
What do you know about modern mathematics?
others in another recent thread to cite some mathematical result where
the notion of potential/actual infinity made a difference. There came no coherent reply (just one from Ross Finlayson I couldn't make head nor
tail of). Potential infinity isn't helpful and isn't needed anymore.
3. The least element of the set of dark numbers, by its very
definition, has been "named", "addressed", "defined", and
"instantiated".
So you counter my proof by silently snipping elements 4, 5 and 6 of it?
That's not a nice thing to do.
They were based on the mistaken 3 and therefore useless.
You didn't point out any mistake in 3. I doubt you can.
Try to remove all numbers individually from the harmonic series such
that none remains. If you can't, find the first one which resists.
Why should I want to do that?
In order to experience that dark numbers exist and can't be manipulated.
Dark numbers don't exist, as Jim and I have proven.
Induction cannot cover all natural numbers but only less than remain
uncovered.
The second part of that sentence is gibberish. Nobody has been talking
about "uncovering" numbers, whatever that might mean. Induction
encompasses all natural numbers. Anything it doesn't cover is not a
natural number, by definition.
On 3/12/2025 4:14 PM, WM wrote:
A single (lossless) exchange cannot delete an O
Finitely.many (lossless) exchanges cannot delete an O
Infinitely.many (lossless) exchanges can delete an O
Infinite is not _only_ bigger. It's different.
Am Wed, 12 Mar 2025 21:14:10 +0100 schrieb WM:
You claim that the lossless exchange of X and O could delete an O. ThatIt actually isn't. We are doing an infinite number of exchanges.
is counter logic.
On 12.03.2025 22:31, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 12.03.2025 18:42, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
Learn what potential infinity is.
I know what it is. It's an outmoded notion of infinity, popular in the >>>> 1880s, but which is entirely unneeded in modern mathematics.
That makes "modern mathematics" worthless.
What do you know about modern mathematics?
I know that it is self-contradictory because it cannot distinguish
potential and actual infinity.
When |ℕ| \ |{1, 2, 3, ..., n}| = ℵo, ....
.... then |ℕ| \ |{1, 2, 3, ..., n+1}| = ℵo. This holds for all elements of the inductive set, i.e., all FISONs F(n) or numbers n which have
more successors than predecessors.
Only those contribute to the inductive set!
Modern mathematics must claim that contrary to the definition ℵo
vanishes to 0 because ℕ \ {1, 2, 3, ...} = { }. That is blatantly
wrong and shows that modern mathematicians believe in miracles.
Matheology.
You may recall me challenging others in another recent thread to cite
some mathematical result where the notion of potential/actual infinity
made a difference. There came no coherent reply (just one from Ross
Finlayson I couldn't make head nor tail of). Potential infinity isn't
helpful and isn't needed anymore.
3. The least element of the set of dark numbers, by its very
definition, has been "named", "addressed", "defined", and
"instantiated".
It is named but has no FISON. That is the crucial condition.
So you counter my proof by silently snipping elements 4, 5 and 6 of it? >>>> That's not a nice thing to do.
They were based on the mistaken 3 and therefore useless.
You didn't point out any mistake in 3. I doubt you can.
I told you that potential infinity has no last element, therefore there
is no first dark number.
Try to remove all numbers individually from the harmonic series such >>>>> that none remains. If you can't, find the first one which resists.
Why should I want to do that?
In order to experience that dark numbers exist and can't be manipulated.
Dark numbers don't exist, as Jim and I have proven.
When |ℕ| \ |{1, 2, 3, ..., n}| = ℵo, then |ℕ| \ |{1, 2, 3, ..., n+1}| =
ℵo. How do the ℵo dark numbers get visible?
Induction cannot cover all natural numbers but only less than remain
uncovered.
The second part of that sentence is gibberish. Nobody has been talking
about "uncovering" numbers, whatever that might mean. Induction
encompasses all natural numbers. Anything it doesn't cover is not a
natural number, by definition.
Every defined number leaves ℵo undefined numbers. Try to find a counterexample. Fail.
Regards, WM
WM pretended :
On 13.03.2025 01:43, Jim Burns wrote:
On 3/12/2025 4:14 PM, WM wrote:
A single (lossless) exchange cannot delete an O
Finitely.many (lossless) exchanges cannot delete an O
Infinitely.many (lossless) exchanges can delete an O
No.
Infinite is not _only_ bigger. It's different.
It obeys logic or it is of no value.
You spelled intuition wrong.
Moebius <invalid@example.invalid> wrote:
Should read: |ℕ \ {1}| = ℵo
Ah. Thanks! If only people could write what they meant.
As you mentioned, "\" (usually) is used for _sets_, but not for cardinal numbers.Should read: and [for all n e ℕ:] if |ℕ \ {1, 2, 3, ..., n}| = ℵo then |ℕ \ {1, 2, 3,
..., n+1}| = ℵo.
OK.
Am 12.03.2025 um 22:31 schrieb Alan Mackenzie:
WM <wolfgang.mueckenheim@tha.de> wrote:
If you were able to learn, then you would have the chance here:
ℕ \ {1} = ℵo
Where do you get that from? You're trying to say a subset of N is
identical to the first transfinite cardinal. That cannot be true.
Should read: |ℕ \ {1}| = ℵo
and if ℕ \ {1, 2, 3, ..., n} = ℵo>> then ℕ \ {1, 2, 3, ..., n+1} = ℵo.
Where do you get that from? It is clearly false - the first of these
sets contains an element, n+1, that the second one doesn't. Therefore
they are distinct sets.
Should read: and if |ℕ \ {1, 2, 3, ..., n}| = ℵo then |ℕ \ {1, 2, 3, ..., n+1}| = ℵo.
Induction cannot cover all natural numbers but only less than remain
uncovered.
Gibberish.
The second part of that sentence is gibberish. Nobody has been talking
about "uncovering" numbers, whatever that might mean. [...]
Of course, his "if ... then ..." claim is "nonsense", since simply
for all n e ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo
as well as
for all n e ℕ: |ℕ| - |{1, 2, 3, ..., n}| = ℵo .
On 13.03.2025 15:56, Moebius wrote:Yes, because they too have inf. many successors.
Of course, his "if ... then ..." claim is "nonsense", since simplyIf all n which have ℵo successors are subtracted from ℕ then the successors vanish?
for all n e ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo
as well as
for all n e ℕ: |ℕ| - |{1, 2, 3, ..., n}| = ℵo .
WM <wolfgang.mueckenheim@tha.de> wrote:
On 12.03.2025 18:42, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
If the numbers are definable.
Meaningless. Or are you admitting that your "dark numbers" aren't
natural numbers after all?
They
They?
Learn what potential infinity is.
I know what it is. It's an outmoded notion of infinity, popular in the
1880s, but which is entirely unneeded in modern mathematics.
That makes "modern mathematics" worthless.
What do you know about modern mathematics? You may recall me challenging others in another recent thread to cite some mathematical result where
the notion of potential/actual infinity made a difference. There came no coherent reply (just one from Ross Finlayson I couldn't make head nor
tail of). Potential infinity isn't helpful and isn't needed anymore.
WM <wolfgang.mueckenheim@tha.de> wrote:
I know that it is self-contradictory because it cannot distinguish
potential and actual infinity.
It can, but doesn't need to. Potential and actual infinity are needless concepts which only serve to confuse and obfuscate. If you disagree,
feel free to cite a standard result in standard mathematics which depends
on these notions.
When |ℕ| \ |{1, 2, 3, ..., n}| = ℵo, ....
The difference operator \
applies to sets, not to cardinal numbers.
.... then |ℕ| \ |{1, 2, 3, ..., n+1}| = ℵo. This holds for all elements >> of the inductive set, i.e., all FISONs F(n) or numbers n which have
more successors than predecessors.
I.e. all natural numbers.
Only those contribute to the inductive set!
The inductive set is all natural numbers. Why must you make such a song
and dance about it?
Modern mathematics must claim that contrary to the definition ℵo
vanishes to 0 because ℕ \ {1, 2, 3, ...} = { }. That is blatantly
wrong and shows that modern mathematicians believe in miracles.
Matheology.
Modern mathematics need not and does not claim such a ridiculous thing.
You didn't point out any mistake in 3. I doubt you can.
I told you that potential infinity has no last element, therefore there
is no first dark number.
The second part of your sentence does not follow clearly from the first, therefore the sentence is false. And even if it were not false, it has
no bearing on my item 3.
But I can agree with you that there is no first "dark number". That is
what I have proven. There is a theorem that every non-empty subset of
the natural numbers has a least member.
Try to remove all numbers individually from the harmonic series such >>>>>> that none remains. If you can't, find the first one which resists.
Why should I want to do that?
In order to experience that dark numbers exist and can't be manipulated.
Dark numbers don't exist, as Jim and I have proven.
When |ℕ \ {1, 2, 3, ..., n}| = ℵo, then |ℕ \ {1, 2, 3, ..., n+1}| =
ℵo. How do the ℵo dark numbers get visible?
There is no such thing as a "dark number". It's a figment of your imagination and faulty intuition.
Induction cannot cover all natural numbers but only less than remain
uncovered.
The second part of that sentence is gibberish. Nobody has been talking
about "uncovering" numbers, whatever that might mean. Induction
encompasses all natural numbers. Anything it doesn't cover is not a
natural number, by definition.
Every defined number leaves ℵo undefined numbers. Try to find a
counterexample. Fail.
What the heck are you talking about? What does it even mean for a number
to "leave" a set of numbers?
Quite aside from the fact that there is no
mathematical definition of a "defined" number. The "definition" you gave
a few posts back was sociological (talking about how people interacted
with eachother) not mathematical.
On 13.03.2025 13:59, Alan Mackenzie wrote:*crickets*
WM <wolfgang.mueckenheim@tha.de> wrote:
I know that it is self-contradictory because it cannot distinguishIt can, but doesn't need to. Potential and actual infinity are
potential and actual infinity.
needless concepts which only serve to confuse and obfuscate. If you
disagree, feel free to cite a standard result in standard mathematics
which depends on these notions.
Indeed, N = {1, 2, 3, ...}.No. All numbers can be subtracted from ℕ such that none remains:When |ℕ| \ |{1, 2, 3, ..., n}| = ℵo, ....I.e. all natural numbers.
.... then |ℕ| \ |{1, 2, 3, ..., n+1}| = ℵo. This holds for all
elements of the inductive set, i.e., all FISONs F(n) or numbers n
which have more successors than predecessors.
ℕ \ {1, 2, 3, ...} = { }, let alone ℵo.
I have no idea what your "definable numbers" are, but there can onlyBecause when only definable numbers are subtracted from ℕ, then ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo infinitely many numbers remain. ThatOnly those contribute to the inductive set!The inductive set is all natural numbers. Why must you make such a
song and dance about it?
is the difference between dark and defiable numbers.
What does this mean?Modern mathematics must claim that contrary to the definition ℵo
vanishes to 0
Which is why they disjunct from the naturals.ℕ \ {1, 2, 3, ...} = { } is wrong?because ℕ \ {1, 2, 3, ...} = { }. That is blatantlyModern mathematics need not and does not claim such a ridiculous thing.
wrong and shows that modern mathematicians believe in miracles.
Matheology.
Try to think better. ℕ_def is a subset of ℕ. If ℕ_def had a last element, the successor would be the first dark number.The second part of your sentence does not follow clearly from theYou didn't point out any mistake in 3. I doubt you can.I told you that potential infinity has no last element, therefore
there is no first dark number.
first, therefore the sentence is false. And even if it were not false,
it has no bearing on my item 3.
But I can agree with you that there is no first "dark number". That isThat theorem is wrong in case of dark numbers.
what I have proven. There is a theorem that every non-empty subset of
the natural numbers has a least member.
They are not dark.Above we have an inductive definition of all elements which haveThere is no such thing as a "dark number". It's a figment of yourWhen |ℕ \ {1, 2, 3, ..., n}| = ℵo, then |ℕ \ {1, 2, 3, ..., n+1}| = >>> ℵo. How do the ℵo dark numbers get visible?Dark numbers don't exist, as Jim and I have proven.In order to experience that dark numbers exist and can't beTry to remove all numbers individually from the harmonic seriesWhy should I want to do that?
such that none remains. If you can't, find the first one which
resists.
manipulated.
imagination and faulty intuition.
infinitely many dark successors.
Why should it.The set ℕ_def defined by induction does not include ℵo undefinedWhat the heck are you talking about? What does it even mean for aEvery defined number leaves ℵo undefined numbers. Try to find aInduction cannot cover all natural numbers but only less than remain >>>>> uncovered.The second part of that sentence is gibberish. Nobody has been
talking about "uncovering" numbers, whatever that might mean.
Induction encompasses all natural numbers. Anything it doesn't cover
is not a natural number, by definition.
counterexample. Fail.
number to "leave" a set of numbers?
numbers.
WMaths does (apparently) have one result that is not a theorem of modern mathematics. In WMaths there sets P and E such that
E in P and P \ {E} = P
WM himself called this a "surprise" but unfortunately he has never been
able to offer a proof.
Am Thu, 13 Mar 2025 17:18:34 +0100 schrieb WM:
Above we have an inductive definition of all elements which haveThey are not dark.
infinitely many dark successors.
Alan Mackenzie <acm@muc.de> writes:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 12.03.2025 18:42, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
If the numbers are definable.
Meaningless. Or are you admitting that your "dark numbers" aren't
natural numbers after all?
They
They?
Learn what potential infinity is.
I know what it is. It's an outmoded notion of infinity, popular in the >>>> 1880s, but which is entirely unneeded in modern mathematics.
That makes "modern mathematics" worthless.
What do you know about modern mathematics? You may recall me challenging
others in another recent thread to cite some mathematical result where
the notion of potential/actual infinity made a difference. There came no
coherent reply (just one from Ross Finlayson I couldn't make head nor
tail of). Potential infinity isn't helpful and isn't needed anymore.
WMaths does (apparently) have one result that is not a theorem of modern mathematics. In WMaths there sets P and E such that
E in P and P \ {E} = P
WM himself called this a "surprise" but unfortunately he has never been
able to offer a proof.
On another occasion he and I came close to another when I defined a
sequence of rationals that he agreed was monotonic, increasing and
bounded above but which (apparently) does not converge to a real in
WMaths. It was "defined enough" to be monotonic and bounded but not
"defined enough" to converge to a real.
But, in general, he hates talking about his WMaths because it gets him
into these sorts of blind alleys.
--
Ben.
On 13.03.2025 13:59, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
I know that it is self-contradictory because it cannot distinguish
potential and actual infinity.
It can, but doesn't need to. Potential and actual infinity are needless
concepts which only serve to confuse and obfuscate. If you disagree,
feel free to cite a standard result in standard mathematics which depends
on these notions.
When |ℕ| \ |{1, 2, 3, ..., n}| = ℵo, ....
The difference operator \ applies to sets, not to cardinal numbers.
I know, but erroneously I had used the sets. I corrected that but
without correcting the sign
.... then |ℕ| \ |{1, 2, 3, ..., n+1}| = ℵo. This holds for all elements >>> of the inductive set, i.e., all FISONs F(n) or numbers n which have
more successors than predecessors.
I.e. all natural numbers.
No. All numbers can be subtracted from ℕ such that none remains:
ℕ \ {1, 2, 3, ...} = { }, let alone ℵo.
Only those contribute to the inductive set!
The inductive set is all natural numbers. Why must you make such a song
and dance about it?
Because when only definable numbers are subtracted from ℕ, ....
.... then ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo infinitely many numbers remain. That is the difference between dark and defiable
numbers.
Modern mathematics must claim that contrary to the definition ℵo
vanishes to 0 because ℕ \ {1, 2, 3, ...} = { }. That is blatantly
wrong and shows that modern mathematicians believe in miracles.
Matheology.
Modern mathematics need not and does not claim such a ridiculous thing.
ℕ \ {1, 2, 3, ...} = { } is wrong?
You didn't point out any mistake in 3. I doubt you can.
I told you that potential infinity has no last element, therefore there
is no first dark number.
The second part of your sentence does not follow clearly from the first,
therefore the sentence is false. And even if it were not false, it has
no bearing on my item 3.
Try to think better. ℕ_def is a subset of ℕ. If ℕ_def had a last element, the successor would be the first dark number.
But I can agree with you that there is no first "dark number". That is
what I have proven. There is a theorem that every non-empty subset of
the natural numbers has a least member.
That theorem is wrong in case of dark numbers.
Try to remove all numbers individually from the harmonic series such >>>>>>> that none remains. If you can't, find the first one which resists.
Why should I want to do that?
In order to experience that dark numbers exist and can't be manipulated.
Dark numbers don't exist, as Jim and I have proven.
When |ℕ \ {1, 2, 3, ..., n}| = ℵo, then |ℕ \ {1, 2, 3, ..., n+1}| = >>> ℵo. How do the ℵo dark numbers get visible?
There is no such thing as a "dark number". It's a figment of your
imagination and faulty intuition.
Above we have an inductive definition of all elements which have
infinitely many dark successors.
Induction cannot cover all natural numbers but only less than remain >>>>> uncovered.
The second part of that sentence is gibberish. Nobody has been talking >>>> about "uncovering" numbers, whatever that might mean. Induction
encompasses all natural numbers. Anything it doesn't cover is not a
natural number, by definition.
Every defined number leaves ℵo undefined numbers. Try to find a
counterexample. Fail.
What the heck are you talking about? What does it even mean for a number
to "leave" a set of numbers?
The set ℕ_def defined by induction does not include ℵo undefined numbers.
Quite aside from the fact that there is no
mathematical definition of a "defined" number. The "definition" you gave
a few posts back was sociological (talking about how people interacted
with eachother) not mathematical.
Mathematics is social, even when talking to oneself. Things which cannot
be represented in any mind cannot be treated.
Regards, WM
On 13.03.2025 01:43, Jim Burns wrote:
A single (lossless) exchange cannot delete an O
Finitely.many (lossless) exchanges cannot delete an O
Infinitely.many (lossless) exchanges can delete an O
No.
Infinite is not _only_ bigger. It's different.
It obeys logic or it is of no value.
On 13.03.2025 01:43, Jim Burns wrote:Yes.
On 3/12/2025 4:14 PM, WM wrote:
A single (lossless) exchange cannot delete an O Finitely.manyNo.
(lossless) exchanges cannot delete an O
Infinitely.many (lossless) exchanges can delete an O
It doesn't obey finite logic.Infinite is not _only_ bigger. It's different.It obeys logic or it is of no value.
On 12.03.2025 22:31, Alan Mackenzie wrote:There is no miracle. N is not a FISON.
WM <wolfgang.mueckenheim@tha.de> wrote:I know that it is self-contradictory because it cannot distinguish
On 12.03.2025 18:42, Alan Mackenzie wrote:What do you know about modern mathematics?
WM <wolfgang.mueckenheim@tha.de> wrote:
If the numbers are definable.Meaningless. Or are you admitting that your "dark numbers" aren't
natural numbers after all?
That makes "modern mathematics" worthless.Learn what potential infinity is.I know what it is. It's an outmoded notion of infinity, popular in
the 1880s, but which is entirely unneeded in modern mathematics.
potential and actual infinity.
When |ℕ| \ |{1, 2, 3, ..., n}| = ℵo, then |ℕ| \ |{1, 2, 3, ..., n+1}| = ℵo. This holds for all elements of the inductive set, i.e., all FISONs
F(n) or numbers n which have more successors than predecessors. Only
those contribute to the inductive set! Modern mathematics must claim
that contrary to the definition ℵo vanishes to 0 because ℕ \ {1, 2, 3, ...} = { }.
That is blatantly wrong and shows that modern mathematicians believe in miracles. Matheology.
Then it is larger than omega.You may recall me challenging
others in another recent thread to cite some mathematical result where
the notion of potential/actual infinity made a difference. There came
no coherent reply (just one from Ross Finlayson I couldn't make head
nor tail of). Potential infinity isn't helpful and isn't needed
anymore.
It is named but has no FISON. That is the crucial condition.3. The least element of the set of dark numbers, by its very
definition, has been "named", "addressed", "defined", and
"instantiated".
Therefore none.I told you that potential infinity has no last element, therefore thereYou didn't point out any mistake in 3. I doubt you can.So you counter my proof by silently snipping elements 4, 5 and 6 ofThey were based on the mistaken 3 and therefore useless.
it? That's not a nice thing to do.
is no first dark number.
By induction.When |ℕ| \ |{1, 2, 3, ..., n}| = ℵo, then |ℕ| \ |{1, 2, 3, ..., n+1}| = ℵo. How do the ℵo dark numbers get visible?Dark numbers don't exist, as Jim and I have proven.In order to experience that dark numbers exist and can't beTry to remove all numbers individually from the harmonic series such >>>>> that none remains. If you can't, find the first one which resists.Why should I want to do that?
manipulated.
They are not undefined.Every defined number leaves ℵo undefined numbers.Induction cannot cover all natural numbers but only less than remainThe second part of that sentence is gibberish. Nobody has been talking
uncovered.
about "uncovering" numbers, whatever that might mean. Induction
encompasses all natural numbers. Anything it doesn't cover is not a
natural number, by definition.
WM <wolfgang.mueckenheim@tha.de> wrote:
"Definable number" has not been defined by you, except in a sociological sense.
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo infinitely many
numbers remain. That is the difference between dark and definable
numbers.
Rubbish! It's just that the set difference between an infinite set and a
one of its finite subsets remains infinite.
That doesn't shed any light
on "dark" or "defi[n]able" numbers.
ℕ_def is a subset of ℕ. If ℕ_def had a last
element, the successor would be the first dark number.
If, if, if, .... "N_def" remains undefined, so it is not sensible to
make assertions about it.
But I can agree with you that there is no first "dark number". That is
what I have proven. There is a theorem that every non-empty subset of
the natural numbers has a least member.
That theorem is wrong in case of dark numbers.
That's a very bold claim. Without further evidence, I think it's fair to
say you are simply mistaken here.
When |ℕ \ {1, 2, 3, ..., n}| = ℵo, then |ℕ \ {1, 2, 3, ..., n+1}| = >>>> ℵo. How do the ℵo dark numbers get visible?
There are no such things as "dark numbers", so talking about their
visibility is not sensible.
There is no such thing as a "dark number". It's a figment of your
imagination and faulty intuition.
Above we have an inductive definition of all elements which have
infinitely many dark successors.
"Dark number" remains undefined, except in a sociological sense. "Dark successor" is likewise undefined.
The set ℕ_def defined by induction does not include ℵo undefined numbers.
The set N doesn't include ANY undefined numbers.
Quite aside from the fact that there is no
mathematical definition of a "defined" number. The "definition" you gave >>> a few posts back was sociological (talking about how people interacted
with eachother) not mathematical.
Mathematics is social, even when talking to oneself. Things which cannot
be represented in any mind cannot be treated.
Natural numbers can be "represented in a mind", in fact in any mathematician's mind.
On 13.03.2025 18:53, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
"Definable number" has not been defined by you, except in a sociological
sense.
Then use numbers defined by induction:
|ℕ \ {1}| = ℵo.
If |ℕ \ {1, 2, 3, ..., n}| = ℵo
then |ℕ \ {1, 2, 3, ..., n+1}| = ℵo.
Here the numbers n belonging to a potentially infinite set are defined.
This set is called ℕ_def.
It strives for ℕ but never reaches it because .....
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo infinitely many
numbers remain. That is the difference between dark and definable
numbers.
Rubbish! It's just that the set difference between an infinite set and a
one of its finite subsets remains infinite.
Yes, just that is the dark part. All definable numbers belong to finite sets.
That doesn't shed any light on "dark" or "defi[n]able" numbers.
Du siehst den Wald vor Bäumen nicht.
[ You can't see the wood for the trees. ]
ℕ_def is a subset of ℕ. If ℕ_def had a last
element, the successor would be the first dark number.
If, if, if, .... "N_def" remains undefined, so it is not sensible to
make assertions about it.
See above. Every inductive set (Zermelo, Peano, v. Neumann) is definable.
But I can agree with you that there is no first "dark number". That
is what I have proven. There is a theorem that every non-empty
subset of the natural numbers has a least member.
That theorem is wrong in case of dark numbers.
That's a very bold claim. Without further evidence, I think it's fair
to say you are simply mistaken here.
The potentially infinite inductive set has no last element. Therefore
its complement has no first element.
When |ℕ \ {1, 2, 3, ..., n}| = ℵo, then |ℕ \ {1, 2, 3, ..., n+1}| = >>>>> ℵo. How do the ℵo dark numbers get visible?
There are no such things as "dark numbers", so talking about their
visibility is not sensible.
But there are ℵo numbers following upon all numbers of ℕ_def.
There is no such thing as a "dark number". It's a figment of your
imagination and faulty intuition.
Above we have an inductive definition of all elements which have
infinitely many dark successors.
"Dark number" remains undefined, except in a sociological sense. "Dark
successor" is likewise undefined.
"Es ist sogar erlaubt, sich die neugeschaffene Zahl ω als Grenze zu
denken, welcher die Zahlen ν zustreben, wenn darunter nichts anderes verstanden wird, als daß ω die erste ganze Zahl sein soll, welche auf
alle Zahlen ν folgt, d. h. größer zu nennen ist als jede der Zahlen ν." E. Zermelo (ed.): "Georg Cantor – Gesammelte Abhandlungen mathematischen und philosophischen Inhalts", Springer, Berlin (1932) p. 195.
[ "It is even permissible to think of the newly created number as a
limit to which the numbers nu tend. If nothing else is understood,
it's held to be the first integer which follows all numbers nu, that
is, is bigger than each of the numbers nu." ]
Between the striving numbers ν and ω lie the dark numbers.
The set ℕ_def defined by induction does not include ℵo undefined numbers.
The set N doesn't include ANY undefined numbers.
ℵo
Quite aside from the fact that there is no mathematical definition
of a "defined" number. The "definition" you gave a few posts back
was sociological (talking about how people interacted with
eachother) not mathematical.
Mathematics is social, even when talking to oneself. Things which cannot >>> be represented in any mind cannot be treated.
Natural numbers can be "represented in a mind", in fact in any
mathematician's mind.
Not those which make the set ℕ empty by subtracting them
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
like the dark numbers can do
ℕ \ {1, 2, 3, ...} = { }.
Regards, WM
WM wrote on 3/14/2025 :
Natural numbers can be "represented in a mind", in fact in any
mathematician's mind.
Not those which make the set ℕ empty by subtracting them
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
like the dark numbers can do
ℕ \ {1, 2, 3, ...} = { }.
The set of natural numbers is never empty. What happens is that you
decide which of its elements complies with the definition of elements of
the new set, and remove the rest from consideration.
Am Thu, 13 Mar 2025 11:45:30 +0100 schrieb WM:
On 13.03.2025 01:43, Jim Burns wrote:
It doesn't obey finite logic.Infinite is not _only_ bigger. It's different.It obeys logic or it is of no value.
WM <wolfgang.mueckenheim@tha.de> wrote:
On 13.03.2025 18:53, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
"Definable number" has not been defined by you, except in a sociological >>> sense.
Then use numbers defined by induction:
|ℕ \ {1}| = ℵo.
If |ℕ \ {1, 2, 3, ..., n}| = ℵo
then |ℕ \ {1, 2, 3, ..., n+1}| = ℵo.
Here the numbers n belonging to a potentially infinite set are defined.
This set is called ℕ_def.
You're confusing yourself with the outdated notion "potentially
infinite". The numbers n in an (?the) inductive set are N, not N_def.
Why do you denote the natural numbers by "N_def" when everybody else just calls them "N"?
It strives for ℕ but never reaches it because .....
It doesn't "strive" for N. You appear to be thinking about a process
taking place in time
"Definable" remains undefined, so there's no point to answer here. Did Zermelo, Peano, or von Neumann use "definable" the way you're trying to
use it, at all?
The potentially infinite inductive set has no last element. Therefore
its complement has no first element.
You're letting "potentially infinite" confuse you again. The inductive
set indeed has no last element. So "its complement" (undefined unless we assume a base set to take the complement in), if somehow defined, is
empty. The empty set has no first element.
But there are ℵo numbers following upon all numbers of ℕ_def.
N_def remains undefined,
|ℕ \ {1}| = ℵo.
If |ℕ \ {1, 2, 3, ..., n}| = ℵo
then |ℕ \ {1, 2, 3, ..., n+1}| = ℵo.
"Dark number" remains undefined, except in a sociological sense. "Dark
successor" is likewise undefined.
"Es ist sogar erlaubt, sich die neugeschaffene Zahl ω als Grenze zu
denken, welcher die Zahlen ν zustreben, wenn darunter nichts anderes
verstanden wird, als daß ω die erste ganze Zahl sein soll, welche auf
alle Zahlen ν folgt, d. h. größer zu nennen ist als jede der Zahlen ν." >> E. Zermelo (ed.): "Georg Cantor – Gesammelte Abhandlungen mathematischen >> und philosophischen Inhalts", Springer, Berlin (1932) p. 195.
[ "It is even permissible to think of the newly created number as a
limit to which the numbers nu tend. If nothing else is understood,
it's held to be the first integer which follows all numbers nu, that
is, is bigger than each of the numbers nu." ]
Between the striving numbers ν and ω lie the dark numbers.
That contradicts the long excerpt from Cantor you've just cited.
According to that, omega is the _first_ number which follows the numbers
nu. I.e., there is nothing between nu (which we can identify with N) and omega. There is no place for "dark numbers".
Natural numbers can be "represented in a mind", in fact in any
mathematician's mind.
Not those which make the set ℕ empty by subtracting them
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
That nonsense has no bearing on the representability of natural numbers
in a mathematician's mind. You're just saying that the complement in N
of a finite subset of N is of infinite size. Yes, and.... ?
like the dark numbers can do
ℕ \ {1, 2, 3, ...} = { }.
Dark numbers remain undefined.
The above identity, more succinctly
written as N \ N = { } holds trivially, and has nothing to say about the mythical "dark numbers".
On 3/13/2025 6:45 AM, WM wrote:
On 13.03.2025 01:43, Jim Burns wrote:
A single (lossless) exchange cannot delete an O
Finitely.many (lossless) exchanges cannot delete an O
Infinitely.many (lossless) exchanges can delete an O
No.
Your "No" responds to infiniteᵂᴹ,
but I wrote infiniteⁿᵒᵗᐧᵂᴹ.
On 13.03.2025 20:41, Jim Burns wrote:...except when the logic of the infinite says otherwise.
On 3/13/2025 6:45 AM, WM wrote:My "No" responds to every finity and every infinity. Logic never ceases
On 13.03.2025 01:43, Jim Burns wrote:Your "No" responds to infiniteᵂᴹ, but I wrote infiniteⁿᵒᵗᐧᵂᴹ.
A single (lossless) exchange cannot delete an O Finitely.manyNo.
(lossless) exchanges cannot delete an O
Infinitely.many (lossless) exchanges can delete an O
to be valid. Lossless exchange is lossless.
On 13.03.2025 22:05, joes wrote:The infinite obeys the logic of the infinite, the finite that of the
Am Thu, 13 Mar 2025 11:45:30 +0100 schrieb WM:
On 13.03.2025 01:43, Jim Burns wrote:
Matheology does not obey logic. Mathematics obeys finite logic. There isIt doesn't obey finite logic.Infinite is not _only_ bigger. It's different.It obeys logic or it is of no value.
no other.
On 14.03.2025 14:35, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 13.03.2025 18:53, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
"Definable number" has not been defined by you, except in a sociological >>>> sense.
Then use numbers defined by induction:
|ℕ \ {1}| = ℵo.
If |ℕ \ {1, 2, 3, ..., n}| = ℵo
then |ℕ \ {1, 2, 3, ..., n+1}| = ℵo.
Here the numbers n belonging to a potentially infinite set are defined.
This set is called ℕ_def.
You're confusing yourself with the outdated notion "potentially
infinite". The numbers n in an (?the) inductive set are N, not N_def.
Why do you denote the natural numbers by "N_def" when everybody else just
calls them "N"?
Perhaps everybody is unable to see that
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo?
It strives for ℕ but never reaches it because .....
It doesn't "strive" for N. You appear to be thinking about a process
taking place in time
Induction and counting are processes. It need not be in time. But it
fails to complete ℕ.
"Definable" remains undefined, so there's no point to answer here. Did
Zermelo, Peano, or von Neumann use "definable" the way you're trying to
use it, at all?
Zermelo claimed that without their construction/proof by induction we
don't know whether infinite sets exist at all.
Um aber die Existenz "unendlicher" Mengen zu sichern, bedürfen wir noch
des folgenden ... Axioms. [Zermelo: Untersuchungen über die Grundlagen
der Mengenlehre I, S. 266] The elements are defined by induction in
order to guarantee the existence of infinite sets.
The potentially infinite inductive set has no last element. Therefore
its complement has no first element.
You're letting "potentially infinite" confuse you again. The inductive
set indeed has no last element. So "its complement" (undefined unless we
assume a base set to take the complement in), if somehow defined, is
empty. The empty set has no first element.
The empty set has not ℵo elements.
But there are ℵo numbers following upon all numbers of ℕ_def.
N_def remains undefined,
|ℕ \ {1}| = ℵo.
If |ℕ \ {1, 2, 3, ..., n}| = ℵo
then |ℕ \ {1, 2, 3, ..., n+1}| = ℵo.
"Dark number" remains undefined, except in a sociological sense. "Dark >>>> successor" is likewise undefined.
"Es ist sogar erlaubt, sich die neugeschaffene Zahl ω als Grenze zu
denken, welcher die Zahlen ν zustreben, wenn darunter nichts anderes
verstanden wird, als daß ω die erste ganze Zahl sein soll, welche auf
alle Zahlen ν folgt, d. h. größer zu nennen ist als jede der Zahlen ν." >>> E. Zermelo (ed.): "Georg Cantor – Gesammelte Abhandlungen mathematischen >>> und philosophischen Inhalts", Springer, Berlin (1932) p. 195.
[ "It is even permissible to think of the newly created number as a
limit to which the numbers nu tend. If nothing else is understood,
it's held to be the first integer which follows all numbers nu, that
is, is bigger than each of the numbers nu." ]
Between the striving numbers ν and ω lie the dark numbers.
That contradicts the long excerpt from Cantor you've just cited.
According to that, omega is the _first_ number which follows the numbers
nu. I.e., there is nothing between nu (which we can identify with N) and
omega. There is no place for "dark numbers".
There is place to strive or tend.
Natural numbers can be "represented in a mind", in fact in any
mathematician's mind.
Not those which make the set ℕ empty by subtracting them
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
That nonsense has no bearing on the representability of natural numbers
in a mathematician's mind. You're just saying that the complement in N
of a finite subset of N is of infinite size. Yes, and.... ?
like the dark numbers can do
ℕ \ {1, 2, 3, ...} = { }.
Dark numbers remain undefined.
Yes, they cannot be determined as individuals.
The above identity, more succinctly written as N \ N = { } holds
trivially, and has nothing to say about the mythical "dark numbers".
n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo proves that definable numbers
are not sufficient.
Regards, WM
On 13.03.2025 18:53, Alan Mackenzie wrote:This set *is* N.
WM <wolfgang.mueckenheim@tha.de> wrote:
"Definable number" has not been defined by you, except in aThen use numbers defined by induction:
sociological sense.
|ℕ \ {1}| = ℵo.
If |ℕ \ {1, 2, 3, ..., n}| = ℵo then |ℕ \ {1, 2, 3, ..., n+1}| = ℵo. Here the numbers n belonging to a potentially infinite set are defined.
This set is called ℕ_def. It strives for ℕ but never reaches it
All naturals do.Yes, just that is the dark part. All definable numbers belong to finite∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo infinitely many numbers >>> remain. That is the difference between dark and definable numbers.Rubbish! It's just that the set difference between an infinite set and
a one of its finite subsets remains infinite.
sets.
As is N.See above. Every inductive set (Zermelo, Peano, v. Neumann) isℕ_def is a subset of ℕ. If ℕ_def had a last element, the successor >>> would be the first dark number.If, if, if, .... "N_def" remains undefined, so it is not sensible to
make assertions about it.
definable.
Because it is empty.The potebtially infinite inductive set has no last element. ThereforeThat's a very bold claim. Without further evidence, I think it's fairBut I can agree with you that there is no first "dark number". ThatThat theorem is wrong in case of dark numbers.
is what I have proven. There is a theorem that every non-empty
subset of the natural numbers has a least member.
to say you are simply mistaken here.
its complement has no first element.
No.But there are ℵo numbers following upon all numbers of ℕ_def.There are no such things as "dark numbers", so talking about theirWhen |ℕ \ {1, 2, 3, ..., n}| = ℵo, then |ℕ \ {1, 2, 3, ..., n+1}| = >>>>> ℵo. How do the ℵo dark numbers get visible?
visibility is not sensible.
Rebutted elsewhere."Es ist sogar erlaubt, sich die neugeschaffene Zahl ω als Grenze zu"Dark number" remains undefined, except in a sociological sense. "DarkThere is no such thing as a "dark number". It's a figment of yourAbove we have an inductive definition of all elements which have
imagination and faulty intuition.
infinitely many dark successors.
successor" is likewise undefined.
denken, welcher die Zahlen ν zustreben, wenn darunter nichts anderes verstanden wird, als daß ω die erste ganze Zahl sein soll, welche auf
alle Zahlen ν folgt, d. h. größer zu nennen ist als jede der Zahlen ν." Between the striving numbers ν and ω lie the dark numbers.
Neither undefined nor in N.ℵoThe set ℕ_def defined by induction does not include ℵo undefinedThe set N doesn't include ANY undefined numbers.
numbers.
Maybe not in your mind.Not those which make the set ℕ empty by subtracting them ∀n ∈ ℕ_def: |ℕNatural numbers can be "represented in a mind", in fact in anyQuite aside from the fact that there is noMathematics is social, even when talking to oneself. Things which
mathematical definition of a "defined" number. The "definition" you
gave a few posts back was sociological (talking about how people
interacted with eachother) not mathematical.
cannot be represented in any mind cannot be treated.
mathematician's mind.
\ {1, 2, 3, ..., n}| = ℵo like the dark numbers can do ℕ \ {1, 2, 3,
...} = { }.
On 14.03.2025 14:35, Alan Mackenzie wrote:That's pretty obvious.
WM <wolfgang.mueckenheim@tha.de> wrote:Perhaps everybody is unable to see that ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ...,
On 13.03.2025 18:53, Alan Mackenzie wrote:You're confusing yourself with the outdated notion "potentially
WM <wolfgang.mueckenheim@tha.de> wrote:
"Definable number" has not been defined by you, except in aThen use numbers defined by induction:
sociological sense.
|ℕ \ {1}| = ℵo.
If |ℕ \ {1, 2, 3, ..., n}| = ℵo then |ℕ \ {1, 2, 3, ..., n+1}| = ℵo.
Here the numbers n belonging to a potentially infinite set are
defined. This set is called ℕ_def.
infinite". The numbers n in an (?the) inductive set are N, not N_def.
Why do you denote the natural numbers by "N_def" when everybody else
just calls them "N"?
n}| = ℵo?
Whatever. They are infinite in any case, which you continually fail to comprehend.Induction and counting are processes. It need not be in time. But itIt strives for ℕ but never reaches it because .....It doesn't "strive" for N. You appear to be thinking about a process
taking place in time
fails to complete ℕ.
ℕ \ {1, 2, 3, ...} = { }.Yes, N = {1, 2, 3, ...}.
The complement (wrt what?) is empty.The potentially infinite inductive set has no last element. Therefore
its complement has no first element.
Come again?You're letting "potentially infinite" confuse you again. The inductiveThe empty set has not ℵo elements.
set indeed has no last element. So "its complement" (undefined unless
we assume a base set to take the complement in), if somehow defined, is
empty. The empty set has no first element.
Wrong. There are no other numbers than the "striving" ones.But there are ℵo numbers following upon all numbers of ℕ_def.N_def remains undefined
"Dark number" remains undefined, except in a sociological sense.[ "It is even permissible to think of the newly created number as a
"Dark successor" is likewise undefined.
limit to which the numbers nu tend. If nothing else is understood,
it's held to be the first integer which follows all numbers nu, that
is, is bigger than each of the numbers nu." ]
Between the striving numbers ν and ω lie the dark numbers.
And it is filled with nu.That contradicts the long excerpt from Cantor you've just cited.There is place to strive or tend.
According to that, omega is the _first_ number which follows the
numbers nu. I.e., there is nothing between nu (which we can identify
with N) and omega. There is no place for "dark numbers".
There is no number that "makes N empty", wtf.Natural numbers can be "represented in a mind", in fact in anyNot those which make the set ℕ empty by subtracting them ∀n ∈ ℕ_def:
mathematician's mind.
|ℕ \ {1, 2, 3, ..., n}| = ℵo
And what?That nonsense has no bearing on the representability of natural numbers
in a mathematician's mind. You're just saying that the complement in N
of a finite subset of N is of infinite size. Yes, and.... ?
I see no dark numbers here.like the dark numbers can do ℕ \ {1, 2, 3, ...} = { }.
It doesn't. Why do you think there is a number k such that N = {1, ...,The above identity, more succinctly written as N \ N = { } holdsn ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo proves that definable numbers are not sufficient.
trivially, and has nothing to say about the mythical "dark numbers".
On 13.03.2025 17:42, joes wrote:Now what?
Am Thu, 13 Mar 2025 17:18:34 +0100 schrieb WM:
Then subtract all definable numbers individually from ℕ with the same result as can be accomplished collectively:Above we have an inductive definition of all elements which haveThey are not dark.
infinitely many dark successors.
ℕ \ {1, 2, 3, ...} = { }.
On 13.03.2025 17:27, Ben Bacarisse wrote:Ah ok, so there are no such sets after all.
WMaths does (apparently) have one result that is not a theorem ofThat is caused by potential infinity. The sets or better collections are
modern mathematics. In WMaths there sets P and E such that
E in P and P \ {E} = P
not fixed.
That sounds very sensible actually.WM himself called this a "surprise" but unfortunately he has never beenMuch more surprising is the idea that all natural numbers can be
able to offer a proof.
subtracted from ℕ with nothing remaining ℕ \ {1, 2, 3, ...} = { }.
But when we explicitly subtract only numbers which have ℵo remainders ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo then by magic spell also all remainders vanish.Doesn't seem magic that... oh, see my sig.
On 13.03.2025 20:41, Jim Burns wrote:
On 3/13/2025 6:45 AM, WM wrote:
On 13.03.2025 01:43, Jim Burns wrote:
A single (lossless) exchange cannot delete an O
Finitely.many (lossless) exchanges cannot delete an O
Infinitely.many (lossless) exchanges can delete an O
No.
Your "No" responds to infiniteᵂᴹ,
but I wrote infiniteⁿᵒᵗᐧᵂᴹ.
My "No" responds to every finity and every infinity.
Logic never ceases to be valid.
Lossless exchange is lossless.
After serious thinking WM wrote :
On 14.03.2025 14:11, FromTheRafters wrote:
WM wrote on 3/14/2025 :
Natural numbers can be "represented in a mind", in fact in any
mathematician's mind.
Not those which make the set ℕ empty by subtracting them
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
like the dark numbers can do
ℕ \ {1, 2, 3, ...} = { }.
The set of natural numbers is never empty. What happens is that you
decide which of its elements complies with the definition of elements
of the new set, and remove the rest from consideration.
The new set defined by individual elements cannot be equal to ℕ.
Why not?
On 3/14/2025 10:33 AM, WM wrote:
My "No" responds to every finity and every infinity.
It doesn't respond to this definition:
WM <wolfgang.mueckenheim@tha.de> wrote:
Perhaps everybody is unable to see that
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo?
Everybody can see that, and everybody but you can see it has nothing to
do with the point it purportedly answers.
Wrong. It is an "instantaneous" definition which completes N.
There are
not various stages of "N" which are in varying stages of completion.
There is place to strive or tend.
The tending takes place, but not in a "place".
That I have to write such
nonsense to answer your point shows the great deterioration which has
taken place in a once vital newsgroup.
Yes, they cannot be determined as individuals.
They don't exist, as I have proven.
On 14.03.2025 16:21, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
Perhaps everybody is unable to see that
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo?
Everybody can see that, and everybody but you can see it has nothing to
do with the point it purportedly answers.
ℕ_def contains all numbers the subtraction of which from ℕ does not result in the empty set.
Obviously the subtraction of all numbers which cannot empty ℕ cannot
empty ℕ. Therefore |ℕ \ ℕ_def| = ℵo. Do you agree?
If not, it is useless to discuss with you.
Wrong. It is an "instantaneous" definition which completes N.
Yes, of course. But ℕ_def is not completed by its definition.
There are not various stages of "N" which are in varying stages of
completion.
ℕ_def is never complete.
There is place to strive or tend.
The tending takes place, but not in a "place".
No? Tending means that hitherto undefined natural numbers become
defined. That takes place on the ordinal line.
That I have to write such nonsense to answer your point shows the
great deterioration which has taken place in a once vital newsgroup.
Hardly to believe that matheology like tending of ordinals outside of
the ordinal line has ever been useful.
Yes, they cannot be determined as individuals.
They don't exist, as I have proven.
You have proven that you are a matheologian with little ability to understand arguments contradicting your matheologial belief.
Regards, WM
On 14.03.2025 18:29, Jim Burns wrote:
On 3/14/2025 10:33 AM, WM wrote:
On 13.03.2025 20:41, Jim Burns wrote:
On 3/13/2025 6:45 AM, WM wrote:
On 13.03.2025 01:43, Jim Burns wrote:
A single (lossless) exchange cannot delete an O
Finitely.many (lossless) exchanges cannot delete an O
Infinitely.many (lossless) exchanges can delete an O
No.
Your "No" responds to infiniteᵂᴹ,
but I wrote infiniteⁿᵒᵗᐧᵂᴹ.
My "No" responds to every finity and every infinity.
It doesn't respond to this definition:
Then it is not worthwhile to read that definition.
Your "No" responds to infiniteᵂᴹ,
but I wrote infiniteⁿᵒᵗᐧᵂᴹ.
on 3/14/2025, WM supposed :
On 14.03.2025 17:09, FromTheRafters wrote:
After serious thinking WM wrote :
On 14.03.2025 14:11, FromTheRafters wrote:
WM wrote on 3/14/2025 :
Natural numbers can be "represented in a mind", in fact in any
mathematician's mind.
Not those which make the set ℕ empty by subtracting them
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
like the dark numbers can do
ℕ \ {1, 2, 3, ...} = { }.
The set of natural numbers is never empty. What happens is that you
decide which of its elements complies with the definition of
elements of the new set, and remove the rest from consideration.
The new set defined by individual elements cannot be equal to ℕ.
Why not?
Because |ℕ \ ℕ_def| = ℵo.
ℕ_def contains all numbers the subtraction of which from ℕ does not
result in the empty set. Obviously the subtraction of all numbers
which cannot empty ℕ cannot empty ℕ. Do you agree?
There you go with the subtraction idea again.
The difference set between
the set of naturals and your imagined set of 'defined naturals' has cardinality aleph_zero because your 'defined naturals' is a finite set.
WM <wolfgang.mueckenheim@tha.de> wrote:
ℕ_def contains all numbers the subtraction of which from ℕ does not
result in the empty set.
What does "which" refer to?
To N_def or to a
member of the "all numbers"?
Assuming the former, then if X is any proper subset of N, N \ X is
non-empty. So by this "definition", N_def is any proper subset of N.
Obviously the subtraction of all numbers which cannot empty ℕ cannot
empty ℕ. Therefore |ℕ \ ℕ_def| = ℵo. Do you agree?
Of course not.
It all depends on the X from which N_def is formed. If
X is N \ {1},
Yes, of course. But ℕ_def is not completed by its definition.
You haven't defined N_def - what appears above is not a coherent
definition.
The tending takes place, but not in a "place".
No? Tending means that hitherto undefined natural numbers become
defined. That takes place on the ordinal line.
"Hitherto" ("bis jetzt" in German) is purely a time based adverb. The natural numbers are not defined in a time based sequence. They are
defined all together.
On 14.03.2025 16:21, Alan Mackenzie wrote:This makes NO SENSE. You either mean N_def=N (no single removed number
WM <wolfgang.mueckenheim@tha.de> wrote:
ℕ_def contains all numbers the subtraction of which from ℕ does not result in the empty set.Perhaps everybody is unable to see that ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, >>> ..., n}| = ℵo?Everybody can see that, and everybody but you can see it has nothing to
do with the point it purportedly answers.
Obviously the subtraction of all numbers whichWhat? Is N_def finite? (don't come at me with "potential").
cannot empty ℕ cannot empty ℕ.
watWrong. It is an "instantaneous" definition which completes N.Yes, of course. But ℕ_def is not completed by its definition.
Then it is not a set. If it were, it would equal N.There areℕ_def is never complete.
not various stages of "N" which are in varying stages of completion.
No. What's undefined doesn't exist. "Tending" is a property of aNo? Tending means that hitherto undefined natural numbers becomeThere is place to strive or tend.The tending takes place, but not in a "place".
defined. That takes place on the ordinal line.
You got that wrong.That I have to write such nonsense to answer your point shows the greatHardly to believe that matheology like tending of ordinals outside of
deterioration which has taken place in a once vital newsgroup.
the ordinal line has ever been useful.
On 15.03.2025 00:11, Alan Mackenzie wrote:It could also have referred to the set. But in this case, N_def=N.
WM <wolfgang.mueckenheim@tha.de> wrote:
It refers to the numbers. An Englishman should comprehend that.ℕ_def contains all numbers the subtraction of which from ℕ does notWhat does "which" refer to?
result in the empty set.
No. N_def is not a number.To N_def or to a member of the "all numbers"?That is one and the same.
As does N.Assuming the former, then if X is any proper subset of N, N \ X isNo, ℕ_def contains only definable numbers.
non-empty. So by this "definition", N_def is any proper subset of N.
Mostly?It all depends on the X from which N_def is formed. If X is N \ {1},Then its elements are mostly undefined as individuals.
Not "therefore".It is coherent enough. Every element has a finite FISON. ℕ is infinite. Therefore it cannot be emptied by the elements of ℕ_def and also not by ℕ_def.Yes, of course. But ℕ_def is not completed by its definition.You haven't defined N_def - what appears above is not a coherent
definition.
Especially those.Not the defined numbers."Hitherto" ("bis jetzt" in German) is purely a time based adverb. TheThe tending takes place, but not in a "place".No? Tending means that hitherto undefined natural numbers become
defined. That takes place on the ordinal line.
natural numbers are not defined in a time based sequence. They are
defined all together.
On 15.03.2025 00:11, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
ℕ_def contains all numbers the subtraction of which from ℕ does not
result in the empty set.
What does "which" refer to?
It refers to the numbers. An Englishman should comprehend that.
To N_def or to a
member of the "all numbers"?
That is one and the same.
Assuming the former, then if X is any proper subset of N, N \ X is
non-empty. So by this "definition", N_def is any proper subset of N.
No, ℕ_def contains only definable numbers.
Obviously the subtraction of all numbers which cannot empty ℕ cannot
empty ℕ. Therefore |ℕ \ ℕ_def| = ℵo. Do you agree?
Of course not.
Then you cannot think logically.
It all depends on the X from which N_def is formed. If
X is N \ {1},
Then its elements are mostly undefined as individuals.
Yes, of course. But ℕ_def is not completed by its definition.
You haven't defined N_def - what appears above is not a coherent
definition.
It is coherent enough.
Every element has a finite FISON. ℕ is infinite. Therefore it cannot
be emptied by the elements of ℕ_def and also not by ℕ_def.
The tending takes place, but not in a "place".
No? Tending means that hitherto undefined natural numbers become
defined. That takes place on the ordinal line.
"Hitherto" ("bis jetzt" in German) is purely a time based adverb. The
natural numbers are not defined in a time based sequence. They are
defined all together.
Not the defined numbers.
Regards, WM
Am Fri, 14 Mar 2025 23:10:11 +0100 schrieb WM:
ℕ_def contains all numbers the subtraction of which from ℕ does notThis makes NO SENSE.
result in the empty set.
Obviously the subtraction of all numbers whichWhat? Is N_def finite? (don't come at me with "potential").
cannot empty ℕ cannot empty ℕ.
Am Sat, 15 Mar 2025 09:56:31 +0100 schrieb WM:
On 15.03.2025 00:11, Alan Mackenzie wrote:It could also have referred to the set. But in this case, N_def=N.
WM <wolfgang.mueckenheim@tha.de> wrote:It refers to the numbers. An Englishman should comprehend that.
ℕ_def contains all numbers the subtraction of which from ℕ does not >>>> result in the empty set.What does "which" refer to?
No. N_def is not a number.To N_def or to a member of the "all numbers"?That is one and the same.
On 15.03.2025 12:26, joes wrote:Like I said:
Am Fri, 14 Mar 2025 23:10:11 +0100 schrieb WM:
There are numbers which can be removed from ℕ without emptying ℕ, for instance {2}. All numbers of that kind can be collected.ℕ_def contains all numbers the subtraction of which from ℕ does notThis makes NO SENSE.
result in the empty set.
No such numbers.You either mean N_def=N (no single removed number
makes the set empty *facepalm*) or N_def={} (subtracting everything
makes the set empty).
Have you ever defined a number by an infinite FISON?Obviously the subtraction of all numbers which cannot empty ℕ cannotWhat? Is N_def finite? (don't come at me with "potential").
empty ℕ.
On 15.03.2025 12:32, joes wrote:For every number in N, if you subtract that number, a nonempty set
Am Sat, 15 Mar 2025 09:56:31 +0100 schrieb WM:ℕ_def contains all numbers the subtraction of which from ℕ does not result in the empty set.
On 15.03.2025 00:11, Alan Mackenzie wrote:It could also have referred to the set. But in this case, N_def=N.
WM <wolfgang.mueckenheim@tha.de> wrote:It refers to the numbers. An Englishman should comprehend that.
ℕ_def contains all numbers the subtraction of which from ℕ does not >>>>> result in the empty set.What does "which" refer to?
ℕ contains all numbers the subtraction of which from ℕ does result inNo element of N, subtracted from N, leaves the empty set (because N
the empty set.
Yes, N_def is a set.ℕ_def contains all numbers the subtraction of which from ℕ does not result in the empty set.No. N_def is not a number.To N_def or to a member of the "all numbers"?That is one and the same.
WM <wolfgang.mueckenheim@tha.de> wrote:
I'm showing you that your "definition" of
"definable numbers" is no definition at all.
Obviously the subtraction of all numbers which cannot empty ℕ cannot >>>> empty ℕ. Therefore |ℕ \ ℕ_def| = ℵo. Do you agree?
Of course not.
Then you cannot think logically.
When confronted with your misguided attempts at mathematics, it is very difficult to follow your "logic", much less agree with it.
It all depends on the X from which N_def is formed. If
X is N \ {1},
Then its elements are mostly undefined as individuals.
"Undefined as individuals" is an undefined notion,
Every element has a finite FISON. ℕ is infinite. Therefore it cannot
be emptied by the elements of ℕ_def and also not by ℕ_def.
A "finite" FISON? What other type is there? What do you mean by
"having" a FISON? What does it mean to "empty" N by a set or elements of
a set? What is the significance, if any, of being able to "empty" a set?
None of these notions are standard mathematical ones. If you want to communicate clearly with mathematicians, you'd do far better if you used
the standard words with their standard meanings. But maybe you don't
want to communicate clearly.
The tending takes place, but not in a "place".
No? Tending means that hitherto undefined natural numbers become
defined. That takes place on the ordinal line.
"Hitherto" ("bis jetzt" in German) is purely a time based adverb. The
natural numbers are not defined in a time based sequence. They are
defined all together.
Not the defined numbers.
"Defined numbers" remains (still) undefined.
"Defined numbers" appears
not to be a coherent mathematical concept.
Am Sat, 15 Mar 2025 17:44:40 +0100 schrieb WM:
On 15.03.2025 12:26, joes wrote:Like I said:
Am Fri, 14 Mar 2025 23:10:11 +0100 schrieb WM:There are numbers which can be removed from ℕ without emptying ℕ, for
ℕ_def contains all numbers the subtraction of which from ℕ does not >>>> result in the empty set.This makes NO SENSE.
instance {2}. All numbers of that kind can be collected.
You either mean N_def=N
or N_def={} (subtracting everything
makes the set empty).
Am Sat, 15 Mar 2025 17:49:49 +0100 schrieb WM:
On 15.03.2025 12:32, joes wrote:For every number in N, if you subtract that number, a nonempty set
Am Sat, 15 Mar 2025 09:56:31 +0100 schrieb WM:ℕ_def contains all numbers the subtraction of which from ℕ does not
On 15.03.2025 00:11, Alan Mackenzie wrote:It could also have referred to the set. But in this case, N_def=N.
WM <wolfgang.mueckenheim@tha.de> wrote:It refers to the numbers. An Englishman should comprehend that.
ℕ_def contains all numbers the subtraction of which from ℕ does not >>>>>> result in the empty set.What does "which" refer to?
result in the empty set.
remains.
Of course, if you subtract all of them, nothing does.
ℕ_def contains all numbers the subtraction of which from ℕ does notYes, N_def is a set.
result in the empty set.
On 15.03.2025 00:17, Jim Burns wrote:
On 3/14/2025 5:55 PM, WM wrote:
On 14.03.2025 18:29, Jim Burns wrote:
On 3/14/2025 10:33 AM, WM wrote:
On 13.03.2025 20:41, Jim Burns wrote:
On 3/13/2025 6:45 AM, WM wrote:
On 13.03.2025 01:43, Jim Burns wrote:
A single (lossless) exchange cannot delete an O
Finitely.many (lossless) exchanges cannot delete an O
Infinitely.many (lossless) exchanges can delete an O
No.
Your "No" responds to infiniteᵂᴹ,
but I wrote infiniteⁿᵒᵗᐧᵂᴹ.
My "No" responds to every finity and every infinity.
It doesn't respond to this definition:
Then it is not worthwhile to read that definition.
Your "No" responds to infiniteᵂᴹ,
but I wrote infiniteⁿᵒᵗᐧᵂᴹ.
My "No" responds to every finity and every infinity.
On 3/15/2025 4:36 AM, WM wrote:
There is a very old anecdote told in which
Galileo Galilei tries to show the moons of Jupiter
to a representative of the pope.
The representative refuses to look at
anything his religion says can't exist.
On 15.03.2025 21:14, Jim Burns wrote:
There is a very old anecdote told in which
Galileo Galilei tries to show the moons of Jupiter
to a representative of the pope.
The representative refuses to look at
anything his religion says can't exist.
That should not seduce anyone to
spend his time with nonsense.
Lossless exchange is lossless for every exchange,
independent of the number of exchanges.
On 3/15/2025 5:58 PM, WM wrote:
On 15.03.2025 21:14, Jim Burns wrote:
There is a very old anecdote told in which
Galileo Galilei tries to show the moons of Jupiter
to a representative of the pope.
The representative refuses to look at
anything his religion says can't exist.
That should not seduce anyone to
spend his time with nonsense.
Lossless exchange is lossless for every exchange,
independent of the number of exchanges.
Define an ovine set to be a set for which
fuller.by.one sets exist which are larger.
On 15.03.2025 12:57, Alan Mackenzie wrote:Those are, by definition, not naturals.
WM <wolfgang.mueckenheim@tha.de> wrote:
I'm showing you that your "definition" of "definable numbers" is noYou are mistaken. Not all numbers have FISONs because ∀n ∈ U(F): |ℕ \ {1, 2, 3, ..., n}| = ℵo.
definition at all.
ℵo nubers have no FISONs.
This is so fucked up. If you take out everything, nothing remains.The subtraction of all numbers which cannot empty ℕ cannot empty ℕ. Simpler logic is hardly possible.When confronted with your misguided attempts at mathematics, it is veryThen you cannot think logically.Obviously the subtraction of all numbers which cannot empty ℕ cannot >>>>> empty ℕ. Therefore |ℕ \ ℕ_def| = ℵo. Do you agree?Of course not.
difficult to follow your "logic", much less agree with it.
Then n is infinite and not in fact natural.No. It says simply that no FISON ending with n can be defined."Undefined as individuals" is an undefined notion,It all depends on the X from which N_def is formed. If X is N \ {1},Then its elements are mostly undefined as individuals.
It does not say what you think it says.Simply try to understand. I have often stated the difference:Every element has a finite FISON. ℕ is infinite. Therefore it cannotA "finite" FISON? What other type is there? What do you mean by
be emptied by the elements of ℕ_def and also not by ℕ_def.
"having" a FISON? What does it mean to "empty" N by a set or elements
of a set? What is the significance, if any, of being able to "empty" a
set?
∀n ∈ U(F): |ℕ \ {1, 2, 3, ..., n}| = ℵo ℕ \ {1, 2, 3, ...} = { }
You can't have a set with undefined elements.None of these notions are standard mathematical ones. If you want toThe set is defined, not its elements. All defined elements
communicate clearly with mathematicians, you'd do far better if you
used the standard words with their standard meanings. But maybe you
don't want to communicate clearly.
"Hitherto" ("bis jetzt" in German) is purely a time based adverb.The tending takes place, but not in a "place".No? Tending means that hitherto undefined natural numbers become
defined. That takes place on the ordinal line.
The natural numbers are not defined in a time based sequence. They
are defined all together.
Yes they can. There are no other numbers.Defined numbers have FISONs ad cannot empty ℕ.Not the defined numbers."Defined numbers" remains (still) undefined.
They are placed on the
ordinal line and can tend to ℕ. This cn happen only on the ordinal line. Your assertion of the contrary is therefore wrong.
"Defined numbers" appears not to be a coherent mathematical concept.The subtraction of all numbers which cannot empty ℕ cannot empty ℕ. The collection of these numbers is ℕ_def.
On 15.03.2025 12:57, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
I'm showing you that your "definition" of
"definable numbers" is no definition at all.
You are mistaken. Not all numbers have FISONs because
∀n ∈ U(F): |ℕ \ {1, 2, 3, ..., n}| = ℵo.
ℵo numbers have no FISONs.
Obviously the subtraction of all numbers which cannot empty ℕ cannot >>>>> empty ℕ. Therefore |ℕ \ ℕ_def| = ℵo. Do you agree?
Of course not.
Then you cannot think logically.
When confronted with your misguided attempts at mathematics, it is very
difficult to follow your "logic", much less agree with it.
The subtraction of all numbers which cannot empty ℕ cannot empty ℕ. Simpler logic is hardly possible.
It all depends on the X from which N_def is formed. If
X is N \ {1},
Then its elements are mostly undefined as individuals.
"Undefined as individuals" is an undefined notion,
No. It says simply that no FISON ending with n can be defined.
Every element has a finite FISON. ℕ is infinite. Therefore it cannot
be emptied by the elements of ℕ_def and also not by ℕ_def.
A "finite" FISON? What other type is there? What do you mean by
"having" a FISON? What does it mean to "empty" N by a set or elements of
a set? What is the significance, if any, of being able to "empty" a set?
Simply try to understand. I have often stated the difference:
∀n ∈ U(F): |ℕ \ {1, 2, 3, ..., n}| = ℵo
ℕ \ {1, 2, 3, ...} = { }
None of these notions are standard mathematical ones. If you want to
communicate clearly with mathematicians, you'd do far better if you used
the standard words with their standard meanings. But maybe you don't
want to communicate clearly.
The tending takes place, but not in a "place".
No? Tending means that hitherto undefined natural numbers become
defined. That takes place on the ordinal line.
"Hitherto" ("bis jetzt" in German) is purely a time based adverb. The >>>> natural numbers are not defined in a time based sequence. They are
defined all together.
The set is defined, not its elements. All defined elements
Not the defined numbers.
"Defined numbers" remains (still) undefined.
Defined numbers have FISONs and cannot empty ℕ.
They are placed on the ordinal line and can tend to ℕ. This can happen
only on the ordinal line. Your assertion of the contrary is therefore
wrong.
"Defined numbers" appears not to be a coherent mathematical concept.
The subtraction of all numbers which cannot empty ℕ cannot
empty ℕ. The collection of these numbers is ℕ_def.
Regards, WM
On 15.03.2025 17:52, joes wrote:So there is a number which when removed from N empties it?
Am Sat, 15 Mar 2025 17:44:40 +0100 schrieb WM:The numbers which can be removed from ℕ without emptying ℕ are not all numbers of ℕ. Subtracting all numbers of ℕ from ℕ would empty ℕ.
On 15.03.2025 12:26, joes wrote:Like I said:
Am Fri, 14 Mar 2025 23:10:11 +0100 schrieb WM:There are numbers which can be removed from ℕ without emptying ℕ, for >>> instance {2}. All numbers of that kind can be collected.
ℕ_def contains all numbers the subtraction of which from ℕ does not >>>>> result in the empty set.This makes NO SENSE.
You either mean N_def=N (no single removed number makes the set empty
*facepalm*)
--ℕ_def contains 2, so it is not empty.or N_def={} (subtracting everything makes the set empty).
No such numbers.Have you ever defined a number by an infinite FISON?Obviously the subtraction of all numbers which cannot empty ℕ cannot >>>>> empty ℕ.What? Is N_def finite? (don't come at me with "potential").
On 16.03.2025 00:42, Jim Burns wrote:
On 3/15/2025 5:58 PM, WM wrote:
On 15.03.2025 21:14, Jim Burns wrote:
There is a very old anecdote told in which
Galileo Galilei tries to show the moons of Jupiter
to a representative of the pope.
The representative refuses to look at
anything his religion says can't exist.
That should not seduce anyone to
spend his time with nonsense.
Lossless exchange is lossless for every exchange,
independent of the number of exchanges.
Define an ovine set to be a set for which
fuller.by.one sets exist which are larger.
All infinitely many lossless exchanges
have finite indices.
Which index has the first lossless exchange
where a loss happens
(by the way one of infinitely many which
are required to empty the matrix of O's)?
Am Sat, 15 Mar 2025 18:13:57 +0100 schrieb WM:
Defined numbers have FISONs and cannot empty ℕ.Yes they can.
On 3/16/2025 6:01 AM, WM wrote:
All infinitely many lossless exchanges
have finite indices.
Which index has the first lossless exchange
where a loss happens
(by the way one of infinitely many which
are required to empty the matrix of O's)?
What do 'finite' and 'infinite' mean to you (WM)?
Am Sat, 15 Mar 2025 18:25:33 +0100 schrieb WM:
The numbers which can be removed from ℕ without emptying ℕ are not all >> numbers of ℕ. Subtracting all numbers of ℕ from ℕ would empty ℕ.So there is a number which when removed from N empties it?
WM <wolfgang.mueckenheim@tha.de> wrote:
On 15.03.2025 12:57, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
I'm showing you that your "definition" of
"definable numbers" is no definition at all.
You are mistaken. Not all numbers have FISONs because
∀n ∈ U(F): |ℕ \ {1, 2, 3, ..., n}| = ℵo.
ℵo numbers have no FISONs.
You haven't said what you mean by F.
All
natural numbers "have" a FISON
If you really think there is a non-empty set of natural numbers which
don't "have" FISONs,
then please say what the least natural number in
that set is, or at the very least, how you'd go about finding it.
The subtraction of all numbers which cannot empty ℕ cannot empty ℕ.
Simpler logic is hardly possible.
You've never said what you
mean by a number "emptying" a set.
It's unclear whether you mean the
subtraction of each number individually, or of all numbers together.
Even "subtraction" is a non-standard word, here. The opposite of "add" (hinzufügen) is "remove", not "subtract".
It all depends on the X from which N_def is formed. If
X is N \ {1},
Then its elements are mostly undefined as individuals.
"Undefined as individuals" is an undefined notion,
No. It says simply that no FISON ending with n can be defined.
A FISON is a set. Sets don't "end" with
anything.
Every element has a finite FISON. ℕ is infinite. Therefore it cannot >>>> be emptied by the elements of ℕ_def and also not by ℕ_def.
A "finite" FISON? What other type is there?
What do you mean by
"having" a FISON? What does it mean to "empty" N by a set or elements of >>> a set? What is the significance, if any, of being able to "empty" a set?
Simply try to understand. I have often stated the difference:
∀n ∈ U(F): |ℕ \ {1, 2, 3, ..., n}| = ℵo
ℕ \ {1, 2, 3, ...} = { }
Which doesn't address my question in the sightest. What do you mean by "emptying" N by a set or by elements of a set?
You haven't said what (if anything) you mean by a number
emptying N. And every natural number "has" a FISON, not just some subset
of them.
They are placed on the ordinal line and can tend to ℕ. This can happen
only on the ordinal line. Your assertion of the contrary is therefore
wrong.
Of the many assertions I've made, the one you're referring to is unclear.
"Defined numbers" appears not to be a coherent mathematical concept.
The subtraction of all numbers which cannot empty ℕ cannot
empty ℕ. The collection of these numbers is ℕ_def.
Incoherent garbage.
You haven't said what you mean by a number
"emptying" a set.
The current state of our discussion is that you have failed to give any coherent definition of "defined numbers";
On 16.03.2025 16:47, Jim Burns wrote:
On 3/16/2025 6:01 AM, WM wrote:
All infinitely many lossless exchanges
have finite indices.
Which index has the first lossless exchange
where a loss happens
(by the way one of infinitely many which
are required to empty the matrix of O's)?
What do 'finite' and 'infinite' mean to you (WM)?
All elements of ℕ,
i.e.,
all indices which Cantor erroneously claimed
could index all fractions,
are finite natural numbers.
The definable natural numbers strive to
the smallest infinite number ω.
On 16.03.2025 13:17, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
ℵo numbers have no FISONs.
You haven't said what you mean by F.
I did in the discussion with JB:
F is the set of FISONs.
All
natural numbers "have" a FISON
Then all natural numbers would be in FISONs.
But because of
∀n ∈ U(F): |ℕ \ {1, 2, 3, ..., n}| = ℵo
all FISONs fail to contain all natural numbers.
On 3/16/2025 12:37 PM, WM wrote:
On 16.03.2025 16:47, Jim Burns wrote:
On 3/16/2025 6:01 AM, WM wrote:
All infinitely many lossless exchanges
have finite indices.
Which index has the first lossless exchange
where a loss happens
(by the way one of infinitely many which
are required to empty the matrix of O's)?
What do 'finite' and 'infinite' mean to you (WM)?
All elements of ℕ,
Without mentioning 'finite' or 'infinite',
what does ℕ mean to you (WM)?
i.e.,
all indices which Cantor erroneously claimed
could index all fractions,
Not a definition, but a claim.
What your claim means
depends upon 'finite', 'infinite', ℕ, etc.
On 16.03.2025 13:17, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 15.03.2025 12:57, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
All natural numbers "have" a FISON
Then all natural numbers would be in FISONs. But because of
∀n ∈ U(F): |ℕ \ {1, 2, 3, ..., n}| = ℵo
all FISONs fail to contain all natural numbers.
If you really think there is a non-empty set of natural numbers which
don't "have" FISONs,
Of course there is such a set. It contains almost all natural numbers.
This has been proven in the OP: All separated definable natural numbers
can be removed from the harmonic series. When only terms containing all definable numbers together remain, then the series diverges. All its
terms are dark.
then please say what the least natural number in that set is, or at
the very least, how you'd go about finding it.
The definable numbers are potentially infinite sequence. With n also
n+1 and n^n^n belong to it.
Regards, WM
On 16.03.2025 18:05, Jim Burns wrote:
On 3/16/2025 12:37 PM, WM wrote:
On 16.03.2025 16:47, Jim Burns wrote:
On 3/16/2025 6:01 AM, WM wrote:
All infinitely many lossless exchanges
have finite indices.
Which index has the first lossless exchange
where a loss happens
(by the way one of infinitely many which
are required to empty the matrix of O's)?
What do 'finite' and 'infinite' mean to you (WM)?
All elements of ℕ,
Without mentioning 'finite' or 'infinite',
what does ℕ mean to you (WM)?
i.e.,
all indices which Cantor erroneously claimed
could index all fractions,
Not a definition, but a claim.
What your claim means
depends upon 'finite', 'infinite', ℕ, etc.
Ridiculous remarks.
Which index has the first lossless exchange
where a loss happens?
On 3/16/2025 12:17 PM, WM wrote:
On 16.03.2025 13:17, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
ℵo numbers have no FISONs.
You haven't said what you mean by F.
I did in the discussion with JB:
F is the set of FISONs.
And a FISON is
a Finite Initial Segment Of Naturals.
However,
if you (WM) have _non.circularly_ said
what 'finite' means, I have missed it.
All
natural numbers "have" a FISON
Then all natural numbers would be in FISONs.
But because of
∀n ∈ U(F): |ℕ \ {1, 2, 3, ..., n}| = ℵo
all FISONs fail to contain all natural numbers.
You (WM) use an unreliable quantifier swap
to arrive at that conclusion.
On 3/16/2025 1:20 PM, WM wrote:
Which index has the first lossless exchange
where a loss happens?
A set larger than
each set without same.sized proper.subsets
is not
any set without same.sized proper.subsets:
That set has same.sized proper.subsets.
WM <wolfgang.mueckenheim@tha.de> wrote:
The definable numbers are potentially infinite sequence. With n also
n+1 and n^n^n belong to it.
So, no answer.
But let's take the implications of your second sentence there, ".... with
n, also n+1 ....". Since I think we'd agree that 0 is a definable
number, then we've just defined the "definable numbers" as the inductive
set.
Thus the set of "definable numbers", N_def is N.
This is another proof that "dark numbers" don't exist, one you cannot disagree with without contradicting what you've previously written.
I doubt I can be bothered any more to address the other falsehoods, contradictions, and delusions in your last post. But it's clear that
what you write about such topics is wrong in general, if not totally.
On 16.03.2025 18:19, Jim Burns wrote:
On 3/16/2025 12:17 PM, WM wrote:
On 16.03.2025 13:17, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
ℵo numbers have no FISONs.
You haven't said what you mean by F.
I did in the discussion with JB:
F is the set of FISONs.
And a FISON is
a Finite Initial Segment Of Naturals.
However,
if you (WM) have _non.circularly_ said
what 'finite' means, I have missed it.
Finite means a natural number.
If you don't know
try to learn what they are.
1 and if n then n+1.
All
natural numbers "have" a FISON
Then all natural numbers would be in FISONs.
But because of
∀n ∈ U(F): |ℕ \ {1, 2, 3, ..., n}| = ℵo
all FISONs fail to contain all natural numbers.
You (WM) use an unreliable quantifier swap
to arrive at that conclusion.
No
it is very reliable to say that
when all FISONs have infinite distance from ω,
then no FISON comes closer.
On 3/16/2025 2:40 PM, WM wrote:
Finite means a natural number.
Circular.
The _emptiest_ set with 1 and if n then n+1
does not hold dark numbers,
On 3/16/2025 3:00 PM, WM wrote:
JB who claims that
lossless exchange causes losses
but refuses to explain this in detail,
Exchanging a superovine set for
a same.sized proper.subset is size.preserving.
On 16.03.2025 18:23, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
The definable numbers are potentially infinite sequence. With n also
n+1 and n^n^n belong to it.
So, no answer.
But let's take the implications of your second sentence there, ".... with
n, also n+1 ....". Since I think we'd agree that 0 is a definable
number, then we've just defined the "definable numbers" as the inductive
set.
Yes.
Thus the set of "definable numbers", N_def is N.
No. The definable natural numbers strive to the smallest infinite number
ω. But they are never there.
This is another proof that "dark numbers" don't exist, one you cannot
disagree with without contradicting what you've previously written.
A proof exists that they are there. ℕ_def contains all numbers the subtraction of which from ℕ does not result in the empty set.
Obviously the subtraction of all numbers which cannot empty ℕ cannot
empty ℕ.
I doubt I can be bothered any more to address the other falsehoods,
contradictions, and delusions in your last post. But it's clear that
what you write about such topics is wrong in general, if not totally.
Obviously both of you are cranks.
JB who claims that lossless exchange causes losses but refuses to
explain this in detail, ....
.... and you who claims that the subtraction of all numbers which
cannot empty ℕ can empty ℕ
Regards, WM
On 16.03.2025 20:17, Jim Burns wrote:
On 3/16/2025 2:40 PM, WM wrote:
Finite means a natural number.
Circular.
Of course.
Every definition is circular
If you don't know
try to learn what they are.
1 and if n then n+1.
Non.circular!
But incomplete.
Add 'emptiest' to it, and you're there.
Every definition is circular
in that it states
what has been known already
in other words.
The _emptiest_ set with 1 and if n then n+1
does not hold dark numbers,
The question here is
the index of the first disappearance of Bob.
On 16.03.2025 18:23, Alan Mackenzie wrote:
JB who claims that
lossless exchange causes losses
but refuses to explain this in detail,
On 16.03.2025 14:09, joes wrote:Oh, you meant that N is not a FISON. Gotcha.
Am Sat, 15 Mar 2025 18:13:57 +0100 schrieb WM:
∀n ∈ U(F): |ℕ \ {1, 2, 3, ..., n}| = ℵo.Defined numbers have FISONs and cannot empty ℕ.Yes they can.
Do you distrust the universal quantifier?
On 16.03.2025 14:12, joes wrote:No, I mean a single number?
Am Sat, 15 Mar 2025 18:25:33 +0100 schrieb WM:
No definable number. Only the collection of dark numbers can empty ℕ.The numbers which can be removed from ℕ without emptying ℕ are not all >>> numbers of ℕ. Subtracting all numbers of ℕ from ℕ would empty ℕ.So there is a number which when removed from N empties it?
On 16.03.2025 13:17, Alan Mackenzie wrote:Not at all. N, equivalent to omega, is not a FISON.
WM <wolfgang.mueckenheim@tha.de> wrote:I did in the discussion with JB: F is the set of FISONs.
On 15.03.2025 12:57, Alan Mackenzie wrote:You haven't said what you mean by F.
WM <wolfgang.mueckenheim@tha.de> wrote:
I'm showing you that your "definition" of "definable numbers" is noYou are mistaken. Not all numbers have FISONs because ∀n ∈ U(F): |ℕ \ >>> {1, 2, 3, ..., n}| = ℵo.
definition at all.
ℵo numbers have no FISONs.
All natural numbers "have" a FISONThen all natural numbers would be in FISONs. But because of ∀n ∈ U(F): |ℕ \ {1, 2, 3, ..., n}| = ℵo all FISONs fail to contain all natural numbers.
No, there are already infinitely many countable ones (=with FISONs).If you really think there is a non-empty set of natural numbers whichOf course there is such a set. It contains almost all natural numbers.
don't "have" FISONs,
Yes, and omega is the least number larger than all those.then please say what the least natural number in that set is, or at theThe definable numbers are potentially infinite sequence. With n also
very least, how you'd go about finding it.
n+1 and n^n^n belong to it.
There is a big difference between subtracting one number from a set and subtracting all numbers.Removing all its elements by subtraction.The subtraction of all numbers which cannot empty ℕ cannot empty ℕ.You've never said what you mean by a number "emptying" a set.
It's unclear whether you mean the subtraction of each numberIf all natural numbers were individually definable, then there would not
individually, or of all numbers together.
be a difference.
imma subtract my latex set iykwimEven "subtraction" is a non-standard word, here. The opposite of "add"The opposite of addition is subtraction. Look for instance: subtraction+of+sets+latex
(hinzufügen) is "remove", not "subtract".
It can be emptied by the infinitely many sets {k} for every k in N.A FISON is a well-ordered set or segment or sequence. It has a largest element.A FISON is a set. Sets don't "end" with anything.No. It says simply that no FISON ending with n can be defined."Undefined as individuals" is an undefined notion,It all depends on the X from which N_def is formed. If X is N \Then its elements are mostly undefined as individuals.
{1},
None, but you should pay attention because ℕ is infinite and therefore cannot be emptied by finite sets.Every element has a finite FISON. ℕ is infinite. Therefore it
cannot be emptied by the elements of ℕ_def and also not by ℕ_def. >>>> A "finite" FISON? What other type is there?
No, you try reformulating.Subtracting a set or elements of a set. See above. Definable elements"emptying" N by a set or by elements of a set?What do you mean by "having" a FISON? What does it mean to "empty" NSimply try to understand. I have often stated the difference:
by a set or elements of a set? What is the significance, if any, of
being able to "empty" a set?
∀n ∈ U(F): |ℕ \ {1, 2, 3, ..., n}| = ℵo ℕ \ {1, 2, 3, ...} = { } >> Which doesn't address my question in the sightest. What do you mean by
can be subtracted individually. Undefinable elements can only be
subtracted collectively.
You haven't said what (if anything) you mean by a number emptying N.You seem unable to learn.
And every natural number "has" a FISON, not just some subset of them.
You said: The tending takes place, but not in a "place".
They are placed on the ordinal line and can tend to ℕ. This can happen >>> only on the ordinal line. Your assertion of the contrary is thereforeOf the many assertions I've made, the one you're referring to is
wrong.
unclear.
You really have problems to comprehend sentences. Try again.
Incoherent garbage."Defined numbers" appears not to be a coherent mathematical concept.The subtraction of all numbers which cannot empty ℕ cannot empty ℕ.
The collection of these numbers is ℕ_def.
It is your responsibility to make yourself understood.You haven't said what you mean by a number "emptying" a set.Even if I had not, an intelligent reader would know it.
And they will all be "defined".The current state of our discussion is that you have failed to give anyA defined number is a number that you can name such that I understand
coherent definition of "defined numbers";
what you mean. In every case you choose almost all numbers will be
greater.
Am Sun, 16 Mar 2025 17:23:00 +0100 schrieb WM:
On 16.03.2025 14:09, joes wrote:Oh, you meant that N is not a FISON.
Am Sat, 15 Mar 2025 18:13:57 +0100 schrieb WM:∀n ∈ U(F): |ℕ \ {1, 2, 3, ..., n}| = ℵo.
Defined numbers have FISONs and cannot empty ℕ.Yes they can.
Do you distrust the universal quantifier?
WM <wolfgang.mueckenheim@tha.de> wrote:
No. The definable natural numbers strive to the smallest infinite number
ω. But they are never there.
Whatever "strive" is supposed to mean.
N is defined as the smallest
inductive set.
A proof exists that they are there. ℕ_def contains all numbers the
subtraction of which from ℕ does not result in the empty set.
That's not a mathematical statement.
It's meaningless gibberish.
Obviously the subtraction of all numbers which cannot empty ℕ cannot
empty ℕ.
"Empty" in this sense is meaningless.
I spent quite a few posts to you
trying to get you to explain what you meant by a number "emptying" a set.
You never gave such an explanation.
JB who claims that lossless exchange causes losses but refuses to
explain this in detail, ....
He's been trying to explain this over quite a few posts.
.... and you who claims that the subtraction of all numbers which
cannot empty ℕ can empty ℕ
How dare you lie about what I have written! I have never claimed
anything involving the crankish notion of subtracting a number from a
set causing "emptying", whatever that might mean.
On 3/16/2025 3:27 PM, WM wrote:
The question here is
the index of the first disappearance of Bob.
The answer requires you to know what 'finite' means.
On 16.03.2025 20:41, Jim Burns wrote:
On 3/16/2025 3:27 PM, WM wrote:
The question here is
the index of the first disappearance of Bob.
The answer requires you to know what 'finite' means.
Answer using your definition of
finite index or finite natural number.
On 3/16/2025 12:00 PM, WM wrote:
No. The definable natural numbers strive to the smallest infinite
number ω. But they are never there.
What is wrong with you? N is all the natural numbers. Saying all does
not mean finite... Name a natural number that cannot be defined?
WM brought next idea :
I informed you several times:
ℕ \ {1, 2, 3, ...} = { }
When two sets are the same set, the difference set is the empty set.
On 3/16/2025 6:22 PM, WM wrote:
On 16.03.2025 20:41, Jim Burns wrote:
On 3/16/2025 3:27 PM, WM wrote:
The question here is
the index of the first disappearance of Bob.
The answer requires you to know what 'finite' means.
Answer using your definition of
finite index or finite natural number.
You misunderstand.
I don't need to find out what 'finite' means.
Do you (WM) know what 'finite' means?
Whatever.
The result is the same: I can't tell you (WM).
Not 'won't', 'can't'.
I try and the answer doesn't reach you.
Bob disappears from a superovine set ==
On 16.03.2025 21:08, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
N is defined as the smallest inductive set.
But that definition is impossible to satisfy. Sets are fixed, inductive "sets" are variable collections.
A proof exists that they are there. ℕ_def contains all numbers the
subtraction of which from ℕ does not result in the empty set.
That's not a mathematical statement.
The numbers 1, 2, 3 are such numbers. They are elements of that set.
It's meaningless gibberish.
You clearly know the meaning of these words.
Obviously the subtraction of all numbers which cannot empty ℕ cannot
empty ℕ.
"Empty" in this sense is meaningless.
You are not unable to understand the meaning. But you are dishonest.
I spent quite a few posts to you trying to get you to explain what you
meant by a number "emptying" a set. You never gave such an
explanation.
I informed you several times:
ℕ \ {1, 2, 3, ...} = { }
The set is emptied collectively by all natural numbers.
It cannot be emptied by definable natural numbers.
.... and you who claims that the subtraction of all numbers which
cannot empty ℕ can empty ℕ
How dare you lie about what I have written! I have never claimed
anything involving the crankish notion of subtracting a number from a
set causing "emptying", whatever that might mean.
You understand very well. You have seen that subtracting is a regular notion.
Regards, WM
WM <wolfgang.mueckenheim@tha.de> wrote:
On 16.03.2025 21:08, Alan Mackenzie wrote:
N is defined as the smallest inductive set.
But that definition is impossible to satisfy. Sets are fixed, inductive
"sets" are variable collections.
Wrong. An inductive set exists by the axiom of infinity.
But just how do you think inductive sets vary? Do they vary by the day
of the week, the phases of the moon, or what? Can you give two
"variations" of an inductive set, and specify an element which is in one
of these variations, but not the other?
[...] your N_def, as you have "defined" it, is satisfied by any proper subset of N.
Or in a different interpretation, N_def = N, since An e N, N\{n} [as well as N\{1, ..., n} --moebius] is non-empty.See?!
Am 17.03.2025 um 12:56 schrieb Alan Mackenzie:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 16.03.2025 21:08, Alan Mackenzie wrote:
N is defined as the smallest inductive set.
Indeed! Or rather, can be defined as such. :-P
But that definition is impossible to satisfy. Sets are fixed, inductive
"sets" are variable collections.
Huh?!
Wrong. An inductive set exists by the axiom of infinity.
Yeah, I already told this fucking asshole full of shit that fact in dsm.
"Das Unendlichkeitsaxiom besagt, dass es eine induktive Menge
gibt." ["The axiom of infinity states that there is an inductive set."]
Source: https://de.wikipedia.org/wiki/Induktive_Menge
But just how do you think inductive sets vary? Do they vary by the day
of the week, the phases of the moon, or what? Can you give two
"variations" of an inductive set, and specify an element which is in one
of these variations, but not the other?
Good question!
Hint@Mückenheim: There are many different inductive sets which we
usually consider as "fixed" (i.e. not varying). for example, IR, Q, Z,
IN, etc.
Hint@Mackenzie: WM calls IN in his ~textbook~ "für die ersten Semester"
a "potentially infinite" set (whatever that may mean). Consequently he
states an "axiom system for IN" which does not even allow to derive that
0 e IN (and for all n e IN: n+1 e IN).
Hmmm... He may now call this IN (mentioned in his book) "IN_def", who
knows?
[...] your N_def, as you have "defined" it, is satisfied by any proper
subset of N.
Indeed!
Or in a different interpretation, N_def = N, since An e N, N\{n} [as
well as N\{1, ..., n} --moebius] is non-empty.
See?!
.
.
.
WM has brought this to us :
On 17.03.2025 00:19, Chris M. Thomasson wrote:
On 3/16/2025 12:00 PM, WM wrote:
No. The definable natural numbers strive to the smallest infinite
number ω. But they are never there.
What is wrong with you? N is all the natural numbers. Saying all does
not mean finite... Name a natural number that cannot be defined?
The number next to ω cannot be defined.
Omega plus one *is* defined. There is no predecessor as it is a limit ordinal.
WM <wolfgang.mueckenheim@tha.de> wrote:
On 16.03.2025 21:08, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
[ .... ]
N is defined as the smallest inductive set.
But that definition is impossible to satisfy. Sets are fixed, inductive
"sets" are variable collections.
Wrong. An inductive set exist by the axiom of infinity.
But just how do you think inductive sets vary? Do they vary by the day
of the week, the phases of the moon, or what? Can you give two
"variations" of an inductive set, and specify an element which is in one
of these variations, but not the other?
A proof exists that they are there. ℕ_def contains all numbers the
subtraction of which from ℕ does not result in the empty set.
That's not a mathematical statement.
The numbers 1, 2, 3 are such numbers. They are elements of that set.
You clearly know the meaning of these words.
And their meaninglessness is clear.
You're doing a quantifier shift
And your N_def, as you have "defined" it, is satisfied by any
proper subset of N.
Or in a different interpretation, N_def = N,
since
An e N, N\{n} is non-empty.
Either you're incapable of writing
mathematically what you mean,
Obviously the subtraction of all numbers which cannot empty ℕ cannot >>>> empty ℕ.
"Empty" in this sense is meaningless.
You are not unable to understand the meaning. But you are dishonest.
I refuse to discuss things expressed in sloppy meaningless language, as "empty" used as a verb here is. A number cannot "empty" a set, because
the number is not an agent; it is not an operator;
it is not a function.
Such sloppy language allows you to reason sloppily, and possibly to
derive falsehoods as if they were facts.
I think you are capable of expressing your thoughts in a mathematical fashion. I wish I could be sure, though.
All you're saying in the above point is that N \ N = { }.
How dare you lie about what I have written! I have never claimed
anything involving the crankish notion of subtracting a number from a
set causing "emptying", whatever that might mean.
You understand very well. You have seen that subtracting is a regular
notion.
Subtraction is a function on two numbers mapping to a number of the same type. What you appear to be talking about is "removal" of an element or subset of a set from that set. Do you really use the word "subtrahieren"
in German for this?
On 17.03.2025 05:23, Jim Burns wrote:
On 3/16/2025 6:22 PM, WM wrote:
On 16.03.2025 20:41, Jim Burns wrote:
On 3/16/2025 3:27 PM, WM wrote:
The question here is
the index of the first disappearance of Bob.
The answer requires you to know what 'finite' means.
Answer using your definition of
finite index or finite natural number.
You misunderstand.
I don't need to find out what 'finite' means.
Do you (WM) know what 'finite' means?
Yes, it means the index is a natural number.
Then apply your [not.WM] version of "finite" and
answer at which finite index
the first loss in lossless exchange happens.
On 3/17/2025 12:58 AM, WM wrote:
On 17.03.2025 05:23, Jim Burns wrote:
On 3/16/2025 6:22 PM, WM wrote:
On 16.03.2025 20:41, Jim Burns wrote:
On 3/16/2025 3:27 PM, WM wrote:
The question here is
the index of the first disappearance of Bob.
The answer requires you to know what 'finite' means.
Answer using your definition of
finite index or finite natural number.
You misunderstand.
I don't need to find out what 'finite' means.
Do you (WM) know what 'finite' means?
Yes, it means the index is a natural number.
That's circular.
You won't or can't answer non.circularly.
Then apply your version of "finite" and
answer at which finite index
the first loss in lossless exchange happens.
After all swaps,
No swap exists immediately before all the swaps.
No visibleᵂᴹ and no darkᵂᴹ.
On 17.03.2025 05:23, Jim Burns wrote:
Do you (WM) know what 'finite' means?
In mathematics, particularly set theory,
a finite set is a set that has
a finite number of elements.
Informally, a finite set is a set which
one could in principle count
and finish counting.
(Wikipedia)
No swap exists immediately before all the swaps.
No visibleᵂᴹ and no darkᵂᴹ.
Therefore:
If Bob disappears, it happens at a finite index.
Name it or
confess that you have used a foolish idea.
On 3/17/2025 12:35 PM, Chris M. Thomasson wrote:
On 3/16/2025 10:10 PM, WM wrote:
On 17.03.2025 00:19, Chris M. Thomasson wrote:
On 3/16/2025 12:00 PM, WM wrote:
No. The definable natural numbers strive to the smallest infinite
number ω. But they are never there.
What is wrong with you? N is all the natural numbers. Saying all
does not mean finite... Name a natural number that cannot be defined?
The number next to ω cannot be defined.
Any natural number can be defined...
ω is outside of the realm of natural numbers...
On 3/17/2025 12:11 PM, WM wrote:
On 17.03.2025 05:23, Jim Burns wrote:
Do you (WM) know what 'finite' means?
In mathematics, particularly set theory,
a finite set is a set that has a finite number of elements.
Informally, a finite set is a set which
one could in principle count
and finish counting.
(Wikipedia)
Therefore,
the set of all finite.sizes does not have
a finite.size.
The set of all finite.sizes is same.sized as
the set of all finite sizes and Bob.
Do you agree that
the set of all finite.sizes has
has fuller.by.one sets which are NOT larger
and emptier.by.one sets which are NOT smaller ?
WM submitted this idea :
On 17.03.2025 20:57, Chris M. Thomasson wrote:
On 3/17/2025 12:35 PM, Chris M. Thomasson wrote:
On 3/16/2025 10:10 PM, WM wrote:
On 17.03.2025 00:19, Chris M. Thomasson wrote:
On 3/16/2025 12:00 PM, WM wrote:The number next to ω cannot be defined.
No. The definable natural numbers strive to the smallest infinite >>>>>>> number ω. But they are never there.
What is wrong with you? N is all the natural numbers. Saying all
does not mean finite... Name a natural number that cannot be defined? >>>>>
Any natural number can be defined...
ω is outside of the realm of natural numbers...
But not the natural number next to it.
There isn't one. It is the ordered set of natural numbers.
on 3/18/2025, WM supposed :
The set is ordered and if it is actually infinite, then all its
elements are there and do not appear from nothing but then it has a
greatest element.
Nope, it is a limit ordinal.
On 18.03.2025 10:32, FromTheRafters wrote:A greatest element is in direct contradiction to the infinity of the set.
WM submitted this idea :The set is ordered and if it is actually infinite, then all its elements
On 17.03.2025 20:57, Chris M. Thomasson wrote:There isn't one. It is the ordered set of natural numbers.
On 3/17/2025 12:35 PM, Chris M. Thomasson wrote:But not the natural number next to it.
On 3/16/2025 10:10 PM, WM wrote:ω is outside of the realm of natural numbers...
On 17.03.2025 00:19, Chris M. Thomasson wrote:Any natural number can be defined...
On 3/16/2025 12:00 PM, WM wrote:The number next to ω cannot be defined.
No. The definable natural numbers strive to the smallest infinite >>>>>>>> number ω. But they are never there.What is wrong with you? N is all the natural numbers. Saying all >>>>>>> does not mean finite... Name a natural number that cannot be
defined?
are there and do not appear from nothing but then it has a greatest
element.
On 17.03.2025 12:56, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 16.03.2025 21:08, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
[ .... ]
N is defined as the smallest inductive set.
But that definition is impossible to satisfy. Sets are fixed, inductive
"sets" are variable collections.
Wrong. An inductive set exist by the axiom of infinity.
The set of FISONs is an inductive set. But it is not ℕ because
∀n ∈ UF: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
The subtraction of the set of all FISONs all of which cannot empty ℕ cannot empty ℕ.
But just how do you think inductive sets vary? Do they vary by the day
of the week, the phases of the moon, or what? Can you give two
"variations" of an inductive set, and specify an element which is in one
of these variations, but not the other?
A proof exists that they are there. ℕ_def contains all numbers the >>>>> subtraction of which from ℕ does not result in the empty set.
That's not a mathematical statement.
The numbers 1, 2, 3 are such numbers. They are elements of that set.
You clearly know the meaning of these words.
And their meaninglessness is clear.
Your following statements prove that you understand the meaning.
You're doing a quantifier shift
Of course. Here it is justified since the subtraction of all FISONs
which cannot empty ℕ cannot empty ℕ.
And your N_def, as you have "defined" it, is satisfied by any
proper subset of N.
No, it is the union of FISONs.
Or in a different interpretation, N_def = N,
No. ℕ_def is a proper subset.
since An e N, N\{n} is non-empty.
Here we use
∀n ∈ UF: |ℕ \ {1, 2, 3, ..., n}| = ℵo
Either you're incapable of writing mathematically what you mean,
I did it frequently:
∀n ∈ UF: |ℕ \ {1, 2, 3, ..., n}| = ℵo
|ℕ \ {1, 2, 3, ...}| = 0
Obviously the subtraction of all numbers which cannot empty ℕ cannot >>>>> empty ℕ.
"Empty" in this sense is meaningless.
You are not unable to understand the meaning. But you are dishonest.
I refuse to discuss things expressed in sloppy meaningless language, as
"empty" used as a verb here is. A number cannot "empty" a set, because
the number is not an agent; it is not an operator;
Subtraction is an operator.
it is not a function. Such sloppy language allows you to reason
sloppily, and possibly to derive falsehoods as if they were facts.
You are unable to read or to understand. You criticise your reading or
your incoherent thinking, not my writing.
I think you are capable of expressing your thoughts in a mathematical
fashion. I wish I could be sure, though.
Simply read what I write. It is ridiculous.
Subtraction is a function on two numbers mapping to a number of the same
type. What you appear to be talking about is "removal" of an element or
subset of a set from that set. Do you really use the word "subtrahieren"
in German for this?
Set subtraction is also used in English.
Regards, WM
On 18.03.2025 10:56, FromTheRafters wrote:
on 3/18/2025, WM supposed :
The set is ordered and if it is actually infinite, then all its
elements are there and do not appear from nothing but then it has a
greatest element.
Nope, it is a limit ordinal.
All elements of ℕ are there. That is the assumption. If no greatest can
be identified, then the reason are dark numbers.
Otherwise only potential infinity could solve the dilemma.
Regards, WM
WM <wolfgang.mueckenheim@tha.de> wrote:
The set of FISONs is an inductive set. But it is not ℕ because
∀n ∈ UF: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
That "because" doesn't hold
The inductive set which the axiom of infinity causes to exist is N.
The subtraction of the set of all FISONs all of which cannot empty ℕ
cannot empty ℕ.
Gibberish.
On 18.03.2025 12:57, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
The set of FISONs is an inductive set. But it is not ℕ because
∀n ∈ UF: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
That "because" doesn't hold
Because you say so? That's not significant.
Even if you can't understand the correct argument, ....
it is decisive: ℕ_def contains all numbers the subtraction of which
from ℕ does not result in the empty set. Obviously the subtraction of
all numbers which cannot empty ℕ cannot empty ℕ.
The subtraction of the set of all FISONs all of which cannot empty ℕ
cannot empty ℕ.
Gibberish.
Try to improve your thinking ability. Then come back. At the present
stage you only waste my time.
Regards, WM
WM <wolfgang.mueckenheim@tha.de> wrote:
On 18.03.2025 12:57, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
The set of FISONs is an inductive set. But it is not ℕ because
∀n ∈ UF: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
That "because" doesn't hold
Because you say so? That's not significant.
There's no connection between the bit before the because and the bit afterwards. You're the one who's lacking a degree in maths,
not me,
and it shows
it is decisive: ℕ_def contains all numbers the subtraction of which
from ℕ does not result in the empty set. Obviously the subtraction of
all numbers which cannot empty ℕ cannot empty ℕ.
I've told you and explained to you, I don't know how many times, that
that is ill-conceived meaningless gibberish.
On 3/18/2025 5:00 AM, WM wrote:
Bob can only disappear in a set of finite size
because all exchanges appear at finite sizes.
Bob disappears from (is swapped from)
each finite set A.
Bob disappears to (is swapped to)
a different finite set
After all the swaps,
without ever disappearing into
anywhere other than a finite set,
Bob disappears out of all finite sets.
The swaps are ⟨n⇄n+1⟩ for all n,
in order by n,
in the emptiest inductive set.
The set of all finite.sizes is same.sized as
the set of all finite sizes and Bob.
Give the index where Bob is lost.
Before all the swaps,
Bob is not lost from all finite indices.
After all the swaps,
Bob is lost from all finite indices.
There is no finite index,
either visible or dark,
from which Bob is last.lost,
or to which Bob is last.lost.
On 17.03.2025 19:13, Jim Burns wrote:
On 3/17/2025 12:11 PM, WM wrote:
On 17.03.2025 05:23, Jim Burns wrote:
Do you (WM) know what 'finite' means?
In mathematics, particularly set theory,
a finite set is
a set that has a finite number of elements.
Informally, a finite set is a set which
one could in principle count
and finish counting.
(Wikipedia)
Therefore,
the set of all finite.sizes does not have
a finite.size.
Bob can only disappear in a set of finite size
because all exchanges appear at finite sizes.
The set of all finite.sizes is same.sized as
the set of all finite sizes and Bob.
Give the index where Bob is lost.
Do you agree that
the set of all finite.sizes has
has fuller.by.one sets which are NOT larger
and emptier.by.one sets which are NOT smaller >
Stop that gobbledegook.
All exchanges happen at finite sizes.
We know what finite site means.
It is quoted above.
Now act accordingly or
confess that you were wrong.
WM <wolfgang.mueckenheim@tha.de> wrote:
On 18.03.2025 12:57, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
The set of FISONs is an inductive set. But it is not ℕ because <bla bla bla>
That "because" doesn't hold
Because you say so? That's not significant.
There's no connection between the bit before the because and the bit afterwards. You're the one who's lacking a degree in maths, not me, and
it shows.
Even if you can't understand the correct argument, ....
Am 18.03.2025 um 17:14 schrieb Alan Mackenzie:
WM <wolfgang.mueckenheim@tha.de> wrote:
wenn die natürlichen Zahlen bzw. IN nach von Neumann definiert sind (was heute "Standard" ist), *ist* IN gleich the set of FISONs.
Am 18.03.2025 um 17:14 schrieb Alan Mackenzie:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 18.03.2025 12:57, Alan Mackenzie wrote:
WM <wolfgang.mueckenheim@tha.de> wrote:
The set of FISONs is an inductive set. But it is not ℕ because <bla >>>>> bla bla>
Mückenheim, ich weiß nicht, wie oft man es Dir schon erklärt hat, aber wenn die natürlichen Zahlen bzw. IN nach von Neumann definiert sind (was heute "Standard" ist), *ist* IN gleich the set of FISONs.
Hint: IN = {{}, {{}}, {{}, {{}}}, ...} = {{}, {0}, {0, 1}, ...} = {F : F
is a FISON} = the_set_of_FISONs.
That "because" doesn't hold
Indeed! Especially, since his claim ("the set of FISONs is not ℕ") is wrong in the context of ZF(C) [with IN defined due to von Neumann].
Because you say so? That's not significant.
Nein, Mückenheim. Sondern, weil es so ist. (Siehe Erklärung weiter
oben.) It seems that you are just too dumb to get it.
There's no connection between the bit before the because and the bit
afterwards. You're the one who's lacking a degree in maths, not me, and
it shows.
Indeed!
Even if you can't understand the correct argument, ....
*lol*
On 18.03.2025 12:57, Alan Mackenzie wrote:There are many such sets: N\{0}, N\{1}, N\{2}, ...
WM <wolfgang.mueckenheim@tha.de> wrote:
The set of FISONs is an inductive set. But it is not ℕ because ∀n ∈ >>> UF: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
That "because" doesn't hold
Because you say so? That's not significant. Even if you can't understand
the correct argument, it is decisive: ℕ_def contains all numbers the subtraction of which from ℕ does not result in the empty set. Obviously
the subtraction of all numbers which cannot empty ℕ cannot empty ℕ.
--The inductive set which the axiom of infinity causes to exist is N.
On 18.03.2025 17:14, Alan Mackenzie wrote:Headline please?
WM <wolfgang.mueckenheim@tha.de> wrote:No.
On 18.03.2025 12:57, Alan Mackenzie wrote:There's no connection between the bit before the because and the bit
WM <wolfgang.mueckenheim@tha.de> wrote:
Because you say so? That's not significant.The set of FISONs is an inductive set. But it is not ℕ because ∀n ∈ >>>>> UF: |ℕ \ {1, 2, 3, ..., n}| = ℵo.That "because" doesn't hold
afterwards. You're the one who's lacking a degree in maths,
Indeed.not me, and it showsthat degrees are not guaranteeing the ability to think mathematically.
This is so stupid. OF COURSE N contains more than one element.Can you imagine a natural number that when subtracted from ℕ does notit is decisive: ℕ_def contains all numbers the subtraction of whichI've told you and explained to you, I don't know how many times, that
from ℕ does not result in the empty set. Obviously the subtraction of
all numbers which cannot empty ℕ cannot empty ℕ.
that is ill-conceived meaningless gibberish.
empty ℕ? Here are some examples: 2, 5, 17. More are available.
Can you imagine a set of natural numbers that when subtracted from ℕHold on. No single one or no FISON of them? Not that it changes anything.
does not empty ℕ? Here are some examples: {4, 9}, {1, 2, 3, 4, 5}. More
are vailable.
Now imagine all FISONs. None empties ℕ when applied one after the other.
Could that change when all are applied together?Yup. Why shouldn't it?
I think I read somewhere that he has a Doctor title in physics.Yes, this is true.*) The funny thing is that he never received a
Perhaps you can help me with something that WM refuses to answer: In
German mathematical texts, is the verb "subtrahieren" used to refer to
the removal of an element or subset of a set from that set?
Thanks!
Am 18.03.2025 um 20:45 schrieb Alan Mackenzie:
I think I read somewhere that he has a Doctor title in physics.Yes, this is true.*) The funny thing is that he never received a
thorough introduction into mathematics, and it shows! (Really!)
Hint: "[WM's] conclusions are based on the sloppiness of his notions,
his inability of giving precise definitions, his fundamental
misunderstanding of elementary mathematical concepts, and sometimes, as
the late Dik Winter remarked [...], on nothing at all."
-- Franz Lemmermeyer**)
_______________________________________________________________________
*) https://de.wikipedia.org/wiki/Wolfgang_M%C3%BCckenheim
**) https://de.wikipedia.org/wiki/Franz_Lemmermeyer
Moebius <invalid@example.invalid> wrote:
Perhaps you can help me with something that WM refuses to answer: In
German mathematical texts, is the verb "subtrahieren" used to refer to
the removal of an element or subset of a set from that set?
Indeed! Especially, since his claim ("the set of FISONs is not ℕ") is
wrong in the context of ZF(C) [with IN defined due to von Neumann].
I've given up arguing with him.
Am Tue, 18 Mar 2025 16:34:21 +0100 schrieb WM:
On 18.03.2025 12:57, Alan Mackenzie wrote:There are many such sets: N\{0}, N\{1}, N\{2}, ...
WM <wolfgang.mueckenheim@tha.de> wrote:
The set of FISONs is an inductive set. But it is not ℕ because ∀n ∈ >>>> UF: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
That "because" doesn't hold
Because you say so? That's not significant. Even if you can't understand
the correct argument, it is decisive: ℕ_def contains all numbers the
subtraction of which from ℕ does not result in the empty set. Obviously
the subtraction of all numbers which cannot empty ℕ cannot empty ℕ.
On 18.03.2025 20:45, Alan Mackenzie wrote:
Moebius <invalid@example.invalid> wrote:
Perhaps you can help me with something that WM refuses to answer: In
German mathematical texts, is the verb "subtrahieren" used to refer to
the removal of an element or subset of a set from that set?
I told you that even in English set subtraction is used. https://math.stackexchange.com/questions/1514758/set-theory-distributive-laws-with-respect-to-the-subtraction-of-sets
Indeed! Especially, since his claim ("the set of FISONs is not ℕ")
is wrong in the context of ZF(C) [with IN defined due to von
Neumann].
Therefore the context of ZF is wrong!
I've given up arguing with him.
Because you have no chance either to make us believe that you can't understand this simple sentence: The subtraction of all FISONs which
cannot empty ℕ cannot empty ℕ. Nor can you reasonably claim that all FISONs fail when applied one after the other but succeed when applied together.
Regards, WM
Am 18.03.2025 um 20:45 schrieb Alan Mackenzie:
Perhaps you can help me with something that WM refuses to answer: In
German mathematical texts, is the verb "subtrahieren" used to refer to
the removal of an element or subset of a set from that set?
Not that I'm aware of. We usually refer to A \ B as the set difference between A and B (dt.: Differenzmenge von A und B).
"subtrahieren" clearly refers to the arithmetical operation of
subtraction.
On the other hand, sometimes (rarely) you will stumble over a reference
to "Subtraktion von Mengen" - just to find an explanation concerning the
_set difference_ (of sets). [...]
What can I say?
"Cranks who contradict some mainstream opinion in some highly technical field, (e.g. mathematics [etc.]) may:
1. exhibit a marked lack of technical ability,
*** 2. misunderstand or not use standard notation and terminology, ***
[...]" (Wikipedia)
Thanks!
N/p.
Man, let's face it, WM is a crank par excellence.
"Discussing" with him is a waste of time.
Am Tue, 18 Mar 2025 17:29:08 +0100 schrieb WM:
Now imagine all FISONs. None empties ℕ when applied one after the other.Hold on. No single one or no FISON of them?
Could that change when all are applied together?Yup. Why shouldn't it?
Am 18.03.2025 um 20:45 schrieb Alan Mackenzie:
Perhaps you can help me with something that WM refuses to answer: In
German mathematical texts, is the verb "subtrahieren" used to refer to
the removal of an element or subset of a set from that set?
Not that I'm aware of.
WM <wolfgang.mueckenheim@tha.de> wrote:
On 18.03.2025 20:45, Alan Mackenzie wrote:
Moebius <invalid@example.invalid> wrote:
[...] Especially, since his claim ("the set of FISONs is not ℕ")
is wrong in the context of ZF(C) [with IN defined due to von
Neumann].
Therefore the context of ZF is wrong! [WM]
I've given up arguing with him.
Because you have no chance either to make us believe that you can't
understand this simple sentence: The subtraction of all FISONs which
cannot empty ℕ cannot empty ℕ. Nor can you reasonably claim that all
FISONs fail when applied one after the other but succeed when applied
together.
WM <wolfgang.mueckenheim@tha.de> wrote:
On 18.03.2025 20:45, Alan Mackenzie wrote:
Moebius <invalid@example.invalid> wrote:
[...] Especially, since his claim ("the set of FISONs is not ℕ")
is wrong in the context of ZF(C) [with IN defined due to von
Neumann].
Therefore the context of ZF is wrong! [WM]
I've given up arguing with him.
Because you have no chance either to make us believe that you can't
understand this simple sentence: The subtraction of all FISONs which
cannot empty ℕ cannot empty ℕ. Nor can you reasonably claim that all
FISONs fail when applied one after the other but succeed when applied
together.
WM <wolfgang.mueckenheim@tha.de> wrote:
On 18.03.2025 20:45, Alan Mackenzie wrote:
Moebius <invalid@example.invalid> wrote:
[...] Especially, since his claim ("the set of FISONs is not ℕ")
is wrong in the context of ZF(C) [with IN defined due to von
Neumann].
Therefore the context of ZF is wrong! [WM]
I've given up arguing with him.
Because you have no chance either to make us believe that you can't
understand this simple sentence: The subtraction of all FISONs which
cannot empty ℕ cannot empty ℕ. Nor can you reasonably claim that all
FISONs fail when applied one after the other but succeed when applied
together.
WM <wolfgang.mueckenheim@tha.de> wrote:
On 18.03.2025 20:45, Alan Mackenzie wrote:
Moebius <invalid@example.invalid> wrote:
Perhaps you can help me with something that WM refuses to answer: In
German mathematical texts, is the verb "subtrahieren" used to refer to
the removal of an element or subset of a set from that set?
I told you that even in English set subtraction is used.
https://math.stackexchange.com/questions/1514758/set-theory-distributive-laws-with-respect-to-the-subtraction-of-sets
Indeed! Especially, since his claim ("the set of FISONs is not ℕ")
is wrong in the context of ZF(C) [with IN defined due to von
Neumann].
Therefore the context of ZF is wrong!
I've given up arguing with him.
Because you have no chance either to make us believe that you can't
understand this simple sentence: The subtraction of all FISONs which
cannot empty ℕ cannot empty ℕ. Nor can you reasonably claim that all
FISONs fail when applied one after the other but succeed when applied
together.
I have already dealt with the things in
that last paragraph.
On 18.03.2025 21:40, joes wrote:It contains only numbers with FISONs.
Am Tue, 18 Mar 2025 16:34:21 +0100 schrieb WM:ℕ_def contains only numbers with FISONs. N\{0} is not a FISON.
On 18.03.2025 12:57, Alan Mackenzie wrote:There are many such sets: N\{0}, N\{1}, N\{2}, ...
WM <wolfgang.mueckenheim@tha.de> wrote:Because you say so? That's not significant. Even if you can't
The set of FISONs is an inductive set. But it is not ℕ because ∀n ∈ >>>>> UF: |ℕ \ {1, 2, 3, ..., n}| = ℵo.That "because" doesn't hold
understand the correct argument, it is decisive: ℕ_def contains all
numbers the subtraction of which from ℕ does not result in the empty
set. Obviously the subtraction of all numbers which cannot empty ℕ
cannot empty ℕ.
WM presented the following explanation :
Obviously the subtraction of the set of all FISONs which as single
FISONs cannot empty ℕ cannot empty ℕ.
Do you believe that the naturals having been extracted from the infinite
set, leaves the still infinite set devoid of naturals?
On 18.03.2025 21:52, joes wrote:Or do you not have a maths degree?
Am Tue, 18 Mar 2025 17:29:08 +0100 schrieb WM:
Headline please?You're the one who's lacking a degree in maths,No.
The union of *all* FISONs (which is an infinite set and not a FISONNo single FISON.Now imagine all FISONs. None empties ℕ when applied one after theHold on. No single one or no FISON of them?
other.
Wrong and not an answer to my question.Obviously the subtraction of the set of all FISONs which as singleCould that change when all are applied together?Yup. Why shouldn't it?
FISONs cannot empty ℕ cannot empty ℕ.
On 18.03.2025 17:34, Jim Burns wrote:
On 3/18/2025 5:00 AM, WM wrote:
Bob can only disappear in a set of finite size
because all exchanges appear at finite sizes.
Bob disappears from (is swapped from)
each finite set A.
Bob disappears to (is swapped to)
a different finite set
Yes.
But he never leaves the matrix!
After all the swaps,
without ever disappearing into
anywhere other than a finitec set,
Bob disappears out of all finite sets.
No, that is impossible.
If there is an "after all swaps",
then all O have settled within the matrix.
Lossless exchanges with losses
are not allowed.
The swaps are ⟨n⇄n+1⟩ for all n,
in order by n,
in the emptiest inductive set.
Yes, for all n reachable by induction.
After all the swaps,
Bob is lost from all finite indices.
No, that is impossible.
Here lies your mistake.
The matrix has no drain!
WM wrote on 3/19/2025 :
On 19.03.2025 13:49, FromTheRafters wrote:
WM presented the following explanation :
Obviously the subtraction of the set of all FISONs which as single
FISONs cannot empty ℕ cannot empty ℕ.
Do you believe that the naturals having been extracted from the
infinite set, leaves the still infinite set devoid of naturals?
I don't believe but have proven: UF = ℕ ==> Ø = ℕ.
FISONs cannit empty ℕ.
Can you answer the question anyway, without the delusions?
Am Wed, 19 Mar 2025 10:39:38 +0100 schrieb WM:
On 18.03.2025 21:40, joes wrote:It contains only numbers with FISONs.
Am Tue, 18 Mar 2025 16:34:21 +0100 schrieb WM:ℕ_def contains only numbers with FISONs. N\{0} is not a FISON.
On 18.03.2025 12:57, Alan Mackenzie wrote:There are many such sets: N\{0}, N\{1}, N\{2}, ...
WM <wolfgang.mueckenheim@tha.de> wrote:Because you say so? That's not significant. Even if you can't
The set of FISONs is an inductive set. But it is not ℕ because ∀n ∈That "because" doesn't hold
UF: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
understand the correct argument, it is decisive: ℕ_def contains all
numbers the subtraction of which from ℕ does not result in the empty >>>> set. Obviously the subtraction of all numbers which cannot empty ℕ
cannot empty ℕ.
Am Wed, 19 Mar 2025 12:55:49 +0100 schrieb WM:
Obviously the subtraction of the set of all FISONs which as singleWrong and not an answer to my question.
FISONs cannot empty ℕ cannot empty ℕ.
On 19.03.2025 16:44, FromTheRafters wrote:
WM wrote on 3/19/2025 :
On 19.03.2025 13:49, FromTheRafters wrote:
WM presented the following explanation :
Obviously
the subtraction of the set of all FISONs which
as single FISONs cannot empty ℕ
cannot empty ℕ.
Do you believe that the naturals
having been extracted from the infinite set,
leaves the still infinite set devoid of naturals?
I don't believe but have proven:
UF = ℕ ==> Ø = ℕ.
FISONs cannit empty ℕ.
Can you answer the question anyway, without the delusions?
Obviously
the subtraction of the set of all FISONs which
as single FISONs cannot empty Cantor's ℕ
cannot empty Cantor's ℕ.
Obviously
On 3/19/2025 11:55 AM, WM wrote:
Obviously
the subtraction of the set of all FISONs which
as single FISONs cannot empty Cantor's ℕ
cannot empty Cantor's ℕ.
All the finite initial segments of ⋂𝒫ⁱⁿᵈ(ℕ)
empty ⋂𝒫ⁱⁿᵈ(ℕ)
⋂𝒫ⁱⁿᵈ(ℕ)\⋃{F} = {}
Obviously
Please prove that it is obvious.
On 3/18/2025 12:58 PM, WM wrote:
⎛ Bob is in.matrix before.swaps, but
⎝ Bob is NOT in.matrix after.swaps.
After all the swaps,
without ever disappearing into
anywhere other than a finitec set,
Bob disappears out of all finite sets.
No, that is impossible.
Then there are subsets larger than their sets.
If there is an "after all swaps",
then all O have settled within the matrix.
That matrix,
with rows and columns indexed by
the emptiest inductive set,
after all the swaps,
that matrix doesn't hold any O.
⎛ Assume otherwise.
⎜ Assume that it's after all swaps,
⎜ and O is in cell <k,1>
After all the swaps,
Bob is lost from all finite indices.
No, that is impossible.
Here lies your mistake.
The matrix has no drain!
If the set of
all set.sizes needing a drain
is a set of a size needing a drain,
then
that set has subsets larger than it, the set.
On 19.03.2025 19:00, Jim Burns wrote:It can't be larger. It contains nothing apart from the inductive set.
On 3/19/2025 11:55 AM, WM wrote:
Of course the inductive subsets of ℕ are eqiuvalent.Obviously the subtraction of the set of all FISONs which as singleAll the finite initial segments of ⋂𝒫ⁱⁿᵈ(ℕ)
FISONs cannot empty Cantor's ℕ cannot empty Cantor's ℕ.
empty ⋂𝒫ⁱⁿᵈ(ℕ) ⋂𝒫ⁱⁿᵈ(ℕ)\⋃{F} = {}
I use Cantor's ℕ which is larger than all its inductive subsets.
Awful wording.ObviouslyPlease prove that it is obvious.
For Cantor's ℕ we have:
|ℕ \ {1}| = ℵo |ℕ \ {1, 2, 3, ..., n}| = ℵo ==>
|ℕ \ {1, 2, 3, ..., n+1}| = ℵo
All FISONs fail to empty ℕ when applied one after the other.
All FISONs fail to empty ℕ when applied at once, because mathematics isOnly you haven't "applied" all of them. That would be
not time-dependent.
Am Wed, 19 Mar 2025 20:30:27 +0100 schrieb WM:
I use Cantor's ℕ which is larger than all its inductive subsets.It can't be larger. It contains nothing apart from the inductive set.
Awful wording.ObviouslyPlease prove that it is obvious.
For Cantor's ℕ we have:
|ℕ \ {1}| = ℵo |ℕ \ {1, 2, 3, ..., n}| = ℵo ==>
|ℕ \ {1, 2, 3, ..., n+1}| = ℵo
All FISONs fail to empty ℕ when applied one after the other.
All FISONs fail to empty ℕ when applied at once, because mathematics isOnly you haven't "applied" all of them.
not time-dependent.
That would be
N \ {1, 2, 3, ...}
On 19.03.2025 21:15, joes wrote:Not a proof, not a valid implication, and nonsense follows.
Am Wed, 19 Mar 2025 20:30:27 +0100 schrieb WM:
Cantor thought so too, but both of you are in error.I use Cantor's ℕ which is larger than all its inductive subsets.It can't be larger. It contains nothing apart from the inductive set.
Proof: UF = ℕ ==> Ø = ℕ.
It's both wrong and awfully worded. Perhaps you meant that no finiteCorrect description of mathematical facts.Awful wording.For Cantor's ℕ we have:ObviouslyPlease prove that it is obvious.
|ℕ \ {1}| = ℵo |ℕ \ {1, 2, 3, ..., n}| = ℵo ==>
|ℕ \ {1, 2, 3, ..., n+1}| = ℵo
All FISONs fail to empty ℕ when applied one after the other.
Derivation of that limit?When applied one after the other, all FISONs fail. Therefore they alsoAll FISONs fail to empty ℕ when applied at once, because mathematicsOnly you haven't "applied" all of them.
is not time-dependent.
fail when being applied at once.
I'll leave you to your physics.That would be N \ {1, 2, 3, ...}You cannot believe in mathematics? Then try another occupation.
On 19.03.2025 16:18, Jim Burns wrote:
On 3/18/2025 12:58 PM, WM wrote:
On 18.03.2025 17:34, Jim Burns wrote:
After all the swaps,
without ever disappearing into
anywhere other than a finite set,
Bob disappears out of all finite sets.
No, that is impossible.
Then there are subsets larger than their sets.
No, there are dark numbers. That is enough.
WM submitted this idea :
On 19.03.2025 16:44, FromTheRafters wrote:
WM wrote on 3/19/2025 :
On 19.03.2025 13:49, FromTheRafters wrote:
WM presented the following explanation :
Obviously the subtraction of the set of all FISONs which as single >>>>>> FISONs cannot empty ℕ cannot empty ℕ.
Do you believe that the naturals having been extracted from the
infinite set, leaves the still infinite set devoid of naturals?
I don't believe but have proven: UF = ℕ ==> Ø = ℕ.
FISONs cannit empty ℕ.
Can you answer the question anyway, without the delusions?
Obviously the subtraction of the set of all FISONs which as single
FISONs cannot empty Cantor's ℕ cannot empty Cantor's ℕ.
A simple no would have been sufficient.
On 3/19/2025 3:19 PM, WM wrote:
On 19.03.2025 16:18, Jim Burns wrote:
On 3/18/2025 12:58 PM, WM wrote:
On 18.03.2025 17:34, Jim Burns wrote:
After all the swaps,
without ever disappearing into
anywhere other than a finite set,
Bob disappears out of all finite sets.
No, that is impossible.
Then there are subsets larger than their sets.
No, there are dark numbers. That is enough.
⎛ A set is WM.sized ==
⎜ membership.change.by.one changes size.
⎜ #A < #Aᣕᵃ
Am Wed, 19 Mar 2025 23:21:30 +0100 schrieb WM:
On 19.03.2025 21:15, joes wrote:Not a proof, not a valid implication,
Am Wed, 19 Mar 2025 20:30:27 +0100 schrieb WM:Cantor thought so too, but both of you are in error.
I use Cantor's ℕ which is larger than all its inductive subsets.It can't be larger. It contains nothing apart from the inductive set.
Proof: UF = ℕ ==> Ø = ℕ.
Perhaps you meant that no finite
number of FISONs "empties" N; but there is an infinite number of them.
WM formulated on Thursday :
On 19.03.2025 23:25, FromTheRafters wrote:
WM submitted this idea :
On 19.03.2025 16:44, FromTheRafters wrote:
WM wrote on 3/19/2025 :
On 19.03.2025 13:49, FromTheRafters wrote:
WM presented the following explanation :
Obviously the subtraction of the set of all FISONs which as
single FISONs cannot empty ℕ cannot empty ℕ.
Do you believe that the naturals having been extracted from the
infinite set, leaves the still infinite set devoid of naturals?
I don't believe but have proven: UF = ℕ ==> Ø = ℕ.
FISONs cannit empty ℕ.
Can you answer the question anyway, without the delusions?
Obviously the subtraction of the set of all FISONs which as single
FISONs cannot empty Cantor's ℕ cannot empty Cantor's ℕ.
A simple no would have been sufficient.
You cannot comprehend this complicated sentence. That does not make my
explanation delusionary.
Correct, what does is the fact that an infinite set is not equal to an
empty set.
WM explained :
I did not prove that an infinite set is equal to an empty set. I
proved that *if* the union of FISONs is ℕ, then ℕ is empty. That is
far from what you think.
I think that you didn't 'prove' anything.
Your notions, based on your
intuition alone,
On 20.03.2025 13:42, FromTheRafters wrote:
WM explained :
lololoThen you cannot follow mathematical thought.I did not prove that an infinite set is equal to an empty set. II think that you didn't 'prove' anything.
proved that *if* the union of FISONs is ℕ, then ℕ is empty. That is
far from what you think.
Your notions, based on your intuition alone,
They are based on induction alone:
Peano: 1 and if n then n+1.
Zermelo: { } and if a then {a}.
WM: ℕ \ {1} = ℵo and if ℕ \ {1, 2, 3, ..., n} = ℵo then ℕ \ {1, 2, 3,
..., n+1} = ℵo.
The infinities defined by induction by these three derivations cannot be distinguished.
On 19.03.2025 23:51, Jim Burns wrote:
On 3/19/2025 3:19 PM, WM wrote:
On 19.03.2025 16:18, Jim Burns wrote:
On 3/18/2025 12:58 PM, WM wrote:
On 18.03.2025 17:34, Jim Burns wrote:
After all the swaps,
without ever disappearing into
anywhere other than a finite set,
Bob disappears out of all finite sets.
No, that is impossible.
Then there are subsets larger than their sets.
No, there are dark numbers. That is enough.
⎛ A set is WM.sized ==
⎜ membership.change.by.one changes size.
⎜ #A < #Aᣕᵃ
If Bob leaves the matrix,
then logic is violated.
If Bob leaves the matrix,
then logic is violated.
Therefore Bob cannot leave the matrix
in reasonable mathematics.
All your waffle cannot change this fact.
Am Thu, 20 Mar 2025 15:57:15 +0100 schrieb WM:v. Neuman: { } and if {0, 1, 2, 3, ..., n} then {{0, 1, 2, 3, ..., n+1}}.
They are based on induction alone:
Peano: 1 and if n then n+1.
Zermelo: { } and if a then {a}.
WM: ℕ \ {1} = ℵo and if ℕ \ {1, 2, 3, ..., n} = ℵo then ℕ \ {1, 2, 3,
..., n+1} = ℵo.
The infinities defined by induction by these three derivations cannot be
distinguished.
One of these is not like the others.
On 3/20/2025 5:10 AM, WM wrote:
If Bob leaves the matrix,
then logic is violated.
That size must change with membership
can be abbreviated as 'WM.logic'.
Our sets do not change.
If Bob leaves the matrix,
then logic is violated.
Therefore Bob cannot leave the matrix
in reasonable mathematics.
Not all sets obey WM.logic.
On 20.03.2025 16:45, Jim Burns wrote:
On 3/20/2025 5:10 AM, WM wrote:
If Bob leaves the matrix,
then logic is violated.
That size must change with membership
can be abbreviated as 'WM.logic'.
Size of sets does not change.
No Bob can vanish.
Our sets do not change.
But you claim that
the set of O's shrinks from infinity to empty.
If Bob leaves the matrix,
then logic is violated.
Therefore Bob cannot leave the matrix
in reasonable mathematics.
Not all sets obey WM.logic.
All manipulations on sets obey logic.
On 3/20/2025 3:52 PM, WM wrote:On 20.03.2025 23:25, Jim Burns wrote:
On 3/20/2025 3:52 PM, WM wrote:
For sets not.having a WM.size,
Bob vanishing isn't a size.change.
Our sets do not change.
But you claim that
the set of O's shrinks from infinity to empty.
I claim that
the set of X's and O's is the same size as
the set of X's,
On 3/21/2025 3:50 AM, WM wrote:
On 20.03.2025 23:25, Jim Burns wrote:
For sets not.having a WM.size,
Bob vanishing isn't a size.change.
Only if reducing isn't reducing.
What you (WM) think is reducing
isn't reducing.
On 20.03.2025 23:25, Jim Burns wrote:
For sets not.having a WM.size,
Bob vanishing isn't a size.change.
Only if reducing isn't reducing.
Our sets do not change.
Only if changing isn't changing.
On 21.03.2025 18:39, Jim Burns wrote:
On 3/21/2025 3:50 AM, WM wrote:
On 20.03.2025 23:25, Jim Burns wrote:
For sets not.having a WM.size,
Bob vanishing isn't a size.change.
Only if reducing isn't reducing.
What you (WM) think is reducing
isn't reducing.
You confuse
the clear fact that
in the reality of
sets vanishing means reducing
with
the foolish claim that
cardinality was a meaningful notion.
Learn that
even Cantor has accepted that
the positive numbers have more reality than
the even positive numbers.
He said that is not in conflict with
the identical cardinality of both sets.
And he was right!
"Counable" is simply
another name for potential infinity.
Therefore
vanishing odd numbers means
reducing the reality of the set.
Therefore the sentence
"What you (WM) think is reducing isn't reducing"
exhibits you as a snooty dilettante who
cannot distinguish between cardinality and reality.
WM pretended :
On 21.03.2025 18:39, Jim Burns wrote:
On 3/21/2025 3:50 AM, WM wrote:
On 20.03.2025 23:25, Jim Burns wrote:
For sets not.having a WM.size,
Bob vanishing isn't a size.change.
Only if reducing isn't reducing.
What you (WM) think is reducing
isn't reducing.
You confuse the clear fact that in the reality of sets vanishing means
reducing with the foolish claim that cardinality was a meaningful notion.
Learn that even Cantor has accepted that the positive numbers have
more reality than the even positive numbers. He said that is not in
conflict with the identical cardinality of both sets. And he was
right! "Counable" is simply another name for potential infinity.
Therefore vanishing odd numbers means reducing the reality of the set.
Therefore the sentence "What you (WM) think is reducing isn't
reducing" exhibits you as a snooty dilettante who cannot distinguish
between cardinality and reality.
What does reality have to do with it?
On 3/21/2025 2:15 PM, WM wrote:
On 21.03.2025 18:39, Jim Burns wrote:
On 3/21/2025 3:50 AM, WM wrote:
On 20.03.2025 23:25, Jim Burns wrote:
For sets not.having a WM.size,
Bob vanishing isn't a size.change.
Only if reducing isn't reducing.
What you (WM) think is reducing
isn't reducing.
You confuse
the clear fact that
in the reality of
sets vanishing means reducing
with
the foolish claim that
cardinality was a meaningful notion.
The set of all sizes.which.WM.considers.sizes
does not have a size.which.WM.considers.a.size.
Learn that
even Cantor has accepted that
the positive numbers have more reality than
the even positive numbers.
Without context, I can't be sure, but
I suspect that
Cantor's "more reality" and
Zermelo's "simplest" serve
the same purpose as my "emptier" and "fuller",
which is to rank infinite sets by
something _other than_ by size.
He said that is not in conflict with
the identical cardinality of both sets.
Thank you.
Two sets of the same cardinality,
one a proper subset of the other,
can be swapped set.wise, one for the other,
and preserve size.
Either swapping all at once,
or swapping in infinitely.many singleton.swaps,
size is preserved,
but reality;simplicity;fullness isn't preserved,
and Bob can disappear without leaving.
On 22.03.2025 06:11, Jim Burns wrote:That a superset contains elements the subset doesn't is trivial.
On 3/21/2025 2:15 PM, WM wrote:It is infinite but nevertheless obeys the logic of lossless exchanges do
On 21.03.2025 18:39, Jim Burns wrote:The set of all sizes.which.WM.considers.sizes does not have a
On 3/21/2025 3:50 AM, WM wrote:
On 20.03.2025 23:25, Jim Burns wrote:
You confuse the clear fact that in the reality of sets vanishing meansWhat you (WM) think is reducing isn't reducing.For sets not.having a WM.size, Bob vanishing isn't a size.change.Only if reducing isn't reducing.
reducing with the foolish claim that cardinality was a meaningful
notion.
size.which.WM.considers.a.size.
not suffer losses.
Cantor recognized that proper subsets have less substance than theirLearn that even Cantor has accepted that the positive numbers haveWithout context, I can't be sure, but I suspect that Cantor's "more
more reality than the even positive numbers.
reality" and Zermelo's "simplest" serve the same purpose as my
"emptier" and "fuller", which is to rank infinite sets by something
_other than_ by size.
sets. That is all and that is simple. Every child could understand it
unless it had been stultified by matheologians with the result that all countable sets have the same cardinality (which is correct) and that
this cardinality is a proof of same number of elements (which is wrong - wrong - wrong!).
That's bullshit. Bijections are "complete".He said that is not in conflict with the identical cardinality of both
sets.
Thank you.
Two sets of the same cardinality,
one a proper subset of the other,
can be swapped set.wise, one for the other,
and preserve size.
Of course. The reason is that all pairs of the bijection proving same cardinality have infinitely many dark successors which cannot be
bijected.
Invisible = gone.Either swapping all at once,No. He can disappear from the visible part but not from the matrix.
or swapping in infinitely.many singleton.swaps, size is preserved,
but reality;simplicity;fullness isn't preserved,
and Bob can disappear without leaving.
On 21.03.2025 21:41, FromTheRafters wrote:Not surprising.
WM pretended :
On 21.03.2025 18:39, Jim Burns wrote:
On 3/21/2025 3:50 AM, WM wrote:
On 20.03.2025 23:25, Jim Burns wrote:What you (WM) think is reducing isn't reducing.
For sets not.having a WM.size, Bob vanishing isn't a size.change.Only if reducing isn't reducing.
You confuse the clear fact that in the reality of sets vanishing means
reducing with the foolish claim that cardinality was a meaningful
notion.
Learn that even Cantor has accepted that the positive numbers have
more reality than the even positive numbers.
This is called a subset.He said that is not in
conflict with the identical cardinality of both sets. And he was
right! "Counable" is simply another name for potential infinity.
Therefore vanishing odd numbers means reducing the reality of the set.
No, sets are immutable.Therefore the sentence "What you (WM) think is reducing isn't
reducing" exhibits you as a snooty dilettante who cannot distinguish
between cardinality and reality.
What does reality have to do with it?
Reality describes that fact that proper subsets have fewer elements than sets. Cardinality describes the fact that potential infinity never can
be finished.
On 22.03.2025 06:11, Jim Burns wrote:
On 3/21/2025 2:15 PM, WM wrote:
On 21.03.2025 18:39, Jim Burns wrote:
On 3/21/2025 3:50 AM, WM wrote:
On 20.03.2025 23:25, Jim Burns wrote:
For sets not.having a WM.size,
Bob vanishing isn't a size.change.
Only if reducing isn't reducing.
What you (WM) think is reducing
isn't reducing.
You confuse
the clear fact that
in the reality of
sets vanishing means reducing
with
the foolish claim that
cardinality was a meaningful notion.
The set of all sizes.which.WM.considers.sizes
does not have a size.which.WM.considers.a.size.
It is infinite
but nevertheless obeys the logic of
lossless exchanges do not suffer losses.
It is infinite
but nevertheless obeys the logic of
lossless exchanges do not suffer losses.
Two sets of the same cardinality,
one a proper subset of the other,
can be swapped set.wise, one for the other,
and preserve size.
Of course.
The reason is that
all pairs of
the bijection proving same cardinality
have
infinitely many dark successors
which cannot be bijected.
On 3/22/2025 5:19 AM, WM wrote:
It is infinite
but nevertheless obeys the logic of
lossless exchanges do not suffer losses.
Infinite sets don't need to obey the logic of
lossless exchanges do not suffer losses.
Of course.
The reason is that
all pairs of
the bijection proving same cardinality
have
infinitely many dark successors
which cannot be bijected.
A bijection which does not biject everything
is not a bijection,
On 22.03.2025 20:33, Jim Burns wrote:
On 3/22/2025 5:19 AM, WM wrote:
Of course.
The reason is that
all pairs of
the bijection proving same cardinality
have
infinitely many dark successors
which cannot be bijected.
A bijection which does not biject everything
is not a bijection,
So it is!
But it is impossible to prove everything
in Cantor's bijections.
Almost all elements follow upon every defined pair.
You (WM) are mistaken about
what 'finite' means and, therefore, about
what 'infinite' means.
Swapping same.sized sets,
even same.sized proper.subsets and supersets,
is not a loss.
On 3/23/2025 1:23 PM, WM wrote:
Almost all elements follow upon every defined pair.
The reason that we know about these pairs
is not
that we come to their end, not in any sense.
The reason that we know about these pairs
is
that we have an indefinite description,
true of each one of infinitely.many pairs,
The reason that we know about these pairs
is that,
if a finite sequence has a false claim,
then it has a first.false claim,
Our ability is the product of long centuries
of intense effort by our best people.
If someone wakes up one morning and announces
"I'm going to repeat that work of centuries today"
and they _fail_ at their project,
they should not conclude, from that,
that all that previous work was in error.
Am Sat, 22 Mar 2025 10:19:13 +0100 schrieb WM:
That a superset contains elements the subset doesn't is trivial.
Whatever your intuition about "number of elements", it isn't cardinality.
Of course. The reason is that all pairs of the bijection proving same
cardinality have infinitely many dark successors which cannot be
bijected.
That's bullshit. Bijections are "complete".
On 23.03.2025 19:23, Jim Burns wrote:Don't let us guess where you think the error lies.
On 3/23/2025 1:23 PM, WM wrote:
Almost all elements follow upon every defined pair.
The reason that we know about these pairs is not that we come to theirSo you believe. Erroneously.
end, not in any sense.
The reason that we know about these pairs is that we have an indefinite
description, true of each one of infinitely.many pairs,
Then it can be disregarded.The reason that we know about these pairs is that,But in case of darkness this cannot be seen.
if a finite sequence has a false claim,
then it has a first.false claim,
You can't contradict most mathematics and still accept it.Our ability is the product of long centuries of intense effort by our
best people.
If someone wakes up one morning and announces "I'm going to repeat that
work of centuries today"
and they _fail_ at their project,
they should not conclude, from that,
that all that previous work was in error.
Most of it concerns potential infinity and is correct.
On 22.03.2025 12:14, joes wrote:(Cardinality is maths.)
Am Sat, 22 Mar 2025 10:19:13 +0100 schrieb WM:
That a superset contains elements the subset doesn't is trivial.It is not intuition but maths, and it isn't cardinality.
Whatever your intuition about "number of elements", it isn't
cardinality.
Right, I forgot you don't believe in bijections. I don't understandThey should be complete. But complete bijecions are easily prove asOf course. The reason is that all pairs of the bijection proving sameThat's bullshit. Bijections are "complete".
cardinality have infinitely many dark successors which cannot be
bijected.
such: They are injective for every surjection. Cantor's "bijections"
fail to stad this test.
On 23.03.2025 19:23, Jim Burns wrote:This is very relevant in a discussion about infinity. And the "exchange"
You (WM) are mistaken about what 'finite' means and, therefore, aboutSince it is completely irrelevant, we need not discuss this. Relevant is
what 'infinite' means.
only that mathematics breaks down and dissolves into matheology, when exchanges defined to be lossless suffer losses.
Yes, but the limit (after inf. many) has no finite index, and is thusSwapping same.sized sets,Swapping at finite indices means that finite logic rules, even in case
even same.sized proper.subsets and supersets,
is not a loss.
of a sequence of swaps with no last swap.
On 23.03.2025 19:23, Jim Burns wrote:
On 3/23/2025 1:23 PM, WM wrote:
Almost all elements follow upon every defined pair.
Yes,
that is a property of sets which
are their.own.only.inductive.subsets.
The reason that we know about these pairs
is not
that we come to their end, not in any sense.
The reason that we know about these pairs
is
that we have an indefinite description,
true of each one of infinitely.many pairs,
So you believe. Erroneously.
The reason that we know about these pairs
is that,
if a finite sequence has a false claim,
then it has a first.false claim,
But in case of darkness this cannot be seen.
Our ability is the product of long centuries
of intense effort by our best people.
If someone wakes up one morning and announces
"I'm going to repeat that work of centuries today"
and they _fail_ at their project,
they should not conclude, from that,
that all that previous work was in error.
Most of it concerns potential infinity
and is correct.
On 22.03.2025 20:33, Jim Burns wrote:Exactly. It does not concern the result, the limit.
On 3/22/2025 5:19 AM, WM wrote:
Every set does because the claim concerns always only one finite term:It is infinite but nevertheless obeys the logic of lossless exchangesInfinite sets don't need to obey the logic of lossless exchanges do not
do not suffer losses.
suffer losses.
One single exchange.
The loss, if happened, had to happe at a finite index.Uh no, then we wouldn't have an infinite sequence. It is your perennial
...which are bijected just the same. Why should that change?So it is! But it is impossible to prove everything in Cantor'sThe reason is that all pairs of the bijection proving same cardinalityA bijection which does not biject everything is not a bijection,
have infinitely many dark successors which cannot be bijected.
bijections. Almost all elements follow upon every defined pair.
Am Sun, 23 Mar 2025 20:36:12 +0100 schrieb WM:
Since it is completely irrelevant, we need not discuss this. Relevant is
only that mathematics breaks down and dissolves into matheology, when
exchanges defined to be lossless suffer losses.
This is very relevant in a discussion about infinity. And the "exchange" which turns the initial state into the one after inf. many other
exchanges has no reason to be "lossless".
Yes, but the limit (after inf. many) has no finite index,Swapping same.sized sets,Swapping at finite indices means that finite logic rules, even in case
even same.sized proper.subsets and supersets,
is not a loss.
of a sequence of swaps with no last swap.
not subject to finite logic.
Perhaps you (WM) haven't noticed that
that is a finite sequence of claims,
each claim of which can, in principle, be printed.
On 24.03.2025 01:46, Jim Burns wrote:
Perhaps you (WM) haven't noticed that
that is a finite sequence of claims,
each claim of which can, in principle, be printed.
That is a wrong claim!
*In fact* at most 10^80 claims can be printed.
On 24.03.2025 01:48, joes wrote:There is the state of all X's, which can only be reached by an
Am Sun, 23 Mar 2025 20:36:12 +0100 schrieb WM:
There is no state after infinitely many. Countable sets have theSince it is completely irrelevant, we need not discuss this. RelevantThis is very relevant in a discussion about infinity. And the
is only that mathematics breaks down and dissolves into matheology,
when exchanges defined to be lossless suffer losses.
"exchange"
which turns the initial state into the one after inf. many other
exchanges has no reason to be "lossless".
property that every element can be indexed by a finite index.
Sure, but the limit is not an element. The sequence converges.There is no limit. The claim states that every element gets a finiteYes, but the limit (after inf. many) has no finite index, and is thusSwapping same.sized sets,Swapping at finite indices means that finite logic rules, even in case
even same.sized proper.subsets and supersets,
is not a loss.
of a sequence of swaps with no last swap.
not subject to finite logic.
index.
On 24.03.2025 01:46, Jim Burns wrote:Didn't you just agree? It wouldn't matter which 10^80 you choose.
Perhaps you (WM) haven't noticed that that is a finite sequence ofThat is a wrong claim! *In fact* at most 10^80 claims can be printed.
claims, each claim of which can, in principle, be printed.
"In principle" certainly more claims can be printed, but who has proved
that each claim can in principle be printed?
On 3/24/2025 3:34 PM, WM wrote:
On 24.03.2025 01:46, Jim Burns wrote:
Perhaps you (WM) haven't noticed that
that is a finite sequence of claims,
each claim of which can, in principle, be printed.
That is a wrong claim!
*In fact* at most 10^80 claims can be printed.
The claims I'm referring to, the claims upon which
my claims to knowledge of the infinite stand,
are many orders of magnitude from 10⁸⁰.many.
How many claims do you (WM) see here?
[ k is in the emptiest inductive set. ]
I see one claim.
On 25.03.2025 00:03, Jim Burns wrote:
On 3/24/2025 3:34 PM, WM wrote:
On 24.03.2025 01:46, Jim Burns wrote:
On 3/23/2025 3:41 PM, WM wrote:
On 23.03.2025 19:23, Jim Burns wrote:
The reason that we know about these pairs
is that,
if a finite sequence has a false claim,
then it has a first.false claim,
But in case of darkness this cannot be seen.
Perhaps you (WM) haven't noticed that
that is a finite sequence of claims,
each claim of which can, in principle, be printed.
Perhaps you (WM) haven't noticed that
that is a finite sequence of claims,
each claim of which can, in principle, be printed.
That is a wrong claim!
*In fact* at most 10^80 claims can be printed.
The claims I'm referring to, the claims upon which
my claims to knowledge of the infinite stand,
are many orders of magnitude from 10⁸⁰.many.
How many claims do you (WM) see here?
[ k is in the emptiest inductive set. ]
I see one claim.
If n ∈ ℕ_def then n+1 ∈ ℕ_def
is one claim.
In fact
it claims infinitely many steps.
Not all steps are true.
It's the same way in which we know that
a square has four corners
-- NOT by counting the corners of squares.
On 25.03.2025 18:46, Jim Burns wrote:
It's the same way in which we know that
a square has four corners
and that lossless exchanges are lossless.
-- NOT by counting the corners of squares.
Simply by definition.
On 3/25/2025 4:17 PM, WM wrote:
On 25.03.2025 18:46, Jim Burns wrote:
It's the same way in which we know that
a square has four corners
and that lossless exchanges are lossless.
and that a set larger than
On 25.03.2025 21:26, Jim Burns wrote:
On 3/25/2025 4:17 PM, WM wrote:
On 25.03.2025 18:46, Jim Burns wrote:
It's the same way in which we know that
a square has four corners
and that losslvvess exchanges are lossless.
and that a set larger than
any set different.sized.by.one from other sets
is not
any set different.sized.by.one from other sets.
Irrelevant.
Your waffle does not change basic logic.
On 3/25/2025 6:12 PM, WM wrote:
On 25.03.2025 21:26, Jim Burns wrote:
On 3/25/2025 4:17 PM, WM wrote:
On 25.03.2025 18:46, Jim Burns wrote:
It's the same way in which we know that
a square has four corners
and that losslvvess exchanges are lossless.
and that a set larger than
any set different.sized.by.one from other sets
is not
any set different.sized.by.one from other sets.
Irrelevant.
Your waffle does not change basic logic.
A set larger than
any set for which my waffle does not change basic WM.logic
is
a set for which my waffle changes basic WM.logic.
On 26.03.2025 07:24, Jim Burns wrote:At finite steps, but not in the limit.
On 3/25/2025 6:12 PM, WM wrote:Nothing of that kind does exist. "WM-logic" says that lossless exchanges
On 25.03.2025 21:26, Jim Burns wrote:A set larger than any set for which my waffle does not change basic
On 3/25/2025 4:17 PM, WM wrote:
On 25.03.2025 18:46, Jim Burns wrote:
Irrelevant. Your waffle does not change basic logic.and that a set larger than any set different.sized.by.one from otherIt's the same way in which we know that a square has four cornersand that losslvvess exchanges are lossless.
sets is not any set different.sized.by.one from other sets.
WM.logic is a set for which my waffle changes basic WM.logic.
at finite steps are lossless and is the foundation of rational thinking.
Am Wed, 26 Mar 2025 19:37:40 +0100 schrieb WM:
On 26.03.2025 07:24, Jim Burns wrote:At finite steps, but not in the limit.
On 3/25/2025 6:12 PM, WM wrote:Nothing of that kind does exist. "WM-logic" says that lossless exchanges
On 25.03.2025 21:26, Jim Burns wrote:A set larger than any set for which my waffle does not change basic
On 3/25/2025 4:17 PM, WM wrote:
On 25.03.2025 18:46, Jim Burns wrote:
Irrelevant. Your waffle does not change basic logic.and that a set larger than any set different.sized.by.one from other >>>>> sets is not any set different.sized.by.one from other sets.It's the same way in which we know that a square has four corners >>>>>> and that losslvvess exchanges are lossless.
WM.logic is a set for which my waffle changes basic WM.logic.
at finite steps are lossless and is the foundation of rational thinking.
On 26.03.2025 07:24, Jim Burns wrote:
On 3/25/2025 6:12 PM, WM wrote:
On 25.03.2025 21:26, Jim Burns wrote:
On 3/25/2025 4:17 PM, WM wrote:
On 25.03.2025 18:46, Jim Burns wrote:
It's the same way in which we know that
a square has four corners
and that lossless exchanges are lossless.
and that a set larger than
any set different.sized.by.one from other sets
is not
any set different.sized.by.one from other sets.
Irrelevant.
Your waffle does not change basic logic.
A set larger than
any set for which my waffle does not change basic WM.logic
is
a set for which my waffle changes basic WM.logic.
Nothing of that kind does exist.
Nothing of that kind does exist.
"WM-logic" says that
lossless exchanges at finite steps
are lossless
and is the foundation of rational thinking.
On 26.03.2025 07:24, Jim Burns wrote:
On 3/25/2025 6:12 PM, WM wrote:
On 25.03.2025 21:26, Jim Burns wrote:
On 3/25/2025 4:17 PM, WM wrote:
On 25.03.2025 18:46, Jim Burns wrote:
It's the same way in which we know that
a square has four corners
and that losslvvess exchanges are lossless.
and that a set larger than
any set different.sized.by.one from other sets
is not
any set different.sized.by.one from other sets.
Irrelevant.
Your waffle does not change basic logic.
A set larger than
any set for which my waffle does not change basic WM.logic
is
a set for which my waffle changes basic WM.logic.
Nothing of that kind does exist.
"WM-logic" says that lossless exchanges at finite steps are lossless
....
.... and is the foundation of rational thinking.
Regards, WM
On 26.03.2025 20:59, joes wrote:I have no idea what you are talking about. What can be proved to be
Am Wed, 26 Mar 2025 19:37:40 +0100 schrieb WM:Countability is proved without limits. Bijections are one-to-one without limit.
On 26.03.2025 07:24, Jim Burns wrote:At finite steps, but not in the limit.
On 3/25/2025 6:12 PM, WM wrote:Nothing of that kind does exist. "WM-logic" says that lossless
On 25.03.2025 21:26, Jim Burns wrote:A set larger than any set for which my waffle does not change basic
On 3/25/2025 4:17 PM, WM wrote:
On 25.03.2025 18:46, Jim Burns wrote:
Irrelevant. Your waffle does not change basic logic.and that a set larger than any set different.sized.by.one fromIt's the same way in which we know that a square has four corners >>>>>>> and that losslvvess exchanges are lossless.
other sets is not any set different.sized.by.one from other sets.
WM.logic is a set for which my waffle changes basic WM.logic.
exchanges at finite steps are lossless and is the foundation of
rational thinking.
Am Wed, 26 Mar 2025 21:09:03 +0100 schrieb WM:
I have no idea what you are talking about. What can be proved to beAt finite steps, but not in the limit.Countability is proved without limits. Bijections are one-to-one without
limit.
countable without resorting to limits?
I didn't say anything about
bijections not being one-to-one, nor about them having limits???
The transformation from the initial to the final matrix can not be
achieved in finitely many steps.
WM <wolfgang.mueckenheim@tha.de> wrote:
On 26.03.2025 07:24, Jim Burns wrote:
"WM-logic" says that lossless exchanges at finite steps are lossless
....
Nobody has contradicted this, that I'm aware of. It's the accumulation
of _all_ of these lossless exchanges where unexpected things happen.
.... and is the foundation of rational thinking.
I think rational thinking has more general foundations than
pontifications about lossless exchanges.
On 3/26/2025 2:37 PM, WM wrote:
"WM-logic" says that
lossless exchanges at finite steps
are lossless
For a set larger than
any set such that
emptier.by.one and smaller.by.one are the same,
emptier.by.one and smaller.by.one aren't the same.
Am 26.03.2025 um 22:34 schrieb Alan Mackenzie:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 26.03.2025 07:24, Jim Burns wrote:
"WM-logic" says that lossless exchanges at finite steps are lossless
....
Nobody has contradicted this, that I'm aware of. It's the accumulation
of _all_ of these lossless exchanges where unexpected things happen.
If they happen, then there is a first instance where they happen. Every
n.e. subset of a countable set has a first element.
.... and is the foundation of rational thinking.
I think rational thinking has more general foundations than
pontifications about lossless exchanges.
But that is one necessary foundation. When it breaks down, then thinking breaks down.
Regards, WM
Am 27.03.2025 um 15:58 schrieb joes:You are writing way too short. There are countable infinities,
Am Wed, 26 Mar 2025 21:09:03 +0100 schrieb WM:
Limits deviate from countability.I have no idea what you are talking about. What can be proved to beAt finite steps, but not in the limit.Countability is proved without limits. Bijections are one-to-one
without limit.
countable without resorting to limits?
There's no contradiction. The limit is not a term of the sequence.I didn't say anything about bijections not being one-to-one, nor aboutThat is true but contradicts Cantor's theory according to which every
them having limits??? The transformation from the initial to the final
matrix can not be achieved in finitely many steps.
pair has a finite index. Every finite index n has only n-1 predecessors.
You [Mückenheim] base your mathematical thinking on faulty intuition. You do not
base it on the axioms and logic which have chrystallised out of a lot of
very clever thinking over the last few centuries.
Am 26.03.2025 um 22:09 schrieb Jim Burns:
On 3/26/2025 2:37 PM, WM wrote:
"WM-logic" says that
lossless exchanges at finite steps
are lossless
For a set larger than
any set such that
emptier.by.one and smaller.by.one are the same,
emptier.by.one and smaller.by.one aren't the same.
Countable sets have only counted elements.
Lossless exchanges
like every other event
can happen only at a finitely indexed step.
But there
it is excluded by logic.
Lossless exchanges
like every other event
can happen only at a finitely indexed step.
But there
it is excluded by logic.
Am 27.03.2025 um 22:34 schrieb Alan Mackenzie:
You [Mückenheim] base your mathematical thinking on
faulty intuition.
You do not base it on the axioms and logic which
have chrystallised out of a lot of very clever thinking
over the last few centuries.
"It is not for nothing, after all,
that set theorists resort to the axiomatic method.
Intuition here /is/ bankrupt."
(W.V.O. Quine, Set Theory and its Logic)
When it breaks down, then thinking breaks down.
WM <invalid@no.org> wrote:
Am 26.03.2025 um 22:34 schrieb Alan Mackenzie:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 26.03.2025 07:24, Jim Burns wrote:
"WM-logic" says that lossless exchanges at finite steps are lossless
....
Nobody has contradicted this, that I'm aware of. It's the accumulation
of _all_ of these lossless exchanges where unexpected things happen.
If they happen, then there is a first instance where they happen. Every
n.e. subset of a countable set has a first element.
The set of integer steps at which a loss occurs is empty.
It thus has
no least member.
It is only in the infinite limit where the loss
occurs.
In the limit, it passes
_all_ places.
and thus is no longer at one of the numbered places.
You base your mathematical thinking on faulty intuition. You do not
base it on the axioms and logic which have chrystallised out of a lot of
very clever thinking over the last few centuries.
Am Thu, 27 Mar 2025 20:50:00 +0100 schrieb WM:
Am 27.03.2025 um 15:58 schrieb joes:You are writing way too short. There are countable infinities,
Am Wed, 26 Mar 2025 21:09:03 +0100 schrieb WM:Limits deviate from countability.
I have no idea what you are talking about. What can be proved to beAt finite steps, but not in the limit.Countability is proved without limits. Bijections are one-to-one
without limit.
countable without resorting to limits?
whether you like that word or not.
There's no contradiction. The limit is not a term of the sequence.I didn't say anything about bijections not being one-to-one, nor aboutThat is true but contradicts Cantor's theory according to which every
them having limits??? The transformation from the initial to the final
matrix can not be achieved in finitely many steps.
pair has a finite index. Every finite index n has only n-1 predecessors.
Aren't you the one?
On 27.03.2025 22:34, Alan Mackenzie wrote:
WM <invalid@no.org> wrote:
Am 26.03.2025 um 22:34 schrieb Alan Mackenzie:
WM <wolfgang.mueckenheim@tha.de> wrote:
On 26.03.2025 07:24, Jim Burns wrote:
"WM-logic" says that lossless exchanges at finite steps are lossless >>>>> ....
Nobody has contradicted this, that I'm aware of. It's the accumulation >>>> of _all_ of these lossless exchanges where unexpected things happen.
If they happen, then there is a first instance where they happen. Every
n.e. subset of a countable set has a first element.
The set of integer steps at which a loss occurs is empty.
There are no other steps at which anything could occur.
It thus has no least member.
Nevertheless all members are finite integers, and afterwards nothing
happens anymore.
It is only in the infinite limit where the loss occurs.
Bijections have no limit.
"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)]
Limits are not determined places.
"such that every element of the set stands at a definite position of
this sequence" [E. Zermelo: "Georg Cantor – Gesammelte Abhandlungen mathematischen und philosophischen Inhalts", Springer, Berlin (1932) p.
152]
In the limit, it passes _all_ places.
Do you think that Cantor's above explanations are wrong?
In informal language, it "disappears off to infinity",
There is no chance to disappear. And never infinity is reached.
and thus is no longer at one of the numbered places.
"such that every element of the set stands at a definite position of
this sequence" [E. Zermelo: "Georg Cantor – Gesammelte Abhandlungen mathematischen und philosophischen Inhalts", Springer, Berlin (1932) p. 152]
You base your mathematical thinking on faulty intuition. You do not
base it on the axioms and logic which have chrystallised out of a lot of
very clever thinking over the last few centuries.
Do you think that Cantor was wrong?
Regards, WM
Do you think that Cantor was wrong?
That has no connection or relevance to my point, which you have evaded addressing. I don't think Cantor was wrong, in general. But I do think
you base your mathematical thinking on faulty intuition, and not on the axioms and logic which have chrystallised out of a lot of very clever thinking over the last few centuries.
Am 03.04.2025 um 22:05 schrieb Alan Mackenzie:
Do you think that Cantor was wrong?
That has no connection or relevance to my point, which you have evaded
addressing. I don't think Cantor was wrong, in general. But I do think >> you base your mathematical thinking on faulty intuition, and not on the
axioms and logic which have chrystallised out of a lot of very clever
thinking over the last few centuries.
[Bolzano, Cantor, Dedekind, Frege, Peano,] Russell, Zermelo, Fraenkel, Skolem, von Neumann, Bernays, Gödel, ... etc.
Am 03.04.2025 um 22:05 schrieb Alan Mackenzie:
Do you think that Cantor was wrong?
That has no connection or relevance to my point, which you have evaded
addressing. I don't think Cantor was wrong, in general. But I do think >> you base your mathematical thinking on faulty intuition, and not on the
axioms and logic which have chrystallised out of a lot of very clever
thinking over the last few centuries.
[Bolzano, Cantor, Dedekind, Frege, Peano,] Russell and Whitehead,
Zermelo, Fraenkel, Skolem, von Neumann, Bernays, Gödel, ... etc.
Do you think that Cantor was wrong?
That has no connection or relevance to my point, which you have evaded addressing. I don't think Cantor was wrong, in general. But I do think
you base your mathematical thinking on faulty intuition, and not on the axioms and logic which have chrystallised out of a lot of very clever thinking over the last few centuries.
"It is not for nothing, after all, that set theorists resort to the
axiomatic method. Intuition here /is/ bankrupt." (W.V.O. Quine, Set
Theory and its Logic)
WM <wolfgang.mueckenheim@tha.de> wrote:table set has a first element.
The set of integer steps at which a loss occurs is empty.
There are no other steps at which anything could occur.
That's your lack of understanding of things infinite.
It thus has no least member.
Nevertheless all members are finite integers, and afterwards nothing
happens anymore.
Eh? Members of what? After what?
It is only in the infinite limit where the loss occurs.
Bijections have no limit.
That has no connection with what I wrote.
Sequences and series may have
limits, not bijections.
What we're talking about is a sequence of
positions the distinguished element is at.
This is a sequence of natural
numbers. At step n, the element is at position n. After an "infinite
number of steps", the distinguished element is not at a naturally
numbered position - it has "disappeared".
"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)]
Irrelevant.
Limits are not determined places.
Meaningless.
"such that every element of the set stands at a definite position of
this sequence" [E. Zermelo: "Georg Cantor – Gesammelte Abhandlungen
mathematischen und philosophischen Inhalts", Springer, Berlin (1932) p.
152]
Irrelevant.
In the limit, it passes _all_ places.
Do you think that Cantor's above explanations are wrong?
I think Cantor would have and did understand the current situation. What
you have quoted from Cantor are not explanations of what we are
discussing.
In informal language, it "disappears off to infinity",
There is no chance to disappear. And never infinity is reached.
Tell us all, then, at which element it ends up at.
"such that every element of the set stands at a definite position of
this sequence" [E. Zermelo: "Georg Cantor – Gesammelte Abhandlungen
mathematischen und philosophischen Inhalts", Springer, Berlin (1932) p. 152]
Irrelevant.
That has no connection or relevance to my point, which you have evaded addressing.
Am 03.04.2025 um 22:05 schrieb Alan Mackenzie: [...]
There's a film called "Hard to Be a God".
I'd say: [It's] hard to "discuss" with a crank.
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