On 21.10.2024 14:41, joes wrote:"yes, but actually no" The numbers ARE fixed, there are just inf. many
Am Mon, 21 Oct 2024 11:59:00 +0200 schrieb WM:Yes, but all the numbers which the set contains are either complete and fixed, or they are variable such that with each request larger numbers
On 21.10.2024 10:21, joes wrote:An infinite set can contain an m>n for every n in it.
Am Mon, 21 Oct 2024 10:14:07 +0200 schrieb WM:No. No set of numbers can include larger numbers.
If the range was complete, the image shows that the range was notIf the range really is complete, it needs to be infinite, so it can
complete.
include the larger numbers.
can be creazed.
The latter case is called potential infinity. In case they are complete--
and fixed we can multiply them by 2 and find larger numbers.
Am Mon, 21 Oct 2024 20:35:59 +0200 schrieb WM:
Yes, but all the numbers which the set contains are either complete and"yes, but actually no" The numbers ARE fixed, there are just inf. many
fixed, or they are variable such that with each request larger numbers
can be created.
of them.
The larger numbers have already all been "created".
On 22.10.2024 11:43, joes wrote:So what?
Am Mon, 21 Oct 2024 20:35:59 +0200 schrieb WM:
All of them are multiplied by 2.Yes, but all the numbers which the set contains are either complete"yes, but actually no" The numbers ARE fixed, there are just inf. many
and fixed, or they are variable such that with each request larger
numbers can be created.
of them.
No. We are taking the ACTUALLY infinite set, which most certainlyThe larger numbers have already all been "created".Not before multiplying them.
Am Tue, 22 Oct 2024 16:11:35 +0200 schrieb WM:
So what?The numbers ARE fixed, there are just inf. manyAll of them are multiplied by 2.
of them.
No. We are taking the ACTUALLY infinite set, which most certainlyThe larger numbers have already all been "created".Not before multiplying them.
also includes the even numbers.
On 22.10.2024 17:03, joes wrote:SO WHAT
Am Tue, 22 Oct 2024 16:11:35 +0200 schrieb WM:Important: _All_ numbers existing in the _invariable_ set.
The numbers ARE fixed, there are just inf. many of them.All of them are multiplied by 2.
Only larger than the one doubled.Nevertheless, doubling creates larger numbers with no doubt.No. We are taking the ACTUALLY infinite set, which most certainly alsoThe larger numbers have already all been "created".Not before multiplying them.
includes the even numbers.
If no larger natural numbers are available, what could be the result?WDYM "available"?
Am Tue, 22 Oct 2024 16:11:35 +0200 schrieb WM:
On 22.10.2024 11:43, joes wrote:
The larger numbers have already all been "created".
Not before multiplying them.
No. We are taking the ACTUALLY infinite set, which most certainly
also includes the even numbers.
On 22.10.2024 11:43, joes wrote:
Am Mon, 21 Oct 2024 20:35:59 +0200 schrieb WM:
Yes, but
all the numbers which the set contains
are either complete and fixed, or
they are variable such that
with each request
larger numbers can be created.
"yes, but actually no"
The numbers ARE fixed,
there are just inf. many of them.
All of them are multiplied by 2.
The larger numbers have already all been "created".
Not before multiplying them.
Am Tue, 22 Oct 2024 18:07:26 +0200 schrieb WM:
If no larger natural numbers are available, what could be the result [of doubling a natural number]
WDYM "available"?
Am Tue, 22 Oct 2024 18:07:26 +0200 schrieb WM:
Nevertheless, doubling creates larger numbers with no doubt.Only larger than the one doubled.
On 10/22/2024 10:11 AM, WM wrote:
The numbers ARE fixed,
there are just inf. many of them.
All of them are multiplied by 2.
The larger numbers have already all been "created".
Not before multiplying them.
"Creation" of these numbers
is confusingly imagined to be
creation of these numbers.
These numbers are described by axioms.
Claims (axioms) are made which are true of
each of these numbers,
No number starts or stops
being described by those axioms.
In that sense,
no number starts or stops
existing.
No number enters or exits
the domain of discourse those axioms describe.
In this discussion,
the numbers being.described.by.those.axioms
== the numbers existing
include, for each such number, double that number.
The numbers and their doubles
always were/will.be being.described.by.those.axioms
always were/will.be existing.
We cannot perform supertasks like
counting infinitely.many.
Am 22.10.2024 um 17:03 schrieb joes:
Am Tue, 22 Oct 2024 16:11:35 +0200 schrieb WM:
It's just a "simple fact" (in the context of set theory), that for all n
e IN: 2*n e IN. ("simple fact" => a provable proposition).
Not before multiplying them.
No. We are taking the ACTUALLY infinite set, which most certainly
also includes the even numbers.
Especially, since we consider the set of ALL natural numbers; and the
natural numbers consist of ALL odd and ALL even natural numbers. :-)
Am Tue, 22 Oct 2024 16:11:35 +0200 schrieb WM:
On 22.10.2024 11:43, joes wrote:
The larger numbers have already all been "created".
Not before multiplying them.
No. We are taking the ACTUALLY infinite set, which most certainly
also includes the even numbers.
On 22.10.2024 19:36, Moebius wrote:There are no such numbers!
Am 22.10.2024 um 17:03 schrieb joes:
Am Tue, 22 Oct 2024 16:11:35 +0200 schrieb WM:
It's just a "simple fact" (in the context of set theory), that for all
n e IN: 2*n e IN. ("simple fact" => a provable proposition).
All its numbers are multiplied.Not before multiplying them.No. We are taking the ACTUALLY infinite set, which most certainly also
includes the even numbers.
Especially, since we consider the set of ALL natural numbers; and theIf so, then the numbers not existing before doubling must be
natural numbers consist of ALL odd and ALL even natural numbers. :-)
non-natural.
Am Tue, 22 Oct 2024 16:11:35 +0200 schrieb WM:
On 22.10.2024 11:43, joes wrote:
The larger numbers have already all been "created".
Not before multiplying them.
No. We are taking the ACTUALLY infinite set, which most certainly
also includes the even numbers.
Of course, this just "shows" that IN is ("actually") infinite.
Especially, since we consider the set of ALL natural numbers; and the
natural numbers consist of ALL odd and ALL even natural numbers.
On 22.10.2024 19:24, Jim Burns wrote:They should, otherwise the set was finite.
On 10/22/2024 10:11 AM, WM wrote:
These numbers did not belong to the set when it was multiplied."Creation" of these numbers is confusingly imagined to be creation ofThe numbers ARE fixed, there are just inf. many of them.All of them are multiplied by 2.
The larger numbers have already all been "created".Not before multiplying them.
these numbers.
That includes the so-called "doubles", i.e. the even numbers.These numbers are described by axioms.By axiom also all numbers of the set are present and available to be multiplied.
Claims (axioms) are made which are true of each of these numbers,which can be multiplied.
No number starts or stops being described by those axioms.The set is complete when its numbers are multiplied.
In that sense,All numbers which are described are in the set when being multiplied.
no number starts or stops existing.
No number enters or exits the domain of discourse those axioms
describe.
Those doubles of doubles are in the original set as well.the numbers being.described.by.those.axioms == the numbers existingOf course. But all these are multiplied.
include, for each such number, double that number.
As are the doubles. N is not limited.The numbers and their doubles always were/will.beThey are existing when multiplied.
being.described.by.those.axioms always were/will.be existing.
Then we get no new numbers.We cannot perform supertasks like counting infinitely.many.But we can use all existing numbers, for instance for mappings.
Am Tue, 22 Oct 2024 20:31:13 +0200 schrieb WM:
All its numbers are multiplied.
There are no such numbers!Especially, since we consider the set of ALL natural numbers; and theIf so, then the numbers not existing before doubling must be
natural numbers consist of ALL odd and ALL even natural numbers. :-)
non-natural.
On 10/22/2024 1:29 PM, WM wrote:
On 22.10.2024 20:38, joes wrote:
Am Tue, 22 Oct 2024 20:31:13 +0200 schrieb WM:
All its numbers are multiplied.There are no such numbers!
Especially, since we consider the set of ALL natural numbers; and the >>>>> natural numbers consist of ALL odd and ALL even natural numbers. :-)If so, then the numbers not existing before doubling must be
non-natural.
Call the result what you like.
2 * any_natural_number = a_natural_number
Got it! God damn. wow.
On 22.10.2024 20:38, joes wrote:I call them the even numbers.
Am Tue, 22 Oct 2024 20:31:13 +0200 schrieb WM:
Call the result what you like.There are no such numbers!Especially, since we consider the set of ALL natural numbers; and theIf so, then the numbers not existing before doubling must be
natural numbers consist of ALL odd and ALL even natural numbers. :-)
non-natural.
On 22.10.2024 11:43, joes wrote:
Am Mon, 21 Oct 2024 20:35:59 +0200 schrieb WM:
Yes, but all the numbers which the set contains are either complete and"yes, but actually no" The numbers ARE fixed, there are just inf. many
fixed, or they are variable such that with each request larger numbers
can be created.
of them.
All of them are multiplied by 2.
The larger numbers have already all been "created".
Not before multiplying them.
Regards, WM
Am Tue, 22 Oct 2024 22:29:52 +0200 schrieb WM:
On 22.10.2024 20:38, joes wrote:I call them the even numbers.
Am Tue, 22 Oct 2024 20:31:13 +0200 schrieb WM:Call the result what you like.
There are no such numbers!Especially, since we consider the set of ALL natural numbers; and the >>>>> natural numbers consist of ALL odd and ALL even natural numbers. :-)If so, then the numbers not existing before doubling must be
non-natural.
Your complete set of the Natural Numbers is not complete.
On 23.10.2024 13:37, Richard Damon wrote:Every single number is greater than its original, but for every number
Your complete set of the Natural Numbers is not complete.I take what is given: the complete set of natural numbers. I double it
an get, according to mathematics 2n > n greater numbers than were given
to me.
On 23.10.2024 13:37, Richard Damon wrote:
Your complete set of the Natural Numbers is not complete.
I take what is given: the complete set of natural numbers. I double it
an get, according to mathematics 2n > n greater numbers than were given
to me.
Regards, WM
Am Wed, 23 Oct 2024 20:03:48 +0200 schrieb WM:
On 23.10.2024 13:37, Richard Damon wrote:Every single number is greater than its original, but for every number
Your complete set of the Natural Numbers is not complete.I take what is given: the complete set of natural numbers. I double it
an get, according to mathematics 2n > n greater numbers than were given
to me.
in the mapped set there is a greater one in the former set.
On 23.10.2024 20:59, joes wrote:
Am Wed, 23 Oct 2024 20:03:48 +0200 schrieb WM:
On 23.10.2024 13:37, Richard Damon wrote:Every single number is greater than its original, but for every number
Your complete set of the Natural Numbers is not complete.I take what is given: the complete set of natural numbers. I double it
an get, according to mathematics 2n > n greater numbers than were given
to me.
in the mapped set there is a greater one in the former set.
Impossible since all have been mapped, even the greater ones.
Regards, WM
On 10/23/24 2:03 PM, WM wrote:
On 23.10.2024 13:37, Richard Damon wrote:
So, your problem is you don't know what you were givenYour complete set of the Natural Numbers is not complete.
I take what is given: the complete set of natural numbers. I double it
an get, according to mathematics 2n > n greater numbers than were
given to me.
On 10/24/24 6:46 AM, WM wrote:
On 23.10.2024 20:59, joes wrote:
Am Wed, 23 Oct 2024 20:03:48 +0200 schrieb WM:
On 23.10.2024 13:37, Richard Damon wrote:Every single number is greater than its original, but for every number
Your complete set of the Natural Numbers is not complete.I take what is given: the complete set of natural numbers. I double it >>>> an get, according to mathematics 2n > n greater numbers than were given >>>> to me.
in the mapped set there is a greater one in the former set.
Impossible since all have been mapped, even the greater ones.
So why can't you get the needed mapped value, since it is in the set.
Infinity isn't just some impossibly big number that twice itself is
bigger than itself, that is a property of the FINITE.
You are just unable to understand the infinite, because you just refuse
to let it be what it is.
On 23.10.2024 20:59, joes wrote:Yes, and those are mapped to even greater ones!
Am Wed, 23 Oct 2024 20:03:48 +0200 schrieb WM:Impossible since all have been mapped, even the greater ones.
On 23.10.2024 13:37, Richard Damon wrote:Every single number is greater than its original, but for every number
Your complete set of the Natural Numbers is not complete.I take what is given: the complete set of natural numbers. I double it
an get, according to mathematics 2n > n greater numbers than were
given to me.
in the mapped set there is a greater one in the former set.
Am Thu, 24 Oct 2024 12:46:05 +0200 schrieb WM:
On 23.10.2024 20:59, joes wrote:Yes, and those are mapped to even greater ones!
Am Wed, 23 Oct 2024 20:03:48 +0200 schrieb WM:Impossible since all have been mapped, even the greater ones.
On 23.10.2024 13:37, Richard Damon wrote:Every single number is greater than its original, but for every number
Your complete set of the Natural Numbers is not complete.I take what is given: the complete set of natural numbers. I double it >>>> an get, according to mathematics 2n > n greater numbers than were
given to me.
in the mapped set there is a greater one in the former set.
WM has brought this to us :
The infinite must adhere to correct mathematics. Otherwise it is only
matheology, to be believed by believers who despise mathematics.
Sets don't change. If
you want to double each element of the naturally ordered set of natural numbers you do it by creating another set with the naturally ordered
doubled elements.
{1,2,3,4,...}
{2,4,6,8,...}
These sets are the same |size|
your swapped Bob is
*always* in the 'next' room since there is no 'last' room at the
infinite hotel.
Scrooge McDuck gets richer and richer as he goes broke.
On 24.10.2024 14:24, FromTheRafters wrote:Please explain how you can count past omega.
WM has brought this to us :
In fact, here "the same size" applies. The change of size between bothThe infinite must adhere to correct mathematics. Otherwise it is onlySets don't change. If you want to double each element of the naturally
matheology, to be believed by believers who despise mathematics.
ordered set of natural numbers you do it by creating another set with
the naturally ordered doubled elements.
{1,2,3,4,...}
{2,4,6,8,...}
These sets are the same |size|
sets is +/- 0. But since the first set contains all natural numbers, the second set contains larger numbers because 2 > n.
No, "countable" means "bijective to (possibly a subset of) the naturals",your swapped Bob is *always* in the 'next' room since there is noBut if all fractions can be counted, then there is a last state, namely
'last' room at the infinite hotel.
when this is counting is complete.
On 24.10.2024 14:10, joes wrote:You have it backwards. Surely the "complete" set should not be missing
Am Thu, 24 Oct 2024 12:46:05 +0200 schrieb WM:The possibility of always even greater ones in natural numbers proves potential infinity.
On 23.10.2024 20:59, joes wrote:Yes, and those are mapped to even greater ones!
Am Wed, 23 Oct 2024 20:03:48 +0200 schrieb WM:Impossible since all have been mapped, even the greater ones.
On 23.10.2024 13:37, Richard Damon wrote:Every single number is greater than its original, but for every
Your complete set of the Natural Numbers is not complete.I take what is given: the complete set of natural numbers. I double
it an get, according to mathematics 2n > n greater numbers than were >>>>> given to me.
number in the mapped set there is a greater one in the former set.
The greater ones have not been doubled because doubling of a completeThey have also been doubled, along with their doubles. The powers of 2 and their multiples form a subset of the naturals. The "size" of this set
set creates a set covering a greater interval than covered before. (Half
the density implies twice the extension.)
On 22.10.2024 19:24, Jim Burns wrote:
On 10/22/2024 10:11 AM, WM wrote:
On 22.10.2024 11:43, joes wrote:
The larger numbers have already
all been "created".
Not before multiplying them.
"Creation" of these numbers
is confusingly imagined to be
creation of these numbers.
These numbers did not belong to the set
when it was multiplied.
These numbers are described by axioms.
By axiom also
all numbers of the set are present and
available to be multiplied.
The numbers and their doubles
always were/will.be being.described.by.those.axioms
always were/will.be existing.
They are existing when multiplied.
We cannot perform supertasks like
counting infinitely.many.
But we can use all existing numbers,
for instance for mappings.
Am Thu, 24 Oct 2024 15:49:48 +0200 schrieb WM:
The possibility of always even greater ones in natural numbers provesYou have it backwards. Surely the "complete" set should not be missing
potential infinity.
those greater numbers that the "potential" set includes.
The greater ones have not been doubled because doubling of a completeThey have also been doubled, along with their doubles. The powers of 2 and their multiples form a subset of the naturals. The "size" of this set
set creates a set covering a greater interval than covered before. (Half
the density implies twice the extension.)
is omega, and 2w=w, regardless of "reality".
The all.the.doubles _claim_
follows the
well.ordered.successored.non.0.predecessored
claim not.first.falsely,
so it can't be false.
Am Thu, 24 Oct 2024 15:42:55 +0200 schrieb WM:
On 24.10.2024 14:24, FromTheRafters wrote:Please explain how you can count past omega.
WM has brought this to us :In fact, here "the same size" applies. The change of size between both
The infinite must adhere to correct mathematics. Otherwise it is onlySets don't change. If you want to double each element of the naturally
matheology, to be believed by believers who despise mathematics.
ordered set of natural numbers you do it by creating another set with
the naturally ordered doubled elements.
{1,2,3,4,...}
{2,4,6,8,...}
These sets are the same |size|
sets is +/- 0. But since the first set contains all natural numbers, the
second set contains larger numbers because 2n > n.
No, "countable" means "bijective to (possibly a subset of) the naturals",your swapped Bob is *always* in the 'next' room since there is noBut if all fractions can be counted, then there is a last state, namely
'last' room at the infinite hotel.
when this is counting is complete.
not that there is a last one. That would imply finiteness, but N is
countably infinite.
Am Thu, 24 Oct 2024 15:42:55 +0200 schrieb WM:
But if all fractions can be counted, then there is a [final] state,
namely when this counting is complete.
[...] "countable" means "bijective to (possibly a subset of) the naturals", not that there is a last one. That would imply finiteness, but N is
countably infinite.
Am Thu, 24 Oct 2024 15:49:48 +0200 schrieb WM:
The [fact] of always even greater ones in [the set of] natural numbers proves
potential infinity.
You have it backwards. Surely the "complete" set should not be missing
those greater [natural] numbers [...]
Am 24.10.2024 um 15:59 schrieb joes:
Am Thu, 24 Oct 2024 15:49:48 +0200 schrieb WM:
The [fact] of always even greater ones in [the set of] natural
numbers proves
potential infinity.
Nonsense. It proves _actual_ infinity.
You have it backwards. Surely the "complete" set should not be missing
those greater [natural] numbers [...]
Indeed.
On 24.10.2024 15:54, joes wrote:If you don't even reach omega, you can forget about larger numbers.
Am Thu, 24 Oct 2024 15:42:55 +0200 schrieb WM:I don't. The required steps are dark. But if the set ℕ is complete, then
On 24.10.2024 14:24, FromTheRafters wrote:Please explain how you can count past omega.
WM has brought this to us :In fact, here "the same size" applies. The change of size between both
The infinite must adhere to correct mathematics. Otherwise it isSets don't change. If you want to double each element of the
only matheology, to be believed by believers who despise
mathematics.
naturally ordered set of natural numbers you do it by creating
another set with the naturally ordered doubled elements.
{1,2,3,4,...}
{2,4,6,8,...}
These sets are the same |size|
sets is +/- 0. But since the first set contains all natural numbers,
the second set contains larger numbers because 2n > n.
it covers a domain at the ordinal axis. If its density is reduced, then
its extension is increased.
No element of the sequence indexes all.We can set up a sequence of sets where in every set one one fractions is indexed.No, "countable" means "bijective to (possibly a subset of) theyour swapped Bob is *always* in the 'next' room since there is noBut if all fractions can be counted, then there is a last state,
'last' room at the infinite hotel.
namely when this is counting is complete.
naturals",
not that there is a last one. That would imply finiteness, but N is
countably infinite.
Only if all fractions have been counted, all fractions have beenAnd infinite.
counted. That means that the sequence is complete.
Hint: For any nonempty set of natural numbers M:
M is infinite iff for each and every element m in M there's an
element m' in M such that m' > m.
(Note that sets don't change.)
Am Thu, 24 Oct 2024 16:17:38 +0200 schrieb WM:
if the set ℕ is complete, then
it covers a domain at the ordinal axis. If its density is reduced, then
its extension is increased.
If you don't even reach omega, you can forget about larger numbers.
The naturals, evens, and powers of 2 all only go up to omega.
WM has brought this to us :
On 24.10.2024 14:24, FromTheRafters wrote:
WM has brought this to us :
The infinite must adhere to correct mathematics. Otherwise it is
only matheology, to be believed by believers who despise mathematics.
Sets don't change. If you want to double each element of the
naturally ordered set of natural numbers you do it by creating
another set with the naturally ordered doubled elements.
{1,2,3,4,...}
{2,4,6,8,...}
These sets are the same |size|
In fact, here "the same size" applies. The change of size between both
sets is +/- 0. But since the first set contains all natural numbers,
the second set contains larger numbers because 2n > n.
So what?
your swapped Bob is *always* in the 'next' room since there is no
'last' room at the infinite hotel.
But if all fractions can be counted, then there is a last state,
namely when this is counting is complete.
Countable does not mean counted.
Scrooge McDuck gets richer and richer as he goes broke.
He does never lose more money than he gets. Therefore he cannot lose
all his money. *That* is dictated by logic.
Neither can he get infinitely rich.
On 10/24/2024 6:42 AM, WM wrote:
On 24.10.2024 14:24, FromTheRafters wrote:
Scrooge McDuck gets richer and richer as he goes broke.
He does never lose more money than he gets.
Therefore he cannot lose all his money. *That* is
On 10/24/2024 12:03 PM, WM wrote:
On 24.10.2024 19:40, FromTheRafters wrote:
Countable does not mean [can be] counted.
It means countable and is a lie if it cannot become counted.
Am 24.10.2024 um 21:36 schrieb Chris M. Thomasson:
On 10/24/2024 6:42 AM, WM wrote:
On 24.10.2024 14:24, FromTheRafters wrote:
Scrooge McDuck gets richer and richer as he goes broke.
Right.
He does never lose more money than he gets.
Right. Actually, he loses exactly the same "amount" of money he gets (got).
On 10/24/2024 10:29 AM, WM wrote:
The set, when existing completely,
covers an interval, namely (0, ω).
When its density is halved
while the number of elements is constant,
then its extension is doubled.
No.
⟨0,1,...,n-1,n,n+1,...,n+n-1,n+n⟩ is finite.
On 24.10.2024 16:05, Jim Burns wrote:
The all.the.doubles _claim_
follows the
well.ordered.successored.non.0.predecessored
claim not.first.falsely,
so it can't be false.
It is false.
The all.the.doubles _claim_
follows the
well.ordered.successored.non.0.predecessored
claim not.first.falsely,
so it can't be false.
It is false.
The set, when existing completely,
covers an interval, namely (0, ω).
When its density is halved
while the number of elements is constant,
then its extension is doubled.
On 25.10.2024 18:54, Jim Burns wrote:
On 10/24/2024 10:29 AM, WM wrote:
The set, when existing completely,
covers an interval, namely (0, ω).
When its density is halved
while the number of elements is constant,
then its extension is doubled.
No.
⟨0,1,...,n-1,n,n+1,...,n+n-1,n+n⟩ is finite.
The extension is doubled in this finite set here
The set,[...] namely (0, ω).
as well as in my infinite set.
'Infinite' does not mean
what you (WM) want it to mean.
On 25.10.2024 20:30, Jim Burns wrote:
'Infinite' does not mean
what you (WM) want it to mean.
Your unfounded claims are irrelevant.
Relevant are only provable mathematical facts:
When the density is halved the covered interval is doubled.
Lossless exchanges do never lose the exchanged.
Am 24.10.2024 um 16:32 schrieb Moebius:
Am 24.10.2024 um 15:59 schrieb joes:
Am Thu, 24 Oct 2024 15:49:48 +0200 schrieb WM:
The [fact] of always even greater ones in [the set of] natural
numbers proves
potential infinity.
Nonsense. It proves _actual_ infinity.
Hint: For any nonempty set of natural numbers M:
M is infinite iff for each and every element m in M there's an
element m' in M such that m' > m.
(Note that sets don't change.)
Am 24.10.2024 um 16:32 schrieb Moebius:
Am 24.10.2024 um 15:59 schrieb joes:
Am Thu, 24 Oct 2024 15:49:48 +0200 schrieb WM:
The [fact] of always even greater ones in [the set of] natural
numbers proves
potential infinity.
Nonsense. It proves _actual_ infinity.
Hint: For any nonempty set of natural numbers M:
M is infinite iff for each and every element m in M there's an
element m' in M such that m' > m.
(Note that sets don't change.)
Am 24.10.2024 um 16:37 schrieb Moebius:
Am 24.10.2024 um 16:32 schrieb Moebius:
Am 24.10.2024 um 15:59 schrieb joes:
[...]
Nonsense. It proves _actual_ infinity.
Hint:
For any nonempty set of natural numbers M:
M is infinite iff
for each and every element m in M
there's an element m' in M such that
m' > m.
(Note that sets don't change.)
The following approach concerning "|-symbols"
might be considered
a "representation" of "potential infinity":
Rule 1:
We may construct the |-symbol |.
Rule 2:
Given any |-symbol
we may construct an |-symbol consisting of
the given |-symbol followed by |.
Rule 3:
All |-symbols have to be constructed by
applying rule 1 and rule 2
(finitely many times).
On 10/25/2024 3:07 PM, WM wrote:
Relevant are only provable mathematical facts:
Provable:
⟦0,n⟧ ⊂ ⟦0,ω⦆ ⇔ ⟦0,n+n⟧ ⊂ ⟦0,ω⦆
When the density is halved the covered interval is doubled.
Provable:
2× : ℕ → 𝔼 : one.to.one
𝔼 = 2×(ℕ) ⊂ ℕ
Lossless exchanges do never lose the exchanged.
Provable:
|𝔼| = |ℕ\𝔼| = |ℕ|
On 26.10.2024 01:18, Jim Burns wrote:
On 10/25/2024 3:07 PM, WM wrote:
Relevant are only provable mathematical facts:
Provable:
⟦0,n⟧ ⊂ ⟦0,ω⦆ ⇔ ⟦0,n+n⟧ ⊂ ⟦0,ω⦆
True for all natnumbers you can conceive of.
When the density is halved
the covered interval is doubled.
Provable:
2× : ℕ → 𝔼 : one.to.one
𝔼 = 2×(ℕ) ⊂ ℕ
Only for natnumbers you can conceive of.
Lossless exchanges do never lose the exchanged.
Provable:
|𝔼| = |ℕ\𝔼| = |ℕ|
Only for natnumbers which have
almost all natnumbers as successors.
On 24.10.2024 15:59, joes wrote:Exactly, they *are* doubled.
Am Thu, 24 Oct 2024 15:49:48 +0200 schrieb WM:
The potentially infinite set does not include them. Then they would be doubled too.The possibility of always even greater ones in natural numbers provesYou have it backwards. Surely the "complete" set should not be missing
potential infinity.
those greater numbers that the "potential" set includes.
The interval is and stays infinite.The complete set covers an interval. When its density is reduced its extension is increased.The greater ones have not been doubled because doubling of a completeThey have also been doubled, along with their doubles. The powers of 2
set creates a set covering a greater interval than covered before.
(Half the density implies twice the extension.)
and their multiples form a subset of the naturals. The "size" of this
set is omega, and 2w=w, regardless of "reality".
On 10/26/2024 11:55 AM, WM wrote:
On 26.10.2024 01:18, Jim Burns wrote:
On 10/25/2024 3:07 PM, WM wrote:
Relevant are only provable mathematical facts:
Provable:
⟦0,n⟧ ⊂ ⟦0,ω⦆ ⇔ ⟦0,n+n⟧ ⊂ ⟦0,ω⦆
True for all natnumbers you can conceive of.
True for all finite ordinal which are finite ordinals.
Each finite ordinal has
almost all finite ordinals as successors.
Am Thu, 24 Oct 2024 16:24:20 +0200 schrieb WM:
The complete set covers an interval. When its density is reduced itsThe interval is and stays infinite.
extension is increased.
On 28.10.2024 17:22, joes wrote:
Am Thu, 24 Oct 2024 16:24:20 +0200 schrieb WM:
The complete set covers an interval. When its density is reduced its
extension is increased.
The interval is and stays infinite.
No discussion possible since you do not accept arguments.
Am 28.10.2024 21:09:38 WM schrieb:
No discussion possible since you do not accept arguments.
@joes: p l e a s e, get that already!
On 10/28/2024 1:09 PM, WM wrote:
On 28.10.2024 17:22, joes wrote:
Am Thu, 24 Oct 2024 16:24:20 +0200 schrieb WM:
The complete set covers an interval. When its density is reduced its
extension is increased.
The interval is and stays infinite.
No discussion possible since you do not accept arguments.
Am 28.10.2024 21:09:38 WM schrieb:
No discussion possible since you do not accept arguments.
@joes: p l e a s e, get that already!
Am 28.10.2024 um 22:15 schrieb Tom Bola:
Am 28.10.2024 21:09:38 WM schrieb:
No discussion possible since you do not accept arguments.
@joes: p l e a s e, get that already!
Agree. P l e a s e stop it now.
On 28.10.2024 17:22, joes wrote:
Am Thu, 24 Oct 2024 16:24:20 +0200 schrieb WM:
The complete set covers an interval. When its density is reduced itsThe interval is and stays infinite.
extension is increased.
No discussion possible since you do not accept arguments.
Regards, WM
On 28.10.2024 13:23, Jim Burns wrote:
On 10/26/2024 11:55 AM, WM wrote:
On 26.10.2024 01:18, Jim Burns wrote:
On 10/25/2024 3:07 PM, WM wrote:
Relevant are only provable mathematical facts:
Provable:
⟦0,n⟧ ⊂ ⟦0,ω⦆ ⇔ ⟦0,n+n⟧ ⊂ ⟦0,ω⦆
True for all natnumbers you can conceive of.
True for all finite ordinal which are finite ordinals.
Relevant is only that
the density in the interval is halved,
the number remains,
the interval is doubled.
On 10/28/2024 4:01 PM, WM wrote:
Relevant is only that
the density in the interval is halved,
the number remains,
the interval is doubled.
For each finite ordinal n
there is a larger finite double n+n
For each finite double n+n
there is a larger finite ordinal n+n+1
The finite ordinal interval and
the finite double interval are the same.
'Infinite' does not mean what you want it to mean.
On 29.10.2024 18:19, Jim Burns wrote:
On 10/28/2024 4:01 PM, WM wrote:
Relevant is only that
the density in the interval is halved,
the number remains,
the interval is doubled.
For each finite ordinal n
there is a larger finite double n+n
For each finite double n+n
there is a larger finite ordinal n+n+1
The finite ordinal interval and
the finite double interval are the same.
'Infinite' does not mean what you want it to mean.
I have no preference.
If infinity is complete,
the we can double all natural numbers with the result
(0, ω)*2 = (0, ω*2).
The some products are in the interval (ω, ω*2).
On 10/30/2024 11:27 AM, WM wrote:
If infinity is complete,
the we can double all natural numbers with the result
(0, ω)*2 = (0, ω*2).
Then some products are in the interval (ω, ω*2).
ω is infinite.
On 30.10.2024 17:52, Jim Burns wrote:
On 10/30/2024 11:27 AM, WM wrote:
If infinity is complete,
the we can double all natural numbers
with the result
(0, ω)*2 = (0, ω*2).
n ∈ [0,ω) ⇒
∃⟨0,1,...,n-1,n⟩ ⇒
∃⟨0,1,...,n-1,n,n+1,...,n+n-1,n+n⟩ ⇒
n+n ∈ [0,ω)
n+n ∈ [0,ω) ⇒
∃⟨0,1,...,n-1,n,n+1,...,n+n-1,n+n⟩ ⇒
∃⟨0,1,...,n-1,n⟩ ⇒
n ∈ [0,ω)
Then some products are in the interval (ω, ω*2).
ω is infinite.
Do all numbers between 0 and ω exist such that
they can be doubled?
There is no n.sequence between 0 and ω
without a n+n.sequence between 0 and ω.
'Infinite' does not mean what you want it to mean.
On 30.10.2024 17:52, Jim Burns wrote:
On 10/30/2024 11:27 AM, WM wrote:
If infinity is complete,
the we can double all natural numbers with the result
(0, ω)*2 = (0, ω*2).
Then some products are in the interval (ω, ω*2).
ω is infinite.
Do all numbers between 0 and ω exist such that they can be doubled?
Regards, WM
On 10/30/24 3:52 PM, WM wrote:
Do all numbers between 0 and ω exist such that they can be doubled?
All Natural Numbers can be doubled and get a number that is in that set.
On 31.10.2024 00:49, Jim Burns wrote:
There is no n.sequence between 0 and ω
without a n+n.sequence between 0 and ω.
'Infinite' does not mean what you want it to mean.
Then infinity means only
Then infinity means only
an interval on the real line that
can be extended by a factor 2
when _all_ its numbers
(including all n+n sequences)
are doubled.
Hilbert, Cantor, and others
call that potential infinite.
On 10/31/2024 4:17 AM, WM wrote:
On 31.10.2024 00:49, Jim Burns wrote:
There is no n.sequence between 0 and ω
without a n+n.sequence between 0 and ω.
'Infinite' does not mean what you want it to mean.
Then infinity means only
'Infinite' means 'not finite'.
Hilbert, Cantor, and others
call that potential infinite.
Does it matter what it's called?
It is what it is.
A finite sequence of only
true.or.not.first.false 𝗰𝗹𝗮𝗶𝗺𝘀
holds a 𝗰𝗹𝗮𝗶𝗺 that,
after all swaps,
Bob is not in any visible or dark room
which he has ever been in.
On 31.10.2024 19:07, Jim Burns wrote:
On 10/31/2024 4:17 AM, WM wrote:
On 31.10.2024 00:49, Jim Burns wrote:
There is no n.sequence between 0 and ω
without a n+n.sequence between 0 and ω.
'Infinite' does not mean what you want it to mean.
Then infinity means only
'Infinite' means 'not finite'.
Potentially infinite means
finite but variable without bound.
Actually infinite means
a fixed quantity larger than every finite quantity.
On 10/31/2024 2:30 PM, WM wrote:
On 31.10.2024 19:07, Jim Burns wrote:
On 10/31/2024 4:17 AM, WM wrote:
On 31.10.2024 00:49, Jim Burns wrote:
There is no n.sequence between 0 and ω
without a n+n.sequence between 0 and ω.
'Infinite' does not mean what you want it to mean.
Then infinity means only
'Infinite' means 'not finite'.
Potentially infinite means
finite but variable without bound.
Our sets do not change.
On 31.10.2024 12:36, Richard Damon wrote:I seriously doubt the sanity of your fictional world where some numbers
On 10/30/24 3:52 PM, WM wrote:
Do all numbers between 0 and ω exist such that they can be doubled?
I read "ALL naturals". Have you considered the inverse of your functionAll Natural Numbers can be doubled and get a number that is in thatBut some of them were not doubled.
set.
On 31.10.2024 19:07, Jim Burns wrote:Potentially infinite means "if you thought it was finite, it's larger
On 10/31/2024 4:17 AM, WM wrote:Potentially infinite means finite but variable without bound. Actually infinite means a fixed quantity larger than every finite quantity.
On 31.10.2024 00:49, Jim Burns wrote:'Infinite' means 'not finite'.
There is no n.sequence between 0 and ω without a n+n.sequence between >>>> 0 and ω.
'Infinite' does not mean what you want it to mean.
He is nowhere.If it is only potentially, then Cantor's theory is wrong.Hilbert, Cantor, and others call that potential infinite.Does it matter what it's called? It is what it is.
A finite sequence of only true.or.not.first.false 𝗰𝗹𝗮𝗶𝗺𝘀 holds a 𝗰𝗹𝗮𝗶𝗺He is not elsewhere either.
that, after all swaps,
Bob is not in any visible or dark room which he has ever been in.
On 31.10.2024 20:25, Jim Burns wrote:All even numbers have a "half" that is natural.
On 10/31/2024 2:30 PM, WM wrote:Then:
On 31.10.2024 19:07, Jim Burns wrote:Our sets do not change.
On 10/31/2024 4:17 AM, WM wrote:
On 31.10.2024 00:49, Jim Burns wrote:
Potentially infinite means finite but variable without bound.'Infinite' means 'not finite'.There is no n.sequence between 0 and ω without a n+n.sequence
between 0 and ω.
'Infinite' does not mean what you want it to mean.
Multiplication of all infinitely many numbers of the open interval (0,
ω) result in some numbers in (ω, ω*2).
Reducing the density increases the interval.
On 31.10.2024 19:07, Jim Burns wrote:
On 10/31/2024 4:17 AM, WM wrote:
On 31.10.2024 00:49, Jim Burns wrote:
There is no n.sequence between 0 and ω
without a n+n.sequence between 0 and ω.
'Infinite' does not mean what you want it to mean.
Then infinity means only
'Infinite' means 'not finite'.
Potentially infinite means finite but variable without bound.
Actually infinite means a fixed quantity larger than every finite
quantity. Die Zahl 0 ist größer als jede endliche Zahl [Cantor].
Hilbert, Cantor, and others
call that potential infinite.
Does it matter what it's called?
It is what it is.
If it is only potentially, then Cantor's theory is wrong.
A finite sequence of only
true.or.not.first.false 𝗰𝗹𝗮𝗶𝗺𝘀
holds a 𝗰𝗹𝗮𝗶𝗺 that,
after all swaps,
Bob is not in any visible or dark room
which he has ever been in.
He is not elsewhere either.
Regards, WM
On 31.10.2024 20:25, Jim Burns wrote:
On 10/31/2024 2:30 PM, WM wrote:
On 31.10.2024 19:07, Jim Burns wrote:
'Infinite' means 'not finite'.
Potentially infinite means
finite but variable without bound.
Our sets do not change.
Then:
Multiplication of
all infinitely many fractions of the open interval (0,1)
results in some fractions in (1, 2).
Multiplication of
all infinitely many numbers of the open interval (0,ω)
result in some numbers in (ω, ω*2).
Multiplication of
all infinitely many numbers of the open interval (0,ω)
result in some numbers in (ω, ω*2).
Reducing the density increases the interval.
Am Thu, 31 Oct 2024 18:36:56 +0100 schrieb WM:
On 31.10.2024 12:36, Richard Damon wrote:I seriously doubt the sanity of your fictional world where some numbers
On 10/30/24 3:52 PM, WM wrote:
Do all numbers between 0 and ω exist such that they can be doubled?
don't exist.
I read "ALL naturals".All Natural Numbers can be doubled and get a number that is in thatBut some of them were not doubled.
set.
that counts the UFs larger than its argument?
Reducing the density increases the interval.All even numbers have a "half" that is natural.
Am Thu, 31 Oct 2024 19:30:00 +0100 schrieb WM:
He is nowhere.Bob is not in any visible or dark room which he has ever been in.He is not elsewhere either.
On 10/31/2024 4:27 PM, WM wrote:
On 31.10.2024 20:25, Jim Burns wrote:
On 10/31/2024 2:30 PM, WM wrote:
On 31.10.2024 19:07, Jim Burns wrote:
'Infinite' means 'not finite'.
Potentially infinite means
finite but variable without bound.
Our sets do not change.
Then:
Multiplication of
all infinitely many fractions of the open interval (0,1)
results in some fractions in (1, 2).
No.
On 10/31/24 4:27 PM, WM wrote:
Reducing the density increases the interval.Nope, becuase there is a difference in nature between an interval of
finite length, and an interval of infinite length.
On 10/31/24 2:30 PM, WM wrote:
'Infinite' means 'not finite'.
Potentially infinite means finite but variable without bound.
Actually infinite means a fixed quantity larger than every finite
quantity. Die Zahl ℵo ist größer als jede endliche Zahl [Cantor].
And, we, as finite beings, can't really perceive the nature of the
actual infinity, and thus can make errors
On 10/31/24 1:36 PM, WM wrote:
On 31.10.2024 12:36, Richard Damon wrote:
On 10/30/24 3:52 PM, WM wrote:
Do all numbers between 0 and ω exist such that they can be doubled?
All Natural Numbers can be doubled and get a number that is in that set.
But some of them were not doubled.
Nope, they all were.
Which one wasn't?
WM pretended :
Our sets do not change.
Then:
Multiplication of all infinitely many fractions of the open interval
(0, 1) results in some fractions in (1, 2).
Multiplication of all infinitely many numbers of the open interval (0,
ω) result in some numbers in (ω, ω*2).
No, there are no finite numbers in the transfinites.
WM used his keyboard to write :
On 31.10.2024 22:53, FromTheRafters wrote:
WM pretended :
Our sets do not change.
Then:
Multiplication of all infinitely many fractions of the open interval
(0, 1) results in some fractions in (1, 2).
Multiplication of all infinitely many numbers of the open interval
(0, ω) result in some numbers in (ω, ω*2).
No, there are no finite numbers in the transfinites.
Numbers like ω + 4 are in the infinite. But if all natural numbers are
doubled, then numbers in the infinite are produced.
What makes you think so?
On 01.11.2024 05:15, Jim Burns wrote:
On 10/31/2024 4:27 PM, WM wrote:
On 31.10.2024 20:25, Jim Burns wrote:
On 10/31/2024 2:30 PM, WM wrote:
Potentially infinite means
finite but variable without bound.
Our sets do not change.
Then:
Multiplication of
all infinitely many fractions of the open interval (0,1)
results in some fractions in (1, 2).
No.
If numbers are doubled, then greater numbers are produced.
That is simple mathematics. There is no doubt about it.
If all natural numbers are doubled,
then not only natural numbers are produced.
On 31.10.2024 22:05, joes wrote:
Am Thu, 31 Oct 2024 19:30:00 +0100 schrieb WM:
On 31.10.2024 19:07, Jim Burns wrote:
A finite sequence of only
true.or.not.first.false 𝗰𝗹𝗮𝗶𝗺𝘀
holds a 𝗰𝗹𝗮𝗶𝗺 that,
after all swaps, Bob is not in
any visible or dark room which he has ever been in.
He is not elsewhere either.
He is nowhere.
My logic prohibits
a loss during lossless exchange,
even if repeated infinitely often.
In my theory I apply this logic.
WM laid this down on his screen :
On 01.11.2024 11:59, FromTheRafters wrote:
WM used his keyboard to write :
On 31.10.2024 22:53, FromTheRafters wrote:
WM pretended :
Our sets do not change.
Then:
Multiplication of all infinitely many fractions of the open
interval (0, 1) results in some fractions in (1, 2).
Multiplication of all infinitely many numbers of the open interval >>>>>> (0, ω) result in some numbers in (ω, ω*2).
No, there are no finite numbers in the transfinites.
Numbers like ω + 4 are in the infinite. But if all natural numbers
are doubled, then numbers in the infinite are produced.
What makes you think so?
Simplest mathematics. Doubling increases the value. If all natnumbers
are existing, then the greatest existing natnumber is existing too,
That is not the nature of the set of natural numbers. There is no last,
that is for finite sets only.
then doubling it does not produce a natural number.
Stemming from a bad assumption, any garbage can be claimed.
Actual infinity is not about variable sets!
Sets don't change!
'Infinity' does not mean what you want it to mean.
On 11/1/2024 6:13 AM, WM wrote:
My logic prohibits
a loss during lossless exchange,
even if repeated infinitely often.
Define
lossless i
Consider the set Lossless of the lossless
Lossless = {i: lossless i}
In my theory I apply this logic.
'Infinite' does not mean what you want it to mean.
WM presented the following explanation :
On 01.11.2024 15:02, Jim Burns wrote:
'Infinity' does not mean what you want it to mean.
Infinity can have two meanings.
More than that, but in this context it simply means not finite.
On 31.10.2024 22:53, FromTheRafters wrote:
WM pretended :
Our sets do not change.
Then:
Multiplication of all infinitely many fractions of the open interval
(0, 1) results in some fractions in (1, 2).
Multiplication of all infinitely many numbers of the open interval
(0, ω) result in some numbers in (ω, ω*2).
No, there are no finite numbers in the transfinites.
Numbers like ω + 4 are in the infinite. But if all natural numbers are doubled, then numbers in the infinite are produced.
Regards, WM
WM was thinking very hard :
On 01.11.2024 19:43, FromTheRafters wrote:
WM presented the following explanation :
On 01.11.2024 15:02, Jim Burns wrote:
'Infinity' does not mean what you want it to mean.
Infinity can have two meanings.
More than that, but in this context it simply means not finite.
Is the elapsing time infinite?
Starting when?
Is the number of real points in the interval (0, 1) infinite?
Uncountably so.
Can you find the difference?
There is no difference up to isomorphism. If time is considered
infinitely divisible
WM expressed precisely :
On 01.11.2024 23:01, FromTheRafters wrote:
WM was thinking very hard :
On 01.11.2024 19:43, FromTheRafters wrote:
WM presented the following explanation :
On 01.11.2024 15:02, Jim Burns wrote:
'Infinity' does not mean what you want it to mean.
Infinity can have two meanings.
More than that, but in this context it simply means not finite.
Is the elapsing time infinite?
Starting when?
Irrelevant. Now or at the big bang.
Is the number of real points in the interval (0, 1) infinite?
Uncountably so.
Can you find the difference?
There is no difference up to isomorphism. If time is considered
infinitely divisible
Time is defined by its smallest unit, for instance a second.
In that case, time is only countably infinite.
On 11/1/24 6:30 AM, WM wrote:
If the complete interval is reduced in density then it is enlarged.Nope. because the "complete interval" has infinite length, and thus
Complete sets remain of same size.
"density" in the finite sense isn't defined.
On 11/1/24 1:28 PM, WM wrote:
if the meaning of invariable sets is used, then every doubling
will produce numbers larger than all doubled elements of the set.
Nope. That is just your finite thinking getting in the way.
You forget that the infinite set has no end,
On 11/1/24 6:28 AM, WM wrote:
On 01.11.2024 00:44, Richard Damon wrote:
On 10/31/24 1:36 PM, WM wrote:
On 31.10.2024 12:36, Richard Damon wrote:
On 10/30/24 3:52 PM, WM wrote:
Do all numbers between 0 and ω exist such that they can be doubled? >>>>All Natural Numbers can be doubled and get a number that is in that
set.
But some of them were not doubled.
Nope, they all were.
If numbers are doubled, then greater numbers are produced.
That is simple mathematics. There is no doubt about it.
If all natural numbers are doubled, then not only natural numbers are
produced.
No;e, becuase those greater numbers were already there.
ω/2 isn't a number
On 11/1/24 6:33 AM, WM wrote:
On 01.11.2024 00:44, Richard Damon wrote:
On 10/31/24 2:30 PM, WM wrote:
'Infinite' means 'not finite'.
Potentially infinite means finite but variable without bound.
Actually infinite means a fixed quantity larger than every finite
quantity. Die Zahl ℵo ist größer als jede endliche Zahl [Cantor].
And, we, as finite beings, can't really perceive the nature of the
actual infinity, and thus can make errors
You do. You forget that logic remains valid always and everywhere.
No, logic remains valid in the field it is defined in.
That two parallel lines will never meet is only valid in some forms of Geometry.
On 31.10.2024 22:07, joes wrote:Those are not even numbers, because they are infinite and not natural.
ω + 2 certainly, but ω*4 + 4?Reducing the density increases the interval.All even numbers have a "half" that is natural.
On 01.11.2024 13:37, FromTheRafters wrote:Why "absorbed"? Do you think some multiple of a power of 2 is not natural?
WM laid this down on his screen :But there are all. No point can escape doubling.
On 01.11.2024 11:59, FromTheRafters wrote:That is not the nature of the set of natural numbers. There is no last,
WM used his keyboard to write :Simplest mathematics. Doubling increases the value. If all natnumbers
On 31.10.2024 22:53, FromTheRafters wrote:What makes you think so?
WM pretended :Numbers like ω + 4 are in the infinite. But if all natural numbers
No, there are no finite numbers in the transfinites.Our sets do not change.Multiplication of all infinitely many fractions of the open
interval (0, 1) results in some fractions in (1, 2).
Multiplication of all infinitely many numbers of the open interval >>>>>>> (0, ω) result in some numbers in (ω, ω*2).
are doubled, then numbers in the infinite are produced.
are existing, then the greatest existing natnumber is existing too,
that is for finite sets only.
Therefore not all doubled elements can be absorbed by the set.Actual infinity is not about variable sets!Sets don't change!
Am Fri, 01 Nov 2024 11:14:59 +0100 schrieb WM:
On 31.10.2024 22:07, joes wrote:Those are not even numbers, because they are infinite and not natural.
ω + 2 certainly, but ω*4 + 4?Reducing the density increases the interval.All even numbers have a "half" that is natural.
--
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.
On 02.11.2024 01:55, Richard Damon wrote:
On 11/1/24 6:30 AM, WM wrote:
If the complete interval is reduced in density then it is enlarged.Nope. because the "complete interval" has infinite length, and thus
Complete sets remain of same size.
"density" in the finite sense isn't defined.
Density does not depend on finiteness. The density of even numbers is
half of the density of integers.
When doubling all natnumbers, the result cannot be that numbers are exorcized.
Regards, WM
On 02.11.2024 01:55, Richard Damon wrote:
On 11/1/24 6:28 AM, WM wrote:
On 01.11.2024 00:44, Richard Damon wrote:
On 10/31/24 1:36 PM, WM wrote:
On 31.10.2024 12:36, Richard Damon wrote:
On 10/30/24 3:52 PM, WM wrote:
Do all numbers between 0 and ω exist such that they can be doubled? >>>>>All Natural Numbers can be doubled and get a number that is in
that set.
But some of them were not doubled.
Nope, they all were.
If numbers are doubled, then greater numbers are produced.
That is simple mathematics. There is no doubt about it.
If all natural numbers are doubled, then not only natural numbers are
produced.
No;e, becuase those greater numbers were already there.
And were doubled.
ω/2 isn't a number
It is not visible.
Regards, WM
Why "absorbed"? Do you think some multiple of a power of 2 is not natural?
WM pretended :
No. It "is not (like every individual transfinite and in general
everything due to an 'idea divina') determined in itself, fixed, and
unchangeable, but a finite in the process of change, having in each of
its actual states a finite size; like, for instance, the time elapsed
after the beginning of the world, which, measured in some time-unit,
for instance a year, is finite in every moment, but always growing
beyond all finite limits, without ever becoming really infinitely
large." [G. Cantor, letter to I. Jeiler (13 Oct 1895)]
Really, aren't you ever going to get beyond the old ways?
On 02.11.2024 13:20, joes wrote:
Why "absorbed"?
Do you think
some multiple of a power of 2 is not natural?
If all multiples of 2 smaller than ω are doubled,
then this doubling results in larger numbers than doubled.
On 11/2/2024 1:24 PM, WM wrote:
On 02.11.2024 13:20, joes wrote:
Why "absorbed"?
Do you think
some multiple of a power of 2 is not natural?
If all multiples of 2 smaller than ω are doubled,
then this doubling results in larger numbers than doubled.
If n is finite ∧ ω ≤ n+n
then ω is finite.
⎜ ⟦0,n⟧ is finite, which means that
⎜ each subset of ⟦0,n⟧ is two.ended.or.{}
⎜
⎜ Each subset of ⟦n,n+n⟧ is two.ended.or.{}
'Finite' means
each subset is two.ended.or.{}
'Infinite' means not finite.
Alan Mackenzie laid this down on his screen :
joes <noreply@example.org> wrote:
Am Fri, 01 Nov 2024 11:14:59 +0100 schrieb WM:
On 31.10.2024 22:07, joes wrote:
Those are not even numbers, because they are infinite and not natural.ω + 2 certainly, but ω*4 + 4?Reducing the density increases the interval.All even numbers have a "half" that is natural.
Mit "even" meinst du "selbst" oder "gerade"? ;-)
Google Translate says:
By "even" do you mean "even" or "even"? ;-)
Not much help there. :)
On 02.11.2024 21:53, Jim Burns wrote:
On 11/2/2024 1:24 PM, WM wrote:
On 02.11.2024 13:20, joes wrote:
Why "absorbed"?
Do you think
some multiple of a power of 2 is not natural?
If all multiples of 2 smaller than ω are doubled,
then this doubling results in larger numbers than doubled.
If n is finite ∧ ω ≤ n+n
then ω is finite.
That might appear so in a set without dark numbers. It is not true when
dark numbers come into play.
⎜ ⟦0,n⟧ is finite, which means that
⎜ each subset of ⟦0,n⟧ is two.ended.or.{}
⎜
⎜ Each subset of ⟦n,n+n⟧ is two.ended.or.{}
Yes, but the order of dark numbers cannot be determined.
Every number n < ω is finite. ω/2 is finite.
'Finite' means
each subset is two.ended.or.{}
'Infinite' means not finite.
Every interval (0, n) is finite because n is finite. But for dark
numbers n this cannot be seen. The dark realm appears as infinite. It
cannot be counted through.
Regards, WM
On 02.11.2024 13:20, joes wrote:Powers of 2, not multiples. Apparently you do think that there is a
Why "absorbed"? Do you think some multiple of a power of 2 is notIf all multiples of 2 smaller than ω are doubled, then this doubling
natural?
results in larger numbers than doubled.
Thanks for expanding on that. I figured that there must be nuance
involved that google translate didn't pick up on.
pparently you do think that there is a
natural n such that 2^n is infinite.
On 11/2/24 5:34 PM, WM wrote:
Remember, 2 times any Natural Number is a Natural Number, provable by mathematics.
Every interval (0, n) is finite because n is finite. But for dark
numbers n this cannot be seen. The dark realm appears as infinite. It
cannot be counted through.
And thus is not finite.
On 03.11.2024 01:17, Richard Damon wrote:
On 11/2/24 5:34 PM, WM wrote:
Remember, 2 times any Natural Number is a Natural Number, provable by
mathematics.
Every doubled interval gets larger. (0, ω)*2 = (0, ω*2).
Every interval (0, n) is finite because n is finite. But for dark
numbers n this cannot be seen. The dark realm appears as infinite. It
cannot be counted through.
And thus is not finite.
Its numbers are finite as we see when a dark number becomes visible. ω
is the first infinite number by definition. ω - 1 is finite but lies in
the dark realm and therefore cannot be counted to. It appears as if it
was infinite.
Regards, WM
On 11/3/24 6:56 AM, WM wrote:
On 03.11.2024 09:50, joes wrote:If a Natural Numbers are there, there is no further needed, as they go without end.
pparently you do think that there is a
natural n such that 2^n is infinite.
If all naturals are there, then no further one is available. But
doubling all yields a greater number than all.
In actual infinity there is no way to avoid this.
On 03.11.2024 09:50, joes wrote:We don't need any further ones because we ALREADY HAVE ALL OF THEM,
pparently you do think that there is a natural n such that 2^n isIf all naturals are there, then no further one is available. But
infinite.
doubling all yields a greater number than all.
In actual infinity there is no way to avoid this.
Am Sun, 03 Nov 2024 12:56:48 +0100 schrieb WM:
On 03.11.2024 09:50, joes wrote:We don't need any further ones because we ALREADY HAVE ALL OF THEM,
pparently you do think that there is a natural n such that 2^n isIf all naturals are there, then no further one is available. But
infinite.
doubling all yields a greater number than all.
In actual infinity there is no way to avoid this.
even including the doubles.
On 03.11.2024 13:13, Richard Damon wrote:
On 11/3/24 6:56 AM, WM wrote:
On 03.11.2024 09:50, joes wrote:If a Natural Numbers are there, there is no further needed, as they go
pparently you do think that there is a
natural n such that 2^n is infinite.
If all naturals are there, then no further one is available. But
doubling all yields a greater number than all.
In actual infinity there is no way to avoid this.
without end.
All are doubled. No remainings cans absorb all products.
Regards, WM
On 03.11.2024 16:55, joes wrote:Yes I have. The set of even numbers is a subset.
Am Sun, 03 Nov 2024 12:56:48 +0100 schrieb WM:But you have not what is done to all of them afterwards.
On 03.11.2024 09:50, joes wrote:We don't need any further ones because we ALREADY HAVE ALL OF THEM,
pparently you do think that there is a natural n such that 2^n isIf all naturals are there, then no further one is available. But
infinite.
doubling all yields a greater number than all.
In actual infinity there is no way to avoid this.
even including the doubles.
You must beI know I will get even numbers.
clairvoyant if you knew in advance whether something is done at all.
On 11/3/2024 3:56 AM, WM wrote:
On 03.11.2024 09:50, joes wrote:
pparently you do think that there is a
natural n such that 2^n is infinite.
If all naturals are there, then no further one is available.
Sigh. There are infinite natural numbers, there is no last largest one.
Why can't you get this! Bone head!
But doubling all yields a greater number than all.
In actual infinity there is no way to avoid this.
Regards, WM
On 11/3/24 9:17 AM, WM wrote:
But there are, because they are infinite.If a Natural Numbers are there, there is no further needed, as they
go without end.
All are doubled. No remainings can absorb all products.
Your logic just doesn't work on infinite sets.
Am Sun, 03 Nov 2024 18:00:18 +0100 schrieb WM:
On 03.11.2024 16:55, joes wrote:Yes I have. The set of even numbers is a subset.
Am Sun, 03 Nov 2024 12:56:48 +0100 schrieb WM:But you have not what is done to all of them afterwards.
On 03.11.2024 09:50, joes wrote:We don't need any further ones because we ALREADY HAVE ALL OF THEM,
pparently you do think that there is a natural n such that 2^n isIf all naturals are there, then no further one is available. But
infinite.
doubling all yields a greater number than all.
In actual infinity there is no way to avoid this.
even including the doubles.
You must beI know I will get even numbers.
clairvoyant if you knew in advance whether something is done at all.
On 11/3/2024 3:56 AM, WM wrote:
If all naturals are there, then no further one is available.
Sigh. There are infinite natural numbers, there is no last largest one.
Why can't you get this!
On 11/3/24 12:00 PM, WM wrote:
On 03.11.2024 16:55, joes wrote:
Am Sun, 03 Nov 2024 12:56:48 +0100 schrieb WM:
On 03.11.2024 09:50, joes wrote:We don't need any further ones because we ALREADY HAVE ALL OF THEM,
pparently you do think that there is a natural n such that 2^n isIf all naturals are there, then no further one is available. But
infinite.
doubling all yields a greater number than all.
In actual infinity there is no way to avoid this.
even including the doubles.
But you have not what is done to all of them afterwards. You must be
clairvoyant if you knew in advance whether something is done at all.
The problem is that if you need to do them in "order" you can't complete
the infinite task.
That is the problem with your finite logic, that it can't actualy DO
things in actual infinity,
We don't need to be clairvoyant to understand what WILL happen with a deterministic operation.
On 03.11.2024 22:21, Richard Damon wrote:
On 11/3/24 9:17 AM, WM wrote:
But there are, because they are infinite.If a Natural Numbers are there, there is no further needed, as they
go without end.
All are doubled. No remainings can absorb all products.
Your logic just doesn't work on infinite sets.
My logic says all are there. Nobody knows whether I will double them
again. If I decide to do so, then larger numbers are created.
Regards, WM
On 03.11.2024 23:12, joes wrote:
Am Sun, 03 Nov 2024 18:00:18 +0100 schrieb WM:
You must be clairvoyant if you knew in advance
whether something is done at all.
I know I will get even numbers.
But you will get
larger even numbers than were multiplied.
On 11/4/2024 12:45 AM, FromTheRafters wrote:
Chris M. Thomasson pretended :
On 11/3/2024 2:40 PM, Chris M. Thomasson wrote:
On 11/3/2024 3:56 AM, WM wrote:
If all naturals are there, then no further one is [bla bla bla].
Sigh. There are infinite natural numbers, there is no last largest one.
I should say infinitely many natural numbers... Sorry! ;^o
Yeah, it is best to use the words 'set of' when 'all' is invoked. The
set of all natural numbers is infinite while each and every one of the
elements is finite. In that sense (the set of) 'all' is different from
'each' and 'every'.
Each and every natural number is in all of them and vise versa? Fair
enough?
Why does WM seem to think that this might hold:
any_natural + 1 = a_strange_weirdo_dark_number_thing ?
I wonder if he has dark hair? ;^)
Am 04.11.2024 um 12:33 schrieb Chris M. Thomasson:
Infinite ways to represent 1:
1 = 4 - 2 - 1
1 = 3*2 - 5
1 = 1 + 1 - 1
1 = ...
?
Right. {4 - 2 - 1, 3*2 - 5, 2 - 1, 3 - 2, 4 - 3, ...} = {1} (though FromTheAfter is too dumb to get that).
Infinite ways to represent 1:
1 = 4 - 2 - 1
1 = 3*2 - 5
1 = 1 + 1 - 1
1 = ...
?
I guess WM would think that taking a gallon of water out of an infinite
pool of water would somehow make it less than.
On 11/4/2024 1:26 AM, Chris M. Thomasson wrote:
On 11/4/2024 12:45 AM, FromTheRafters wrote:
Yeah, it is best to use the words 'set of'
when 'all' is invoked.
The set of all natural numbers is infinite
while
each and every one of the elements is finite.
In that sense (the set of) 'all'
is different from 'each' and 'every'.
Each and every natural number is in
all of them and vise versa?
Fair enough?
For instance 2 = 2 wrt the vise versa comment.
On 11/4/2024 6:19 AM, WM wrote:
On 03.11.2024 23:12, joes wrote:
Am Sun, 03 Nov 2024 18:00:18 +0100 schrieb WM:
You must be clairvoyant if you knew in advance
whether something is done at all.
I know I will get even numbers.
But you will get
larger even numbers than were multiplied.
No.
ℕ ⊇ 𝔼 := 2×ᵉᵃᶜʰ ℕ
Our sets do not change.
Yes, we know if you will double them again, as we started with the
assumption that we double ALL the number, and any even number will be
reached from doubling the number that is one half of it and will be
doubled again.
On 02.11.2024 21:53, Jim Burns wrote:
On 11/2/2024 1:24 PM, WM wrote:
If all multiples of 2 smaller than ω are doubled,
then this doubling results in larger numbers than doubled.
If n is finite ∧ ω ≤ n+n
then ω is finite.
That might appear so in a set without dark numbers.
It is not true when
dark numbers come into play.
On 11/4/2024 8:52 AM, Moebius wrote:
Am 04.11.2024 um 12:34 schrieb Chris M. Thomasson:
I guess WM would think that taking a gallon of water out of an
infinite pool of water would somehow make it less than.
If you repeat that infinitely many times, it might. :-P
The infinite pool is infinite so it can never run out? You can take out
an infinite number of gallons, and the pool is still there in its
infinite glory filled to the rim with brim ;^)... In the magic land of
port foozle in the land of Zork?
On 04.11.2024 13:26, Richard Damon wrote:
Yes, we know if you will double them again, as we started with the
assumption that we double ALL the number, and any even number will be
reached from doubling the number that is one half of it and will be
doubled again.
Please mark the sentence which you suspect to be wrong:
- All natural numbers exist.
- The even numbers have only half of the substance(*) of the integers.
- Multiplication of all elements of a set does not change the substance.
(*) For every interval (0, 2n] the number E of even natnumbers is half
of the number N of natnumbers: E/N = 1/2. For (0, oo] we obtain the
limit of the sequence 1/2, 1/2, 1/2, ... which is 1/2. This is the ratio
of the substances or, as Cantor called it, the reality.
Regards, WM
On 11/4/24 12:24 PM, WM wrote:
On 04.11.2024 13:26, Richard Damon wrote:
Yes, we know if you will double them again, as we started with the
assumption that we double ALL the number, and any even number will be
reached from doubling the number that is one half of it and will be
doubled again.
Please mark the sentence which you suspect to be wrong:
- All natural numbers exist.
- The even numbers have only half of the substance(*) of the integers.
Not a valid statement, as half of infinity is still infinity.
- Multiplication of all elements of a set does not change the substance.
(*) For every interval (0, 2n] the number E of even natnumbers is half
of the number N of natnumbers: E/N = 1/2. For (0, oo] we obtain the
limit of the sequence 1/2, 1/2, 1/2, ... which is 1/2. This is the
ratio of the substances or, as Cantor called it, the reality.
In other words, your logic can only deal with finite subsets of the
infinite set
On 03.11.2024 23:12, joes wrote:Not if you really multiply all numbers.
Am Sun, 03 Nov 2024 18:00:18 +0100 schrieb WM:It has only half of the reality of the natnumbers. But when doubling
On 03.11.2024 16:55, joes wrote:Yes I have. The set of even numbers is a subset.
Am Sun, 03 Nov 2024 12:56:48 +0100 schrieb WM:But you have not what is done to all of them afterwards.
On 03.11.2024 09:50, joes wrote:We don't need any further ones because we ALREADY HAVE ALL OF THEM,
pparently you do think that there is a natural n such that 2^n isIf all naturals are there, then no further one is available. But
infinite.
doubling all yields a greater number than all.
In actual infinity there is no way to avoid this.
even including the doubles.
them, their full reality is maintained.
But you will get larger even numbers than were multiplied.
You must be clairvoyant if you knew in advance whether something isI know I will get even numbers.
done at all.
Am Mon, 04 Nov 2024 12:19:33 +0100 schrieb WM:
Not if you really multiply all numbers.But you will get larger even numbers than were multiplied.You must be clairvoyant if you knew in advance whether something isI know I will get even numbers.
done at all.
On 05.11.2024 04:08, Richard Damon wrote:
On 11/4/24 12:24 PM, WM wrote:
On 04.11.2024 13:26, Richard Damon wrote:
Yes, we know if you will double them again, as we started with the
assumption that we double ALL the number, and any even number will
be reached from doubling the number that is one half of it and will
be doubled again.
Please mark the sentence which you suspect to be wrong:
- All natural numbers exist.
- The even numbers have only half of the substance(*) of the integers.
Not a valid statement, as half of infinity is still infinity.
Substance means number obtained from limit: 1/2, 1/2, 1/2, ... --> 1/2.
- Multiplication of all elements of a set does not change the substance. >>>
(*) For every interval (0, 2n] the number E of even natnumbers is
half of the number N of natnumbers: E/N = 1/2. For (0, oo] we obtain
the limit of the sequence 1/2, 1/2, 1/2, ... which is 1/2. This is
the ratio of the substances or, as Cantor called it, the reality.
In other words, your logic can only deal with finite subsets of the
infinite set
It deals with limits.
Regards, WM
On 03.11.2024 23:40, Chris M. Thomasson wrote:Yes, and you can't seem to imagine the infinite whole.
On 11/3/2024 3:56 AM, WM wrote:
Because Cantor applies all with no exception for enumerating purposes.If all naturals are there, then no further one is available.Sigh. There are infinite natural numbers, there is no last largest one.
Why can't you get this!
On 05.11.2024 20:39, joes wrote:In every finite case.
Am Mon, 04 Nov 2024 12:19:33 +0100 schrieb WM:
In every case.Not if you really multiply all numbers.But you will get larger even numbers than were multiplied.You must be clairvoyant if you knew in advance whether something isI know I will get even numbers.
done at all.
Am Mon, 04 Nov 2024 12:28:44 +0100 schrieb WM:
On 03.11.2024 23:40, Chris M. Thomasson wrote:Yes, and you can't seem to imagine the infinite whole.
On 11/3/2024 3:56 AM, WM wrote:Because Cantor applies all with no exception for enumerating purposes.
If all naturals are there, then no further one is available.Sigh. There are infinite natural numbers, there is no last largest one.
Why can't you get this!
On 11/5/24 11:43 AM, WM wrote:
- All natural numbers exist.
- The even numbers have only half of the substance(*) of the integers.
Not a valid statement, as half of infinity is still infinity.
Substance means number obtained from limit: 1/2, 1/2, 1/2, ... --> 1/2.
So, all you are doing is proving that half of infinity is still
Am Wed, 06 Nov 2024 11:41:47 +0100 schrieb WM:
On 05.11.2024 20:39, joes wrote:In every finite case.
Am Mon, 04 Nov 2024 12:19:33 +0100 schrieb WM:In every case.
Not if you really multiply all numbers.But you will get larger even numbers than were multiplied.You must be clairvoyant if you knew in advance whether something is >>>>>> done at all.I know I will get even numbers.
If I start with 2 and keep doubling, I never leave N or reach omega.
WM brought next idea :
On 06.11.2024 12:46, Richard Damon wrote:
On 11/5/24 11:43 AM, WM wrote:
So, all you are doing is proving that half of infinity is still- All natural numbers exist.
- The even numbers have only half of the substance(*) of the
integers.
Not a valid statement, as half of infinity is still infinity.
Substance means number obtained from limit: 1/2, 1/2, 1/2, ... --> 1/2. >>>
half of infinity.
Nope, a ray is the same size as a line.
On 11/6/2024 7:14 AM, WM wrote:
The mathe[maticians] who claim that after every natural number there are
almost all natural numbers
sum [series],
1+2+3+4+5+6+...
It still equals a natural number at every iteration.
You are so wrong it's a bit scary... ;^o
On 06.11.2024 12:46, Richard Damon wrote:
On 11/5/24 11:43 AM, WM wrote:
- All natural numbers exist.Not a valid statement, as half of infinity is still infinity.
- The even numbers have only half of the substance(*) of the integers. >>>>
Substance means number obtained from limit: 1/2, 1/2, 1/2, ... --> 1/2.
So, all you are doing is proving that half of infinity is still
half of infinity.
Regards, WM
Am 06.11.2024 um 22:00 schrieb Chris M. Thomasson:
On 11/6/2024 7:14 AM, WM wrote:
The mathe[maticians] who claim that after every natural number there
are almost all natural numbers
are quite right.
Hint: If n is a natural number, then it is "followed" by the infinitely
many natural numbers n+1, n+2, n+3, ...
On 07.11.2024 01:03, Moebius wrote:What the fuck are you talking about?
Am 06.11.2024 um 22:00 schrieb Chris M. Thomasson:Almost all remain outside of the bijection with the rationals? How many rationals remain outside too?
On 11/6/2024 7:14 AM, WM wrote:are quite right.
The mathe[maticians] who claim that after every natural number there
are almost all natural numbers
Hint: If n is a natural number, then it is "followed" by the infinitely
many natural numbers n+1, n+2, n+3, ...
Am Thu, 07 Nov 2024 10:07:19 +0100 schrieb WM:
On 07.11.2024 01:03, Moebius wrote:
On 11/6/2024 7:14 AM, WM wrote:
The mathe[maticians] who claim that after every natural number there >>>>> are almost all natural numbers
are quite right.
Hint: If n is a natural number, then it is "followed" by the infinitely
many natural numbers n+1, n+2, n+3, ...
Almost all remain outside of the bijection with the rationals? How many
rationals remain outside too? [WM]
What the fuck are you talking about?
On 11/6/2024 4:07 PM, Moebius wrote:
Am 06.11.2024 um 22:04 schrieb Chris M. Thomasson:
sum [series],
1+2+3+4+5+6+...
is not a natural number. :-P
Since there is no largest natural number, the sum is a natural number?
See: https://en.wikipedia.org/wiki/Series_(mathematics)
It still equals a natural number at every iteration.
I guess, what you mean is:
For all n e IN:
1 + ... + n
is a natural number (i.e. in IN).
Yeah. (Simple proof by induction.)
On 11/7/2024 1:34 PM, Moebius wrote:
Am 07.11.2024 um 22:08 schrieb Chris M. Thomasson:
On 11/6/2024 4:07 PM, Moebius wrote:
Am 06.11.2024 um 22:04 schrieb Chris M. Thomasson:
sum [series],
1+2+3+4+5+6+...
is not a natural number. :-P
Since there is no largest natural number, the sum is a natural number?
Nope. There simply is no "sum".
The [partial] sum keeps getting bigger and bigger
However, each iteration is a sum and is a natural number?
For all n e IN:
1 + ... + n
is a natural number (i.e. in IN).
For all n e IN:
1 + ... + n
is a natural number (i.e. in IN).
(Simple proof by induction.)
For all n e IN:
1 + ... + n
is a natural number (i.e. in IN).
(Simple proof by induction.)
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