Does ℕ = {1, 2, 3, ...} contain all natural numbers such that none cn be
added?

Wären in der Menge {2, 4, 6, ...} genau so viele Zahlen wie in ℕ, dann
lieferte die Vervollständigung {1, 2, 3, 4, 5, 6, ...} "mehr Realität
wie" ℕ = {1, 2, 3, ...}. Dann gäbe es also in der Realität mehr
natürliche Zahlen als ℕ enthält.

On 3/24/2024 1:00 PM, WM wrote:
> Does ℕ = {1, 2, 3, ...} contain all natural numbers such that none cn be
> added?
>
>
> Wären in der Menge {2, 4, 6, ...} genau so viele Zahlen wie in ℕ, dann
> lieferte die Vervollständigung {1, 2, 3, 4, 5, 6, ...} "mehr Realität
> wie" ℕ = {1, 2, 3, ...}. Dann gäbe es also in der Realität mehr
> natürliche Zahlen als ℕ enthält.
>
>

adding one to infinity = infinity. Give me your largest number. I say,
that plus one... ;^) Take the natural numbers:

Think of an infinite waterfall pouring into a magic sack that can hold
infinity... It has a boundary, yet can hold infinity... The infinite
waterfall pours into it for infinity. The sack stays the same size.

Disney did a little animation about it. Humm.... Let me try to find the
scene...

https://youtu.be/Tb75RjpvBIk

;^D

Something that has a border, yet is infinite in and of itself... The
Mandelbrot set. ;^) lol.

Does ℕ = {1, 2, 3, ...} contain all natural numbers such that none can
be added?

If so, then the bijection of ℕ with E = {2, 4, 6, ...} would prove that
both sets have the same number of elements. Then the completion of E
resulting in E = {1, 2, 3, 4, 5, 6, ...} would double the number of its
elements. Then there are more natural numbers than were originally in ℕ.

Regards, WM

Le 24/03/2024 à 21:11, "Chris M. Thomasson" a écrit :

> adding one to infinity = infinity.

If ℕ is complete no natural number can be added.

Regards, WM

On 3/24/2024 1:13 PM, WM wrote:
> Le 24/03/2024 à 21:11, "Chris M. Thomasson" a écrit :
>
>> adding one to infinity = infinity.
>
> If ℕ is complete no natural number can be added.

Sigh. You are misunderstanding infinity... I think wrt infinity, when
you hear the word "complete", your mind instantly thinks, "finite".
Well, this is wrong....................

Infinity is not finite...

;^D

Le 24/03/2024 à 21:16, "Chris M. Thomasson" a écrit :
> On 3/24/2024 1:13 PM, WM wrote:
>> Le 24/03/2024 à 21:11, "Chris M. Thomasson" a écrit :
>>
>>> adding one to infinity = infinity.
>>
>> If ℕ is complete no natural number can be added.
>
> Sigh. You are misunderstanding infinity... I think wrt infinity, when
> you hear the word "complete", your mind instantly thinks, "finite".

Can you add a natural number to the set of all natura, numbers?

> Infinity is not finite...

But logic has to be observed.

Regards, WM

WM schrieb:
>
> Does ℕ = {1, 2, 3, ...} contain all natural numbers such that none
> can be added?
>

Sure - adding a number that is already contained in ℕ, doesn't change
the cardinality of the set, since equal elements are counted as one
element.

But you can do the following described by Cantor:

"That ℵo is a transfinite number, that is to say, is
not equal to any finite number μ, follows from the
simple fact that, if to the aggregate {ν} is added a
new element e_0, the union-aggregate ({ν}, e_0 ) is
equivalent to the original aggregate {ν}. For we
can think of this reciprocally univocal correspondence
between them: to the element e_0 of the first
corresponds the element 1 of the second, and to the
element ν of the first corresponds the element ν + 1 of
the other. By §3 we thus have

(2) ℵo + 1 = ℵo "

(Georg Cantor:
Contributions to the founding of the theory of tranfinite numbers.
Dover Publications, 1915; p.104)

> If so, then the bijection of ℕ with E = {2, 4, 6, ...} would prove
> that both sets have the same number of elements.

Infinite sets don't have a "number of elements". This concept (which can
only be used for finite sets) is generalized for infinite sets by the
concept of "transfinite cardinal numbers".

And indeed: there is a bijection from the set of natural numbers ℕ
to the set of even natural numbers 𝔼 = {2, 4, 6, ..}.

f: ℕ → 𝔼 , n ↦ 2n

This function is both injective (or one-to-one) and surjective (or
onto), thus it is bijective.

> Then the completion of E resulting in E = {1, 2, 3, 4, 5, 6, ...}
> would double the number of its elements. Then there are more natural
> numbers than were originally in ℕ.

Rubbish. The cardinality of an infinite set is described by an
transfinite cardinal number and not by a finite "number of elements".
Your problem is: You try to apply facts, that hold for finite sets,
on infinite sets. That doesn't work.

Dieter Heidorn

WM was thinking very hard :
> Does ℕ = {1, 2, 3, ...} contain all natural numbers such that none cn be
> added?

Yes, it is the set of natural numbers.

> Wären in der Menge {2, 4, 6, ...} genau so viele Zahlen wie in ℕ, dann
> lieferte die Vervollständigung {1, 2, 3, 4, 5, 6, ...} "mehr Realität wie" ℕ
> = {1, 2, 3, ...}. Dann gäbe es also in der Realität mehr natürliche Zahlen
> als ℕ enthält.

I don't read German.

On 3/24/24 4:11 PM, WM wrote:
>
> Does ℕ = {1, 2, 3, ...} contain all natural numbers such that none can
> be added?
>
> If so, then the bijection of ℕ with E = {2, 4, 6, ...} would prove that
> both sets have the same number of elements. Then the completion of E
> resulting in E = {1, 2, 3, 4, 5, 6, ...} would double the number of its
> elements. Then there are more natural numbers than were originally in ℕ.
>
> Regards, WM
>
>

Yep, because "infinity" just doesn't obey the logic you are used to, and
insist on using, even in cases it can't be appled to, making your logic
system just all blown up to smithereens.

Turns out that aleph0 = aleph0 - 1, = aleph0 + 1 = aleph0 / 2 =
aleph0 * 2 = aleph0 ^ 2 = square root (aleph0)

despite that breaking most of you concepts of the logic of numbers.

FromTheRafters schrieb:
> WM was thinking very hard :
>> Does ℕ = {1, 2, 3, ...} contain all natural numbers such that none cn
>> be added?
>
> Yes, it is the set of natural numbers.
>
>> Wären in der Menge {2, 4, 6, ...} genau so viele Zahlen wie in ℕ, dann
>> lieferte die Vervollständigung {1, 2, 3, 4, 5, 6, ...} "mehr Realität
>> wie" ℕ = {1, 2, 3, ...}. Dann gäbe es also in der Realität mehr
>> natürliche Zahlen als ℕ enthält.
>
> I don't read German.

It means:

"If there were exactly as many numbers in the set {2, 4, 6, ...} as in
N, then the completion would yield {1, 2, 3, 4, 5, 6, ...} 'more
reality like' N = {1, 2, 3, ...}. In reality, there would be more
natural numbers than N contains."

Dieter Heidorn

WM explained :
> Does ℕ = {1, 2, 3, ...} contain all natural numbers such that none can be
> added?
>
> If so, then the bijection of ℕ with E = {2, 4, 6, ...} would prove that both
> sets have the same number of elements.

Actually, same size set. "Number of elements" is better suited to
finite sets.

> Then the completion of E resulting in
> E = {1, 2, 3, 4, 5, 6, ...} would double the number of its elements.

That is not a completion of E. But still the same size set. By your
sense of 'complete' the set of even numbers was already 'complete'
because no more even numbers could be 'added'.

> Then there are more natural numbers than were originally in ℕ.

Nope.

Dieter Heidorn wrote :
> FromTheRafters schrieb:
>> WM was thinking very hard :
>>> Does ℕ = {1, 2, 3, ...} contain all natural numbers such that none cn be
>>> added?
>>
>> Yes, it is the set of natural numbers.
>>
>>> Wären in der Menge {2, 4, 6, ...} genau so viele Zahlen wie in ℕ, dann
>>> lieferte die Vervollständigung {1, 2, 3, 4, 5, 6, ...} "mehr Realität wie"
>>> ℕ = {1, 2, 3, ...}. Dann gäbe es also in der Realität mehr natürliche
>>> Zahlen als ℕ enthält.
>>
>> I don't read German.
>
> It means:
>
> "If there were exactly as many numbers in the set {2, 4, 6, ...} as in
> N, then the completion would yield {1, 2, 3, 4, 5, 6, ...} 'more
> reality like' N = {1, 2, 3, ...}. In reality, there would be more
> natural numbers than N contains."
>
> Dieter Heidorn

Thank you, or danke.

On 3/24/2024 1:20 PM, WM wrote:
> Le 24/03/2024 à 21:16, "Chris M. Thomasson" a écrit :
>> On 3/24/2024 1:13 PM, WM wrote:
>>> Le 24/03/2024 à 21:11, "Chris M. Thomasson" a écrit :
>>>
>>>> adding one to infinity = infinity.
>>>
>>> If ℕ is complete no natural number can be added.
>>
>> Sigh. You are misunderstanding infinity... I think wrt infinity, when
>> you hear the word "complete", your mind instantly thinks, "finite".
>
> Can you add a natural number to the set of all natura, numbers?

ℕ is not finite! Did you notice my infinite waterfall example?

>> Infinity is not finite...
>
> But logic has to be observed.

Right. So any natural number you give me, I can say that plus one. So,
its unbounded in that sense... ;^)

[...]

On 3/24/2024 1:20 PM, WM wrote:
> Le 24/03/2024 à 21:16, "Chris M. Thomasson" a écrit :
>> On 3/24/2024 1:13 PM, WM wrote:
>>> Le 24/03/2024 à 21:11, "Chris M. Thomasson" a écrit :
>>>
>>>> adding one to infinity = infinity.
>>>
>>> If ℕ is complete no natural number can be added.
>>
>> Sigh. You are misunderstanding infinity... I think wrt infinity, when
>> you hear the word "complete", your mind instantly thinks, "finite".
>
> Can you add a natural number to the set of all natura, numbers?

https://youtu.be/Tb75RjpvBIk

;^D

>
>> Infinity is not finite...
>
> But logic has to be observed.
[...]

Le 24/03/2024 à 22:02, Dieter Heidorn a écrit :
> WM schrieb:
>>
>> Does ℕ = {1, 2, 3, ...} contain all natural numbers such that none
>> can be added?
>>
> Sure

Of course.
>
> But you can do the following described by Cantor:

Irrelevant. Important is only this: You cannot add a natural number to the
set ℕ.
>
>> If so, then the bijection of ℕ with E = {2, 4, 6, ...} would prove
>> that both sets have the same number of elements.
>
> Infinite sets don't have a "number of elements".

This cannot be denied: A bijection, if really existing, proves that one of
both sets has not one element more nor less than the other!
>
> And indeed: there is a bijection from the set of natural numbers ℕ
> to the set of even natural numbers 𝔼 = {2, 4, 6, ..}.
>
> f: ℕ → 𝔼 , n ↦ 2n
>
> This function is both injective (or one-to-one) and surjective (or
> onto), thus it is bijective.

If so, that would result in: The set 𝔼 has not one element more nor
less than the set ℕ.
>
>> Then the completion of 𝔼 resulting in E = {1, 2, 3, 4, 5, 6, ...}
>> would double the number of its elements. Then there are more natural
>> numbers than were originally in ℕ.
>
> Rubbish. The cardinality of an infinite set is described by an
> transfinite cardinal number and not by a finite "number of elements".

Here we do not use the rubbish of cardinality but the definition of
bijection proving that one of both sets has not one element more or less
than the other!

> Your problem is: You try to apply facts,

I apply logic which is universally valid.

If the set 𝔼 = {2, 4, 6, ..} has not one element more or less than the
set ℕ = {1, 2, 3, ...}, then adding an element to 𝔼 destroys this
state.

> that hold for finite sets,
> on infinite sets. That doesn't work.

Your problem is you deny logic which is universally valid.

Regards, WM

Le 24/03/2024 à 22:09, Richard Damon a écrit :
> On 3/24/24 4:11 PM, WM wrote:
>>
>> Does ℕ = {1, 2, 3, ...} contain all natural numbers such that none can
>> be added?
>>
>> If so, then the bijection of ℕ with E = {2, 4, 6, ...} would prove that
>> both sets have the same number of elements. Then the completion of E
>> resulting in E = {1, 2, 3, 4, 5, 6, ...} would double the number of its
>> elements. Then there are more natural numbers than were originally in ℕ.
>>
>
> Yep, because "infinity" just doesn't obey the logic you are used to

Then do not talk about a bijection. A bijection between thwo sets proves
that one of both sets has not one element more nor less than the other!
>
> Turns out that aleph0 = aleph0 - 1, = aleph0 + 1 = aleph0 / 2 =
> aleph0 * 2 = aleph0 ^ 2 = square root (aleph0)

This shows that card is in contradiction with basic logic.
>
> despite that breaking most of you concepts of the logic of numbers.

It breaks the definition of bijection. Note: A bijection between thwo sets
proves that one of both sets has not one element more nor less than the
other!

Regards, WM

Le 24/03/2024 à 22:10, FromTheRafters a écrit :
> WM explained :
>> Does ℕ = {1, 2, 3, ...} contain all natural numbers such that none can be
>> added?
>>
>> If so, then the bijection of ℕ with E = {2, 4, 6, ...} would prove that both
>> sets have the same number of elements.
>
> Actually, same size set. "Number of elements" is better suited to
> finite sets.

Insted of number of elements say: A bijection between thwo sets proves
that one of both sets has not one element more nor less than the other!
>
>> Then the completion of E resulting in
>> E = {1, 2, 3, 4, 5, 6, ...} would double the number of its elements.
>
> That is not a completion of E. But still the same size set.

It has more elements than before, namely all odd numbers.

> By your
> sense of 'complete' the set of even numbers was already 'complete'
> because no more even numbers could be 'added'.

But the odd numbers could be added.
>
>> Then there are more natural numbers than were originally in ℕ.
>
> Nope.

Try logic: If E = {2, 4, 6, ...} has as many (not more and not less)
elements as ℕ = {1, 2, 3, ...}, then addition of natural numbers yields
more natural numbers than are in ℕ = {1, 2, 3, ...}.

Regards, WM

Le 24/03/2024 à 22:33, "Chris M. Thomasson" a écrit :
> On 3/24/2024 1:20 PM, WM wrote:
>> Le 24/03/2024 à 21:16, "Chris M. Thomasson" a écrit :
>>> On 3/24/2024 1:13 PM, WM wrote:
>>>> Le 24/03/2024 à 21:11, "Chris M. Thomasson" a écrit :
>>>>
>>>>> adding one to infinity = infinity.
>>>>
>>>> If ℕ is complete no natural number can be added.
>>>
>>> Sigh. You are misunderstanding infinity... I think wrt infinity, when
>>> you hear the word "complete", your mind instantly thinks, "finite".
>>
>> Can you add a natural number to the set of all natura, numbers?
>
> ℕ is not finite!

But it is complete. Sets are complete in ZF.
>
>>> Infinity is not finite...
>>
>> But logic has to be observed.
>
> Right. So any natural number you give me, I can say that plus one.

And that is also already in ℕ.

Regards, WM

On 3/25/24 7:10 AM, WM wrote:
> Le 24/03/2024 à 22:02, Dieter Heidorn a écrit :
>> WM schrieb:
>>>
>>> Does ℕ = {1, 2, 3, ...} contain all natural numbers such that none
>>> can be added?
>>>
>> Sure
>
> Of course.
>>
>> But you can do the following described by Cantor:
>
> Irrelevant. Important is only this: You cannot add a natural number to
> the set ℕ.
>>
>>> If so, then the bijection of ℕ with E = {2, 4, 6, ...} would prove
>>> that both sets have the same number of elements.
>>
>> Infinite sets don't have a "number of elements".
>
> This cannot be denied: A bijection, if really existing, proves that one
> of both sets has not one element more nor less than the other!

Nope, just that they have the "same size", since for infinite sets, one
more or less is still the "same size", even twice the number of elements
can be the "same size", or the square of the number of elements is the
"same size"

That is just a property of infinite sets,

>>
>> And indeed: there is a bijection from the set of natural numbers ℕ
>> to the set of even natural numbers 𝔼 = {2, 4, 6, ..}.
>>
>>     f: ℕ → 𝔼 ,  n ↦ 2n
>>
>> This function is both injective (or one-to-one) and surjective (or
>> onto), thus it is bijective.
>
> If so, that would result in: The set 𝔼 has not one element more nor less
> than the set ℕ.

Nope, you are just AGAIN using the wrong definition of "same size"
because of your mind being stuck in "finite thinking".

>>
>>> Then the completion of 𝔼 resulting in E = {1, 2, 3, 4, 5, 6, ...}
>>> would double the number of its elements. Then there are more natural
>>> numbers than were originally in ℕ.
>>
>> Rubbish. The cardinality of an infinite set is described by an
>> transfinite cardinal number and not by a finite "number of elements".
>
> Here we do not use the rubbish of cardinality but the definition of
> bijection proving that one of both sets has not one element more or less
> than the other!

Nope, it proves thay have the "same size". "Number of Elements" is not a
defined term for infinite sets, in the same way it is defined for finite
sets.

You are just stuck in your finite thinking.

>
>> Your problem is: You try to apply facts,
>
> I apply logic which is universally valid.

Nope, you think it is, so you break it.

>
> If the set 𝔼 = {2, 4, 6, ..} has not one element more or less than the
> set ℕ = {1, 2, 3, ...}, then adding an element to 𝔼 destroys this state.

Nope.

>
>> that hold for finite sets,
>> on infinite sets. That doesn't work.
>
> Your problem is you deny logic which is universally valid.

Nope, your problem is you THINK your logic is universally valid, which
makes it go BOOM and destroys itself.

>
> Regards, WM
>
>

On 3/25/24 7:15 AM, WM wrote:
> Le 24/03/2024 à 22:09, Richard Damon a écrit :
>> On 3/24/24 4:11 PM, WM wrote:
>>>
>>> Does ℕ = {1, 2, 3, ...} contain all natural numbers such that none
>>> can be added?
>>>
>>> If so, then the bijection of ℕ with E = {2, 4, 6, ...} would prove
>>> that both sets have the same number of elements. Then the completion
>>> of E resulting in E = {1, 2, 3, 4, 5, 6, ...} would double the number
>>> of its elements. Then there are more natural numbers than were
>>> originally in ℕ.
>>>
>>
>> Yep, because "infinity" just doesn't obey the logic you are used to
>
> Then do not talk about a bijection. A bijection between thwo sets proves
> that one of both sets has not one element more nor less than the other!

Nope, it proves they have the same size.

For FINITE sets (which is all your logic handles) that is the same
"number" of elements, but not for infinite sets.

>>
>> Turns out that aleph0 = aleph0 - 1, = aleph0 + 1 = aleph0 / 2 =
>>     aleph0 * 2 = aleph0 ^ 2 = square root (aleph0)
>
> This shows that card is in contradiction with basic logic.

Nope, it shows that YOUR logic is not UNIVERSALLY VALID as you try to claim.

>>
>> despite that breaking most of you concepts of the logic of numbers.
>
> It breaks the definition of bijection. Note: A bijection between thwo
> sets proves that one of both sets has not one element more nor less than
> the other!
>

Nope, the definition of Bijection shows that the two sets have the same
size. "Size" of an infinite set isn't a normal "number" if the set is
"infinite".

> Regards, WM
>
>

on 3/25/2024, WM supposed :
> Le 24/03/2024 à 22:10, FromTheRafters a écrit :
>> WM explained :
>>> Does ℕ = {1, 2, 3, ...} contain all natural numbers such that none can be
>>> added?
>>>
>>> If so, then the bijection of ℕ with E = {2, 4, 6, ...} would prove that
>>> both sets have the same number of elements.
>>
>> Actually, same size set. "Number of elements" is better suited to finite
>> sets.
>
> Insted of number of elements say: A bijection between thwo sets proves that
> one of both sets has not one element more nor less than the other!

No, it shows that two sets are the same size. For such finite sets, it
also shows they have the same number of elements.

>>> Then the completion of E resulting in E = {1, 2, 3, 4, 5, 6, ...} would
>>> double the number of its elements.
>>
>> That is not a completion of E. But still the same size set.
>
> It has more elements than before, namely all odd numbers.

SETS DON'T CHANGE!!!

>> By your sense of 'complete' the set of even numbers was already 'complete'
>> because no more even numbers could be 'added'.
>
> But the odd numbers could be added.

SETS DON'T CHANGE!!!

You can however make a *NEW* set.

>>> Then there are more natural numbers than were originally in ℕ.
>>
>> Nope.
>
> Try logic: If E = {2, 4, 6, ...} has as many (not more and not less) elements
> as ℕ = {1, 2, 3, ...}, then addition of natural numbers yields more natural
> numbers than are in ℕ = {1, 2, 3, ...}.

SETS DON'T CHANGE!!!

On 3/25/2024 7:10 AM, WM wrote:
> Le 24/03/2024 à 22:02, Dieter Heidorn a écrit :

>> [...]
>
> Important is only this:
> You cannot add a natural number to the set ℕ.

not.exists finiteⁿᵒᵗᐧᵂᴹ ordinal n not.in ℕ

not.exists finiteⁿᵒᵗᐧᵂᴹ ordinal n such that
⟦0,n⟧ not.fits ℕ
== not.exists 1.to.1.map ⟦0,n⟧ ⇉ ℕ

for each finiteⁿᵒᵗᐧᵂᴹ set S
exists finiteⁿᵒᵗᐧᵂᴹ ordinal n such that
⟦0,n⟧ not.fits S
== not.exists 1.to.1.map ⟦0,n⟧ ⇉ S

ℕ is not finiteⁿᵒᵗᐧᵂᴹ.

for each finiteⁿᵒᵗᐧᵂᴹ set S
not.exists 1.to.1.map S∪{Bob} > S

ℕ is not finiteⁿᵒᵗᐧᵂᴹ

for not.finiteⁿᵒᵗᐧᵂᴹ ℕ
exists 1.to.1.map ℕ∪{Bob} ⇉ ℕ

You can add Bob to ℕ

Le 25/03/2024 à 12:25, Richard Damon a écrit :
> On 3/25/24 7:10 AM, WM wrote:
>> This cannot be denied: A bijection, if really existing, proves that one
>> of both sets has not one element more nor less than the other!
>
> Nope,

You deny the definition of what you use.

> just that they have the "same size", since for infinite sets, one
> more or less is still the "same size", even twice the number of elements
> can be the "same size", or the square of the number of elements is the
> "same size"
>
> That is just a property of infinite sets,

It is not a property of a bijection.

Regards, WM

Le 25/03/2024 à 12:29, Richard Damon a écrit :
> On 3/25/24 7:15 AM, WM wrote:

>> Then do not talk about a bijection. A bijection between thwo sets proves
>> that one of both sets has not one element more nor less than the other!
>
> Nope, it proves they have the same size.

Cantor claims true bijections.
>
> Nope, it shows that YOUR logic is not UNIVERSALLY VALID as you try to claim.

My logic is universally valid, and your "logic" is simply rubbish.

> Nope, the definition of Bijection shows that the two sets have the same
> size.

Yes, if it exists. That means it is surjective and injective.

Regards, WM

Le 25/03/2024 à 19:31, FromTheRafters a écrit :
> on 3/25/2024, WM supposed :

>> Insted of number of elements say: A bijection between thwo sets proves that
>> one of both sets has not one element more nor less than the other!
>
> No, it shows that two sets are the same size.

by showing injectivity and surjectivity. That means one-to-one
correspondence: one of both sets has not one element more nor less than
the other!

>> It has more elements than before, namely all odd numbers.
>
> SETS DON'T CHANGE!!!

Exactly! so you have been taught shit - and you have swallowed it.

Regards, WM