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WM
2024-03-24 20:00:52 UTC
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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.
Chris M. Thomasson
2024-03-24 20:11:43 UTC
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Post by WM
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...



;^D

Something that has a border, yet is infinite in and of itself... The
Mandelbrot set. ;^) lol.
WM
2024-03-24 20:13:43 UTC
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Post by Chris M. Thomasson
adding one to infinity = infinity.
If ℕ is complete no natural number can be added.

Regards, WM
Chris M. Thomasson
2024-03-24 20:16:21 UTC
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Post by WM
Post by Chris M. Thomasson
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
WM
2024-03-24 20:20:40 UTC
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Post by Chris M. Thomasson
Post by WM
Post by Chris M. Thomasson
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?
Post by Chris M. Thomasson
Infinity is not finite...
But logic has to be observed.

Regards, WM
Chris M. Thomasson
2024-03-24 21:33:54 UTC
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Post by WM
Post by Chris M. Thomasson
Post by WM
Post by Chris M. Thomasson
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?
Post by WM
Post by Chris M. Thomasson
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... ;^)

[...]
WM
2024-03-25 11:23:07 UTC
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Post by Chris M. Thomasson
Post by WM
Post by Chris M. Thomasson
Post by WM
Post by Chris M. Thomasson
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.
Post by Chris M. Thomasson
Post by WM
Post by Chris M. Thomasson
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
Chris M. Thomasson
2024-03-26 03:40:07 UTC
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Post by WM
Post by Chris M. Thomasson
Post by WM
Post by Chris M. Thomasson
Post by WM
Post by Chris M. Thomasson
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.
I think you are confusing complete with finite. This is not true of ℕ.
Post by WM
Post by Chris M. Thomasson
Post by WM
Post by Chris M. Thomasson
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.
This is why ℕ + 1 = ℕ

It's infinite and unbounded. ℕ is just a symbol to represent it...
WM
2024-03-26 13:11:29 UTC
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Post by Chris M. Thomasson
Post by WM
Post by Chris M. Thomasson
Post by WM
Post by Chris M. Thomasson
Post by WM
Post by Chris M. Thomasson
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.
I think you are confusing complete with finite. This is not true of ℕ.
Try to think better. Complete means complete, i.e., no element of ℕ is
available to be added to ℕ.
Post by Chris M. Thomasson
Post by WM
Post by Chris M. Thomasson
Post by WM
Post by Chris M. Thomasson
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.
This is why ℕ + 1 = ℕ
No. ℕ + 1 is undefined.

Regards, WM
Chris M. Thomasson
2024-03-26 19:56:25 UTC
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Post by WM
Post by Chris M. Thomasson
Post by WM
Post by Chris M. Thomasson
Post by WM
Post by Chris M. Thomasson
Post by WM
Post by Chris M. Thomasson
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.
I think you are confusing complete with finite. This is not true of ℕ.
Try to think better. Complete means complete, i.e., no element of ℕ is
available to be added to ℕ.
infinity + 1 = infinity

[...]
WM
2024-03-27 13:50:19 UTC
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Post by Chris M. Thomasson
Post by WM
Try to think better. Complete means complete, i.e., no element of ℕ is
available to be added to ℕ.
infinity + 1 = infinity
But not the same set. What natnumber could be added to ℕ?

Regards, WM
Chris M. Thomasson
2024-03-27 20:49:22 UTC
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Post by WM
Post by Chris M. Thomasson
Post by WM
Try to think better. Complete means complete, i.e., no element of ℕ
is available to be added to ℕ.
infinity + 1 = infinity
But not the same set. What natnumber could be added to ℕ?
Any natural can be added to ℕ, for ℕ + ℕ = ℕ,

ℕ + any natural number = ℕ

?
WM
2024-03-28 19:53:57 UTC
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Post by Chris M. Thomasson
Post by WM
Post by Chris M. Thomasson
Post by WM
Try to think better. Complete means complete, i.e., no element of ℕ
is available to be added to ℕ.
infinity + 1 = infinity
But not the same set. What natnumber could be added to ℕ?
Any natural can be added to ℕ, for ℕ + ℕ = ℕ,
What natnumber could be added to ℕ such that the result is larger?

Regards, WM
Chris M. Thomasson
2024-03-28 20:31:52 UTC
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Post by WM
Post by Chris M. Thomasson
Post by WM
Post by Chris M. Thomasson
Post by WM
Try to think better. Complete means complete, i.e., no element of ℕ
is available to be added to ℕ.
infinity + 1 = infinity
But not the same set. What natnumber could be added to ℕ?
Any natural can be added to ℕ, for ℕ + ℕ = ℕ,
What natnumber could be added to ℕ such that the result is larger?
None. Adding a natural number to ℕ means that it was already there.
Therefore, it equals itself...
Chris M. Thomasson
2024-03-28 20:34:59 UTC
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Post by Chris M. Thomasson
Post by WM
Post by Chris M. Thomasson
Post by WM
Post by Chris M. Thomasson
Post by WM
Try to think better. Complete means complete, i.e., no element of
ℕ is available to be added to ℕ.
infinity + 1 = infinity
But not the same set. What natnumber could be added to ℕ?
Any natural can be added to ℕ, for ℕ + ℕ = ℕ,
What natnumber could be added to ℕ such that the result is larger?
None. Adding a natural number to ℕ means that it was already there.
Therefore, it equals itself...
Think of, give me a natural number A. B = A + 1, A and B are already in
ℕ. ℕ is all of them...
Chris M. Thomasson
2024-03-26 20:00:20 UTC
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Post by WM
Post by Chris M. Thomasson
Post by WM
Post by Chris M. Thomasson
Post by WM
Post by Chris M. Thomasson
Post by WM
Post by Chris M. Thomasson
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.
I think you are confusing complete with finite. This is not true of ℕ.
Try to think better. Complete means complete, i.e., no element of ℕ is
available to be added to ℕ.
ℕ is all the natural numbers, so adding one to it equals itself.

You keep confusing complete with finite.
Post by WM
Post by Chris M. Thomasson
Post by WM
Post by Chris M. Thomasson
Post by WM
Post by Chris M. Thomasson
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.
This is why ℕ + 1 = ℕ
No. ℕ + 1 is undefined.
Regards, WM
Chris M. Thomasson
2024-03-26 20:19:18 UTC
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On 3/26/2024 6:11 AM, WM wrote:
[...]

You should argue with an AI for fun here. Ask it the following question:

______________
So adding one to the set of natural numbers, equals the set of natural
numbers. Fair enough?
______________

Leave us alone with your nonsense. Sigh.
Chris M. Thomasson
2024-03-26 20:29:03 UTC
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Post by Chris M. Thomasson
[...]
______________
So adding one to the set of natural numbers, equals the set of natural
numbers. Fair enough?
______________
Leave us alone with your nonsense. Sigh.
Then ask it:
______________
Rather, we're saying that adding one more element to an infinite set
doesn't change its size; it's still infinite? Fair enough?
______________

;^D lol.
Chris M. Thomasson
2024-03-24 21:37:37 UTC
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Post by WM
Post by Chris M. Thomasson
Post by WM
Post by Chris M. Thomasson
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?
http://youtu.be/Tb75RjpvBIk

;^D
Post by WM
Post by Chris M. Thomasson
Infinity is not finite...
But logic has to be observed.
[...]
WM
2024-03-24 20:11:45 UTC
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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
Dieter Heidorn
2024-03-24 21:02:50 UTC
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Post by WM
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)
Post by WM
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.
Post by WM
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
2024-03-25 11:10:53 UTC
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Post by Dieter Heidorn
Post by WM
Does ℕ = {1, 2, 3, ...} contain all natural numbers such that none
can be added?
Sure
Of course.
Irrelevant. Important is only this: You cannot add a natural number to the
set ℕ.
Post by Dieter Heidorn
Post by WM
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!
Post by Dieter Heidorn
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.
Post by Dieter Heidorn
Post by WM
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!
Post by Dieter Heidorn
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.
Post by Dieter Heidorn
that hold for finite sets,
on infinite sets. That doesn't work.
Your problem is you deny logic which is universally valid.

Regards, WM
Richard Damon
2024-03-25 11:25:38 UTC
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Post by WM
Post by Dieter Heidorn
Post by WM
Does ℕ = {1, 2, 3, ...} contain all natural numbers such that none
can be added?
Sure
Of course.
Irrelevant. Important is only this: You cannot add a natural number to
the set ℕ.
Post by Dieter Heidorn
Post by WM
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,
Post by WM
Post by Dieter Heidorn
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".
Post by WM
Post by Dieter Heidorn
Post by WM
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.
Post by WM
Post by Dieter Heidorn
Your problem is: You try to apply facts,
I apply logic which is universally valid.
Nope, you think it is, so you break it.
Post by WM
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.
Post by WM
Post by Dieter Heidorn
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.
Post by WM
Regards, WM
WM
2024-03-25 21:45:58 UTC
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Post by Richard Damon
Post by WM
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.
Post by Richard Damon
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
Jim Burns
2024-03-25 19:38:01 UTC
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Permalink
[...]
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
WM
2024-03-25 22:01:15 UTC
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Post by Jim Burns
[...]
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
WM
2024-03-25 22:11:27 UTC
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Post by Jim Burns
for not.finiteⁿᵒᵗᐧᵂᴹ ℕ
exists 1.to.1.map ℕ∪{Bob} ⇉ ℕ
A silly and wrong claim.

The infinite bijection ℕ <--> ℕ consists of infinitely many finite
bijections. At least one of them must be destroyed when introducing Bob.
Post by Jim Burns
You can add Bob to ℕ
Irrelevant. You cannot add a natural number to ℕ. But a bijection of ℕ
with |E = {2, 4, 6, ...} would prove that both sets have the same number
of elements. Adding an element to |E destroys this state and shows ℕ is
larger than ℕ. Contradiction!

Regards, WM
Jim Burns
2024-03-26 00:04:30 UTC
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Post by WM
[...]
You cannot add a natural number to ℕ.
You cannot add a natural number to ℕ ==
ℕ holds all sizes of sets for which
changing by one element changes the set's size.

ℕ doesn't have any of those sizes.

ℕ isn't a set for which
changing by one element changes the set's size.
Post by WM
But a bijection of ℕ with |E = {2, 4, 6, ...}
would prove that both sets have
the same number of elements.
ℕ and 𝔼 aren't sets for which
changing by one element changes the set's size.
Post by WM
Adding an element to |E destroys this state
1 ⟼ 2
n ⟼ n+2
𝔼∪{1} ⇉ 𝔼
Post by WM
and shows ℕ is larger than ℕ.
Contradiction!
| CAESAR (recovering his self-possession):
| Pardon him, Theodotus:
| he is a barbarian, and thinks that
| the customs of his tribe and island are
| the laws of nature.
|
George Bernard Shaw, "Caesar and Cleopatra" (1898)
Chris M. Thomasson
2024-03-26 04:43:28 UTC
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Post by Jim Burns
Post by WM
[...]
You cannot add a natural number to ℕ.
You cannot add a natural number to ℕ ==
ℕ holds all sizes of sets for which
changing by one element changes the set's size.
The set of odd numbers is infinite. There are an infinite number of
natural numbers. infinity = infinity. Density is another matter.

.01->.001 can represent infinity... Its denser, so to speak.
Post by Jim Burns
ℕ doesn't have any of those sizes.
ℕ isn't a set for which
changing by one element changes the set's size.
Post by WM
But a bijection of ℕ with |E = {2, 4, 6, ...}
would prove that both sets have
the same number of elements.
ℕ and 𝔼 aren't sets for which
changing by one element changes the set's size.
Post by WM
Adding an element to |E destroys this state
1 ⟼ 2
n ⟼ n+2
𝔼∪{1} ⇉ 𝔼
Post by WM
and shows ℕ is larger than ℕ.
Contradiction!
| he is a barbarian, and thinks that
| the customs of his tribe and island are
| the laws of nature.
|
George Bernard Shaw, "Caesar and Cleopatra" (1898)
Jim Burns
2024-03-26 15:40:05 UTC
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Post by Chris M. Thomasson
Post by Jim Burns
Post by WM
You cannot add a natural number to ℕ.
You cannot add a natural number to ℕ ==
ℕ holds all sizes of sets for which
changing by one element changes the set's size.
The set of odd numbers is infinite.
There are an infinite number of natural numbers.
infinity = infinity.
Density is another matter.
I prefer
finity = finity, and
everything else = everything else.

"Everything else" sounds vague, which is my point.

"Finite" has clear, specific descriptions.
"Other than finite" has what?

It seems to me that
a specific role will have an associated infinity
-- though, even here, other infinities can be
shoehorned into or spackled onto.

It's complicated.
Post by Chris M. Thomasson
Density is another matter.
.01->.001 can represent infinity...
Its denser, so to speak.
Ah, well. It's complicated.

ℕ is discrete. ℚ is dense.
ℕ and ℚ have the same infinity.

ℝ has a larger infinity.
ℝ is denser than dense ℚ? I guess?

For any set S, its power set 𝒫(S) is larger.
Even for infinite sets.

|ℕ| = |ℚ| < |ℝ| ≤ |𝒫(ℚ)| < |𝒫(𝒫(ℚ))| < ...

𝒫(𝒫(𝒫(ℚ))) is even more denser than dense?
I suspect that this is not a fruitful path
which we're on.

Just spitballing, what if
graduating to a new infinity involves
a semi.arbitrary description inserted somewhere?
Like ??? in
𝒫(ℚ) ⊇ {S ⊆ ℚ: S is ??? } ∈ 𝒫(𝒫(ℚ))

Perhaps someone else has thought about this
and pointed out its strengths and weaknesses.
I confess that I haven't looked for them, yet.
Spitballing is easier.
WM
2024-03-27 13:38:47 UTC
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Post by Jim Burns
ℕ and ℚ have the same infinity.
Only if logic (every lossless exchange is lossless) is violated and
damaged, i.e., only in matheology.

Regards, WM
Jim Burns
2024-03-27 18:54:37 UTC
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Post by WM
Post by Jim Burns
ℕ and ℚ have the same infinity.
Only if
logic (every lossless exchange is lossless)
is violated and damaged, i.e.,
only in matheology.
ℕ and ℚᶠʳᵃᶜ are larger than each set which is
not the same size as any subset or superset.

Therefore,
ℕ and ℚᶠʳᵃᶜ are not sets which are
not the same size as any subset or superset.

k ∈ ℕ
k ⟼ iₖ/jₖ
sₖ = max{h: (h-1)(h-2/2 < k }
iₖ = k-(sₖ-1)(sₖ-2)/2
jₖ = sₖ-iₖ
iₖ/jₖ ∈ ℚᶠʳᵃᶜ

iₖ/jₖ ∈ ℚᶠʳᵃᶜ
iₖ/jₖ ⟼ kᵢₖⱼₖ
sᵢₖⱼₖ = iₖ+jₖ
kᵢₖⱼₖ = (sᵢₖⱼₖ-1)(sᵢₖⱼₖ-2)/2+iₖ
kᵢₖⱼₖ ∈ ℕ

kᵢₖⱼₖ = k

ℕ and ℚᶠʳᵃᶜ are the same size.
WM
2024-03-28 20:07:29 UTC
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Post by Jim Burns
Post by WM
Post by Jim Burns
ℕ and ℚ have the same infinity.
Only if
logic (every lossless exchange is lossless)
is violated and damaged, i.e.,
only in matheology.
ℕ and ℚᶠʳᵃᶜ are the same size.
ℕ and ℕ are the same size.
ℚ and ℚ are the same size.

Removing a proper fraction decreases ℚ but leaves it larger than ℕ.
When the size changes it cannot remain the same.

Showing a not bijection proves different sizes of sets.
Why is that more meaningful than Cantor's bijections?

Between infinite sets there cannot exist any mapping because most elements
are dark. But we can assume that very simple mappings like f(x) = x are
true even for dark elements.

Therefore between the rational numbers and the natural numbers f(n) = n/1
can be accepted, also f(n) = 1/n, but not f(n) = 2n.

Regards, WM


Regards, WM
Chris M. Thomasson
2024-03-28 20:33:26 UTC
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Post by WM
Post by Jim Burns
Post by WM
Post by Jim Burns
ℕ and ℚ have the same infinity.
Only if
logic (every lossless exchange is lossless)
is violated and damaged, i.e.,
only in matheology.
ℕ and ℚᶠʳᵃᶜ are the same size.
ℕ and ℕ are the same size.
ℚ and ℚ are the same size.
Removing a proper fraction decreases ℚ but leaves it larger than ℕ. When
the size changes it cannot remain the same.
[...]

How do you "remove" a number from ℕ? Then it would no longer be ℕ. See?
Richard Damon
2024-03-29 13:29:42 UTC
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Permalink
Post by WM
Post by Jim Burns
Post by WM
Post by Jim Burns
ℕ and ℚ have the same infinity.
Only if
logic (every lossless exchange is lossless)
is violated and damaged, i.e.,
only in matheology.
ℕ and ℚᶠʳᵃᶜ are the same size.
ℕ and ℕ are the same size.
ℚ and ℚ are the same size.
And, it can be shown that ℕ and ℚ are the same size.
Post by WM
Removing a proper fraction decreases ℚ but leaves it larger than ℕ. When
the size changes it cannot remain the same.
Nope, Removing a single element from an infinite set doesn't change its
size.
Post by WM
Showing a not bijection proves different sizes of sets.
Why is that more meaningful than Cantor's bijections?
Nope, showing NO BIJECTION CAN EXIST proves different sizes.

Having just a failed bijection shows nothing for infinite sets.
Post by WM
Between infinite sets there cannot exist any mapping because most
elements are dark. But we can assume that very simple mappings like f(x)
= x are true even for dark elements.
Only because you logic doesn't handle the infinite sets.

Your "Darkness" is just your logic system saying "No" to its misuse.
Post by WM
Therefore between the rational numbers and the natural numbers f(n) =
n/1 can be accepted, also f(n) = 1/n, but not f(n) = 2n.
Sure it can, you just don't understand it because you are using
imporoper logic for such a system.

Like trying to argue about colors in a black and white photo.
Post by WM
Regards, WM
Regards, WM
WM
2024-03-30 16:01:12 UTC
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Post by Richard Damon
And, it can be shown that ℕ and ℚ are the same size.
That is believed by some stupids who are too dense to understand logic.
The Os in the matrix
XOOO...
XOOO...
XOOO...
XOOO...
..
cannot disappear by exchange with Xs.
Post by Richard Damon
Nope, Removing a single element from an infinite set doesn't change its
size.
That is believed by some stupids who are too dense to understand logic.
Removing one element makes the set having one element less than before.

Regrads, WM
Richard Damon
2024-03-30 17:47:21 UTC
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Post by WM
Post by Richard Damon
And, it can be shown that ℕ and ℚ are the same size.
That is believed by some stupids who are too dense to understand logic.
The Os in the matrix
XOOO...
XOOO...
XOOO...
XOOO...
..
cannot disappear by exchange with Xs.
So?

That isn't the Bijection we are looking for.
Post by WM
Post by Richard Damon
Nope, Removing a single element from an infinite set doesn't change
its size.
That is believed by some stupids who are too dense to understand logic.
Removing one element makes the set having one element less than before.
Regrads, WM
Which doesn't affect the "size" of infinity.

DEFINITIONS you know.
Richard Damon
2024-03-28 00:52:42 UTC
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Post by WM
Post by Jim Burns
ℕ and ℚ have the same infinity.
Only if logic (every lossless exchange is lossless) is violated and
damaged, i.e., only in matheology.
Regards, WM
Since you can only validly and consistantly talk about infinite sets in
a matheology, that seems right.

YOUR logic goes BOOM when it touches infinite sets, but since you got
blown up with it, you don't seem to notice.
WM
2024-03-26 13:24:19 UTC
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Post by Jim Burns
ℕ and 𝔼 aren't sets for which
changing by one element changes the set's size.
Adding an element to one of the sets in bijection destroys this bijection.

Regards, WM
Jim Burns
2024-03-26 17:04:35 UTC
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Post by WM
Post by Jim Burns
ℕ and 𝔼 aren't sets for which
changing by one element changes the set's size.
Adding an element to one of the sets in bijection
destroys this bijection.
Adding an element "destroys" the set.
Sets do not change.

There are other bijections for other sets.

----
Back up a little.
Consider sets A and B with 1.to.1 f: A ⇉ B
A ∋ a′ and B ∋ b′

From f can be defined
1.to.1 f⤨: A\{a′} ⇉ B\{b′}

f is 1.to.1
f(a′) = b″
f(a″) = b′

Define
f⤨(a′) = b′
f⤨(a″) = b″
and otherwise f⤨(y) = f(y)

f⤨ is also 1.to.1
and A\{a′} doesn't map to {b′}

1.to.1 f⤨: A\{a′} ⇉ B\{b′}
for
f⤨(f⁻¹(b′)) = f(a′)
and otherwise f⤨(y) = f(y)

----
There is a 1.to.1.map ℕ ⇉ ℕ\{1}
Therefore,
there is a 1.to.1.map ℕ\{1} ⇉ ℕ\{1,2}
And ℕ\{1,2} ⇉ ℕ\{1,2,3}
Et cetera, ad infinitum.

ℕ ⇉ ℕ\{1} ⇉ ℕ\{1,2} ⇉ ℕ\{1,2,3} ⇉ ...

An end not.exists.
Visibleᵂᴹ or darkᵂᴹ, an end not.exists.
ℕ is infiniteⁿᵒᵗᐧᵂᴹ.
WM
2024-03-27 13:46:03 UTC
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Post by Jim Burns
Post by WM
Post by Jim Burns
ℕ and 𝔼 aren't sets for which
changing by one element changes the set's size.
Adding an element to one of the sets in bijection
destroys this bijection.
Adding an element "destroys" the set.
Sets do not change.
It makes another set.
Post by Jim Burns
There are other bijections for other sets.
Bijections that tolerate additional elements cannot be complete. They are
only potentially infinite. All + 1 is more than all.
Post by Jim Burns
And ℕ\{1,2} ⇉ ℕ\{1,2,3}
Et cetera, ad infinitum.
When infinitely many numbers have been subtracted, then only finitely many
can remain. There are not two consecutive infinite sets in ℕ possible.

Regards, WM
Jim Burns
2024-03-27 19:10:46 UTC
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Post by WM
Post by Jim Burns
ℕ and 𝔼 aren't sets for which
changing by one element changes the set's size.
All + 1 is more than all.
There are not two
consecutive infinite sets in ℕ possible.
Infiniteᵂᴹ is only
really.really.humongously finiteⁿᵒᵗᐧᵂᴹ

ℕ can't be any size such that
size+1 > size
because
N holds a finiteⁿᵒᵗᐧᵂᴹ ordinal larger than that size.

Therefore,
size(ℕ)+1 = size(ℕ)
Chris M. Thomasson
2024-03-27 20:57:53 UTC
Reply
Permalink
Post by Jim Burns
Post by WM
Post by Jim Burns
ℕ and 𝔼 aren't sets for which
changing by one element changes the set's size.
All + 1 is more than all.
There are not two
consecutive infinite sets in ℕ possible.
Infiniteᵂᴹ is only
really.really.humongously finiteⁿᵒᵗᐧᵂᴹ
ℕ can't be any size such that
size+1 > size
because
N holds a finiteⁿᵒᵗᐧᵂᴹ ordinal larger than that size.
Therefore,
size(ℕ)+1 = size(ℕ)
ℕ + 1 = ℕ

So adding one to the set of all natural numbers means that number was
already there, therefore it equals itself?
Jim Burns
2024-03-27 22:50:28 UTC
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Post by Chris M. Thomasson
Post by Jim Burns
Post by WM
Post by Jim Burns
ℕ and 𝔼 aren't sets for which
changing by one element changes the set's size.
All + 1 is more than all.
There are not two
consecutive infinite sets in ℕ possible.
Infiniteᵂᴹ is only
really.really.humongously finiteⁿᵒᵗᐧᵂᴹ
ℕ can't be any size such that
size+1 > size
because
N holds a finiteⁿᵒᵗᐧᵂᴹ ordinal larger than that size.
Therefore,
size(ℕ)+1 = size(ℕ)
ℕ + 1 = ℕ
|ℕ| + 1 = |ℕ|
Post by Chris M. Thomasson
So
adding one to the set of all natural numbers
means
that number was already there,
therefore it equals itself?
The set of all natural number being
the set of all natural numbers
means that
that natural number was already there.

Inserting NaN into ℕ yields ℕ∪{NaN}

|ℕ∪{NaN}| ≤ |ℕ|

...which means
1.to.1.map f: ℕ∪{NaN} ⇉ ℕ exists

...which it does
f(NaN) = 0
f(j) = j+1 otherwise

Also
|ℕ| ≤ |ℕ∪{NaN}|

f⁻¹: ℕ ⇉ ℕ∪{NaN}
f⁻¹(0) = NaN
f⁻¹(k) = k-1

Thus,
|ℕ∪{NaN}| = |ℕ|

Define
|ℕ|+1 := |ℕ∪{NaN}|


|ℕ|+1 = |ℕ|
Richard Damon
2024-03-27 01:19:07 UTC
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Post by WM
Post by Jim Burns
ℕ and 𝔼 aren't sets for which
changing by one element changes the set's size.
Adding an element to one of the sets in bijection destroys this bijection.
Regards, WM
But creates another that shows that they are still the same size.

The existance of ANY bijection between the sets, proves them to be of
the same size.

Of course, you can't understand this, because none of it make sense in
your finite logic that blew itself up when you pushed it past what it
could handle.
WM
2024-03-27 13:57:37 UTC
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Post by Richard Damon
Post by WM
Post by Jim Burns
ℕ and 𝔼 aren't sets for which
changing by one element changes the set's size.
Adding an element to one of the sets in bijection destroys this bijection.
But creates another that shows that they are still the same size.
Only if the bijection was between potentially infinite sets.
Post by Richard Damon
The existance of ANY bijection between the sets, proves them to be of
the same size.
The existence of any injective but not surjective mapping shows that the
sets are not of same size. The existence of any surjective but not
injective mapping shows that the sets are not of same size.
Post by Richard Damon
Of course, you can't understand this,
I understand that your claims can only be satisfied by potential infinity.

Regards, WM
Richard Damon
2024-03-28 00:57:07 UTC
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Post by WM
Post by Richard Damon
Post by WM
Post by Jim Burns
ℕ and 𝔼 aren't sets for which
changing by one element changes the set's size.
Adding an element to one of the sets in bijection destroys this bijection.
But creates another that shows that they are still the same size.
Only if the bijection was between potentially infinite sets.
Which N and E are/
Post by WM
Post by Richard Damon
The existance of ANY bijection between the sets, proves them to be of
the same size.
The existence of any injective but not surjective mapping shows that the
sets are not of same size. The existence of any surjective but not
injective mapping shows that the sets are not of same size.
Right, but you need to prove that those really don't exist, but
Post by WM
Post by Richard Damon
Of course, you can't understand this,
I understand that your claims can only be satisfied by potential infinity.
Regards, WM
That is all your logic seems to allow, but then it can't actually handle
those infinite sets.

We are talking about ACTUAL infinite sets, not just "potential", but
perhaps your definition are that squirely,

All the elements of N are finite numbers, so none are actually infinite,
but their values are unbounded, and thus don't have a maximum.

That might be what you are calling "potential infinity"

The set N itself, is infinite in size, since no finite number can
describe it.

These "Countable Infinite" sets are on this strange boundry between
finite and infinite that cause a lot of "common sense" to break.
WM
2024-03-28 20:00:19 UTC
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Post by Richard Damon
Post by WM
Only if the bijection was between potentially infinite sets.
*********************
Post by Richard Damon
Which N and E are/
*********************
Post by Richard Damon
Post by WM
I understand that your claims can only be satisfied by potential infinity.
*******************************************************************
Post by Richard Damon
We are talking about ACTUAL infinite sets, not just "potential",
*******************************************************************

Regards, WM
Richard Damon
2024-03-29 13:31:37 UTC
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Post by WM
Post by Richard Damon
Post by WM
Only if the bijection was between potentially infinite sets.
*********************
Post by Richard Damon
Which N and E are/
*********************
No, N and E are ACTUALLY infinite sets of potentially infinite numbers
(if I am understanding the meaning of your not well defined terms)
Post by WM
Post by Richard Damon
Post by WM
I understand that your claims can only be satisfied by potential infinity.
*******************************************************************
Post by Richard Damon
We are talking about ACTUAL infinite sets, not just "potential",
*******************************************************************
Right, so FINITE logic doesn't apply to them.
Post by WM
Regards, WM
Richard Damon
2024-03-24 21:09:24 UTC
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Post by WM
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.
WM
2024-03-25 11:15:24 UTC
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Post by Richard Damon
Post by WM
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!
Post by Richard Damon
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.
Post by Richard Damon
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
Richard Damon
2024-03-25 11:29:19 UTC
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Post by WM
Post by Richard Damon
Post by WM
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.
Post by WM
Post by Richard Damon
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.
Post by WM
Post by Richard Damon
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".
Post by WM
Regards, WM
WM
2024-03-25 21:49:33 UTC
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Post by Richard Damon
Post by WM
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.
Post by Richard Damon
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.
Post by Richard Damon
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
FromTheRafters
2024-03-24 21:10:34 UTC
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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.
WM
2024-03-25 11:20:44 UTC
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Post by FromTheRafters
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!
Post by FromTheRafters
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.
Post by FromTheRafters
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.
Post by FromTheRafters
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
FromTheRafters
2024-03-25 18:31:26 UTC
Reply
Permalink
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!!!
WM
2024-03-25 21:53:59 UTC
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Post by FromTheRafters
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!
Post by FromTheRafters
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
FromTheRafters
2024-03-25 22:49:19 UTC
Reply
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Post by WM
Post by FromTheRafters
Post by WM
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!
For finite sets.
Post by WM
Post by FromTheRafters
Post by WM
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.
Wrong again.
WM
2024-03-26 13:18:38 UTC
Reply
Permalink
Post by FromTheRafters
Post by WM
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!
For finite sets.
For all bijections which deserve this name.

Regards, WM
FromTheRafters
2024-03-26 13:39:50 UTC
Reply
Permalink
Post by WM
Post by FromTheRafters
Post by WM
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!
For finite sets.
For all bijections which deserve this name.
What name?
WM
2024-03-26 13:47:34 UTC
Reply
Permalink
Post by FromTheRafters
Post by WM
Post by FromTheRafters
Post by WM
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!
For finite sets.
For all bijections which deserve this name.
What name?
The name bijection.

Regards, WM
FromTheRafters
2024-03-26 18:04:34 UTC
Reply
Permalink
Post by WM
Post by FromTheRafters
Post by WM
Post by FromTheRafters
Post by WM
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!
For finite sets.
For all bijections which deserve this name.
What name?
The name bijection.
All bijections deserve the name bijection.
Phil Carmody
2024-03-26 21:12:30 UTC
Reply
Permalink
Post by FromTheRafters
Post by WM
Post by FromTheRafters
Post by WM
Post by FromTheRafters
Post by WM
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!
For finite sets.
For all bijections which deserve this name.
What name?
The name bijection.
All bijections deserve the name bijection.
Even the ones that are just bicuriousjections?

Phil
--
We are no longer hunters and nomads. No longer awed and frightened, as we have
gained some understanding of the world in which we live. As such, we can cast
aside childish remnants from the dawn of our civilization.
-- NotSanguine on SoylentNews, after Eugen Weber in /The Western Tradition/
FromTheRafters
2024-03-26 21:40:10 UTC
Reply
Permalink
Post by Phil Carmody
Post by FromTheRafters
Post by WM
Post by FromTheRafters
Post by WM
Post by FromTheRafters
Post by WM
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!
For finite sets.
For all bijections which deserve this name.
What name?
The name bijection.
All bijections deserve the name bijection.
Even the ones that are just bicuriousjections?
Phil
Yes. Math is an equivalence class opportunity employer. :D
Chris M. Thomasson
2024-03-27 02:43:16 UTC
Reply
Permalink
Post by FromTheRafters
Post by Phil Carmody
Post by FromTheRafters
Post by WM
Post by FromTheRafters
Post by WM
Post by FromTheRafters
Post by WM
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!
For finite sets.
For all bijections which deserve this name.
What name?
The name bijection.
All bijections deserve the name bijection.
Even the ones that are just bicuriousjections?
lol.
Post by FromTheRafters
Post by Phil Carmody
Phil
Yes. Math is an equivalence class opportunity employer. :D
ROFL!!!!
Phil Carmody
2024-03-27 11:35:45 UTC
Reply
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Post by FromTheRafters
Post by Phil Carmody
Post by FromTheRafters
Post by WM
Post by FromTheRafters
Post by WM
Post by FromTheRafters
Post by WM
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!
For finite sets.
For all bijections which deserve this name.
What name?
The name bijection.
All bijections deserve the name bijection.
Even the ones that are just bicuriousjections?
Yes. Math is an equivalence class opportunity employer. :D
As it should be - I marched for homeomorphic rights when I was younger.

Phil
--
We are no longer hunters and nomads. No longer awed and frightened, as we have
gained some understanding of the world in which we live. As such, we can cast
aside childish remnants from the dawn of our civilization.
-- NotSanguine on SoylentNews, after Eugen Weber in /The Western Tradition/
WM
2024-03-27 13:48:29 UTC
Reply
Permalink
Post by FromTheRafters
All bijections deserve the name bijection.
Not those which violate logic in that lossless exchange between X and O
deletes O.

Regards, WM
FromTheRafters
2024-03-27 15:23:40 UTC
Reply
Permalink
Post by WM
Post by FromTheRafters
All bijections deserve the name bijection.
Not those which violate logic in that lossless exchange between X and O
deletes O.
Not showing a bijection is not the same as not having a bijection.
WM
2024-03-27 21:50:46 UTC
Reply
Permalink
Post by FromTheRafters
Post by WM
Post by FromTheRafters
All bijections deserve the name bijection.
Not those which violate logic in that lossless exchange between X and O
deletes O.
Not showing a bijection is not the same as not having a bijection.
Showing a not bijection proves different sizes of sets.
Why is that more meaningful than Cantor's bijections?

Between infinite sets there cannot exist any mapping because most elements
are dark. But we can assume that very simple mappings like f(x) = x are
true even for dark elements.

Therefore between the rational numbers and the natural numbers f(n) = n/1
can be accepted, also f(n) = 1/n, but not f(n) = 2n.

Regards, WM
FromTheRafters
2024-03-27 23:37:40 UTC
Reply
Permalink
Post by WM
Post by FromTheRafters
Post by WM
Post by FromTheRafters
All bijections deserve the name bijection.
Not those which violate logic in that lossless exchange between X and O
deletes O.
Not showing a bijection is not the same as not having a bijection.
Showing a not bijection proves different sizes of sets.
No it doesn't.
WM
2024-03-28 19:55:33 UTC
Reply
Permalink
Post by FromTheRafters
Post by WM
Showing a not bijection proves different sizes of sets.
No it doesn't.
Why not?

Regards, WM
FromTheRafters
2024-03-28 22:11:55 UTC
Reply
Permalink
Post by WM
Post by FromTheRafters
Post by WM
Showing a not bijection proves different sizes of sets.
No it doesn't.
Why not?
Because not all sets are finite and failing to show a bijection is not
the same as showing that a bijection is an impossiblity.
Richard Damon
2024-03-29 17:05:36 UTC
Reply
Permalink
Post by WM
Post by FromTheRafters
Post by WM
Showing a not bijection proves different sizes of sets.
No it doesn't.
Why not?
Regards, WM
Because it is showable that you CAN make non-working attempts at
bijections on infinite sets that you can also show working bijections,
so clearly, failed bijections do not show that infinite sets are of a
different size.

That is just one of the interesting properties of infinity and
transfinite math.

Some things that might seem "obvious" in the finite world, just don't work.
WM
2024-03-30 16:05:08 UTC
Reply
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Post by Richard Damon
Post by WM
Post by FromTheRafters
Post by WM
Showing a not bijection proves different sizes of sets.
No it doesn't.
Why not?
Because it is showable that you CAN make non-working attempts at
bijections on infinite sets that you can also show working bijections,
Only if the Os in the matrix
XOOO...
XOOO...
XOOO...
XOOO...
..
could disappear by exchange with Xs which is ipossible.

Regards, WM
Richard Damon
2024-03-30 17:48:57 UTC
Reply
Permalink
Post by WM
Post by Richard Damon
Post by WM
Post by FromTheRafters
Post by WM
Showing a not bijection proves different sizes of sets.
No it doesn't.
Why not?
Because it is showable that you CAN make non-working attempts at
bijections on infinite sets that you can also show working bijections,
Only if the Os in the matrix
XOOO...
XOOO...
XOOO...
XOOO...
..
could disappear by exchange with Xs which is ipossible.
Regards, WM
Failed bijections do not prove anything for infinite sets.

Since we CAN create a bijection between the sets:


XXXXXXXX....

and

OOO...
OOO...
OOO...
...

We can show they they are the same size, and that you are just ignorant.
Richard Damon
2024-03-28 00:59:13 UTC
Reply
Permalink
Post by WM
Post by FromTheRafters
Post by WM
Post by FromTheRafters
All bijections deserve the name bijection.
Not those which violate logic in that lossless exchange between X and
O deletes O.
Not showing a bijection is not the same as not having a bijection.
Showing a not bijection proves different sizes of sets.
Why is that more meaningful than Cantor's bijections?
Nope. Cantor shows that there may well be "failed" attempts at a
bijection that actually don't tell us anything about the sizes of two
infinite sets.
Post by WM
Between infinite sets there cannot exist any mapping because most
elements are dark. But we can assume that very simple mappings like f(x)
= x are true even for dark elements.
Nope, that just shows that you logic system just can't handle these systems.
Post by WM
Therefore between the rational numbers and the natural numbers f(n) =
n/1 can be accepted, also f(n) = 1/n, but not f(n) = 2n.
Nope. Just shows the limits of your logic.
Post by WM
Regards, WM
WM
2024-03-28 19:57:54 UTC
Reply
Permalink
Post by Richard Damon
Post by WM
Showing a not bijection proves different sizes of sets.
Why is that more meaningful than Cantor's bijections?
Nope. Cantor shows that there may well be "failed" attempts at a
bijection that actually don't tell us anything about the sizes of two
infinite sets.
Why do you think so?

Regards, WM
Richard Damon
2024-03-29 17:07:42 UTC
Reply
Permalink
Post by WM
Post by Richard Damon
Post by WM
Showing a not bijection proves different sizes of sets.
Why is that more meaningful than Cantor's bijections?
Nope. Cantor shows that there may well be "failed" attempts at a
bijection that actually don't tell us anything about the sizes of two
infinite sets.
Why do you think so?
Regards, WM
Because it is easy to come up with failed bijections for sets that a
succefful bijection exists.

Thus failed bijections for infinite sets do not prove different sizes.

This is just one of the interesting properties of infinite/transfinite
logic and math. Something you likely can't understand since you are
stuck in finite logic.
Jim Burns
2024-03-28 21:25:39 UTC
Reply
Permalink
Post by WM
Post by FromTheRafters
Not showing a bijection is not the same as
not having a bijection.
Showing a not bijection proves
different sizes of sets.
I'll call that a Mückenheim.set.

For each Mückenheim.set S,
for each set T,
if exists 1.to.1 f: S ⇉ T: f(S) ≠ T
then not.exists 1.to.1 g: S ⇉ T: g(S) = T
and S and T are different sizes.

ℕᴶᵛᴺ is the set of Mückenheim von.Neumann ordinals.
For each n ∈ ℕᴶᵛᴺ:
n = ⟦0,n⦆ ∧
∃f:⟦0,n⦆ ⇉ T: f⟦0,n⦆ ≠ T ⟹
¬∃g:⟦0,n⦆ ⇉ T: g⟦0,n⦆ = T

ℕᴶᵛᴺ is not a Mückenheim.set.

∃f:ℕᴶᵛᴺ ⇉ ℕᴶᵛᴺ: fℕᴶᵛᴺ ≠ ℕᴶᵛᴺ ∧
∃g:ℕᴶᵛᴺ ⇉ ℕᴶᵛᴺ: gℕᴶᵛᴺ = ℕᴶᵛᴺ

f(n) = n+1
g(n) = n

Mückenheim.sets are finiteⁿᵒᵗᐧᵂᴹ sets.
Post by WM
But we can assume that
very simple mappings like f(x) = x are true
even for dark elements.
Assume what you say.
Then there is no Mückenheim.set B which contains
a not.Mückenheim.set U as a subset.
Only not.Mückenheim.sets are supersets of
not.Mückenheim.sets.
Post by WM
But we can assume that
very simple mappings like f(x) = x are true
even for dark elements.
Therefore
between the rational numbers and the natural numbers
f(n) = n/1 can be accepted,
also f(n) = 1/n,
but not f(n) = 2n.
For k,m ∈ ℕᴶᵛᴺ define addition such that
k+m = nₖₘ ⟺
exists Mückenheim.3.tuple.sequence ⟨⟨k,0,k⟩,…,⟨k,m,nₖₘ⟩⟩
such that
⟨k,0,k⟩ ∈ ⟨⟨k,0,k⟩,…,⟨k,m,nₖₘ⟩⟩ ∧
⟨k,m,nₖₘ⟩ ∈ ⟨⟨k,0,k⟩,…,⟨k,m,nₖₘ⟩⟩ ∧
∀i ∈ m: ∃!⟨k,i,j⟩ ∈ ⟨⟨k,0,k⟩,…,⟨k,m,nₖₘ⟩⟩ ∧
∀⟨k,i,j⟩ ∈ ⟨⟨k,0,k⟩,…,⟨k,m,nₖₘ⟩⟩ ∋ ⟨k,i⁺¹,j⁺¹⟩

nₖₘ ∈ ℕᴶᵛᴺ

For k,m ∈ ℕᴶᵛᴺ define multiplication such that
k⋅m = n′ₖₘ ⟺
exists Mückenheim.3.tuple.sequence ⟨⟨k,0,0⟩,…,⟨k,m,n′ₖₘ⟩⟩
such that
⟨k,0,0⟩ ∈ ⟨⟨k,0,0⟩,…,⟨k,m,n′ₖₘ⟩⟩ ∧
⟨k,m,n′ₖₘ⟩ ∈ ⟨⟨k,0,0⟩,…,⟨k,m,n′ₖₘ⟩⟩ ∧
∀i ∈ m: ∃!⟨k,i,j⟩ ∈ ⟨⟨k,0,0⟩,…,⟨k,m,n′ₖₘ⟩⟩ ∧
∀⟨k,i,j⟩ ∈ ⟨⟨k,0,0⟩,…,⟨k,m,n′ₖₘ⟩⟩ ∋ ⟨k,i⁺¹,j+k⟩

n′ₖₘ ∈ ℕᴶᵛᴺ

For k ∈ ℕᴶᵛᴺ define fraction iₖ/jₖ such that
sₖ = max{h: (h-1)(h-2)/2 < k }
iₖ = k-(sₖ-1)(sₖ-2)/2
jₖ = sₖ-iₖ

iₖ,jₖ ∈ ℕᴶᵛᴺ

For i,j ∈ ℕᴶᵛᴺ define index kᵢⱼ such that
sᵢⱼ = i+j
kᵢⱼ = (sᵢⱼ-1)(sᵢⱼ)-2)/2+i

kᵢⱼ ∈ ℕᴶᵛᴺ

kᵢⱼ = k <-> <i,j> = <iₖ,jₖ>
Post by WM
But we can assume that
very simple mappings like f(x) = x are true
even for dark elements.
Therefore
between the rational numbers and the natural numbers
f(n) = n/1 can be accepted,
also f(n) = 1/n,
but not f(n) = 2n.
Where are darkᵂᴹ numbers introduced?
WM
2024-03-28 21:40:44 UTC
Reply
Permalink
Post by Jim Burns
Post by WM
Post by FromTheRafters
Not showing a bijection is not the same as
not having a bijection.
Showing a not bijection proves
different sizes of sets.
I'll call that a Mückenheim.set.
Mückenheim.sets are finiteⁿᵒᵗᐧᵂᴹ sets.
No. The set {1, 2, 3, 4, 5, ..., ω} is infinite, although there is a
last element.
Under f(x) = 2x we get the image {2, 4, 6, 8, 10, ..., 2ω]. The mapping
restricted to the natural numbers shows less evens than naturals.
Post by Jim Burns
Post by WM
But we can assume that
very simple mappings like f(x) = x are true
even for dark elements.
Therefore
between the rational numbers and the natural numbers
f(n) = n/1 can be accepted,
also f(n) = 1/n,
but not f(n) = 2n.
Where are darkᵂᴹ numbers introduced?
They are the places where Bob rests when the mapping is finished.

Regards, WM
Jim Burns
2024-03-28 22:30:04 UTC
Reply
Permalink
Post by Jim Burns
Post by WM
Post by FromTheRafters
Not showing a bijection is not the same as
not having a bijection.
Showing a not bijection proves
different sizes of sets.
I'll call that a Mückenheim.set.
Mückenheim.sets are finiteⁿᵒᵗᐧᵂᴹ sets.
No.
The set {1, 2, 3, 4, 5, ..., ω} is infinite,
although there is a last element.
The set {1, 2, 3, 4, 5, ..., ω} is not.Mückenheim.

Consider not.bijection
g: {1,2,3,4,5,...,ω} ⇉ {1,2,3,4,5,...,ω}
g(ω) = 1
g(n) = n+1 otherwise

{1,2,3,4,5,...,ω} is not a different size than
{1,2,3,4,5,...,ω}

finiteⁿᵒᵗᐧᵂᴹ does not mean what you think it means.
Under f(x) = 2x
we get the image {2, 4, 6, 8, 10, ..., 2ω].
The mapping restricted to the natural numbers shows
less evens than naturals.
s/less/fewer

No.
f(x) = 2x
not.bijection f: ℕ ⇉ ℕ
f(ℕ) = 𝔼 ≠ ℕ

If Mückenheim ℕ then |ℕ| ≠ |ℕ|

not.Mückenheim ℕ

One can't draw valid conclusions from
the assertion that ℕ is Mückenheim.
Post by Jim Burns
Post by WM
But we can assume that
very simple mappings like f(x) = x are true
even for dark elements.
Therefore
between the rational numbers and the natural numbers
f(n) = n/1 can be accepted,
also f(n) = 1/n,
but not f(n) = 2n.
Where are darkᵂᴹ numbers introduced?
They are the places where
Bob rests when the mapping is finished.
Where in these definitions
did I introduce darkᵂᴹ numbers?


For k,m ∈ ℕᴶᵛᴺ define addition such that
k+m = nₖₘ ⟺
exists Mückenheim.3.tuple.sequence ⟨⟨k,0,k⟩,…,⟨k,m,nₖₘ⟩⟩
such that
⟨k,0,k⟩ ∈ ⟨⟨k,0,k⟩,…,⟨k,m,nₖₘ⟩⟩ ∧
⟨k,m,nₖₘ⟩ ∈ ⟨⟨k,0,k⟩,…,⟨k,m,nₖₘ⟩⟩ ∧
∀i ∈ m: ∃!⟨k,i,j⟩ ∈ ⟨⟨k,0,k⟩,…,⟨k,m,nₖₘ⟩⟩ ∧
∀⟨k,i,j⟩ ∈ ⟨⟨k,0,k⟩,…,⟨k,m,nₖₘ⟩⟩ ∋ ⟨k,i⁺¹,j⁺¹⟩

lemma. nₖₘ ∈ ℕᴶᵛᴺ

For k,m ∈ ℕᴶᵛᴺ define multiplication such that
k⋅m = n′ₖₘ ⟺
exists Mückenheim.3.tuple.sequence ⟨⟨k,0,0⟩,…,⟨k,m,n′ₖₘ⟩⟩
such that
⟨k,0,0⟩ ∈ ⟨⟨k,0,0⟩,…,⟨k,m,n′ₖₘ⟩⟩ ∧
⟨k,m,n′ₖₘ⟩ ∈ ⟨⟨k,0,0⟩,…,⟨k,m,n′ₖₘ⟩⟩ ∧
∀i ∈ m: ∃!⟨k,i,j⟩ ∈ ⟨⟨k,0,0⟩,…,⟨k,m,n′ₖₘ⟩⟩ ∧
∀⟨k,i,j⟩ ∈ ⟨⟨k,0,0⟩,…,⟨k,m,n′ₖₘ⟩⟩ ∋ ⟨k,i⁺¹,j+k⟩

lemma. n′ₖₘ ∈ ℕᴶᵛᴺ

For k ∈ ℕᴶᵛᴺ define fraction iₖ/jₖ such that
sₖ = max{h: (h-1)(h-2)/2 < k }
iₖ = k-(sₖ-1)(sₖ-2)/2
jₖ = sₖ-iₖ

lemma. iₖ,jₖ ∈ ℕᴶᵛᴺ

For i,j ∈ ℕᴶᵛᴺ define index kᵢⱼ such that
sᵢⱼ = i+j
kᵢⱼ = (sᵢⱼ-1)(sᵢⱼ)-2)/2+i

lemma. kᵢⱼ ∈ ℕᴶᵛᴺ

lemma. kᵢⱼ = k ⟺ ⟨i,j⟩ = ⟨iₖ,jₖ⟩
Chris M. Thomasson
2024-03-29 00:54:46 UTC
Reply
Permalink
Post by Jim Burns
Post by WM
Post by FromTheRafters
Not showing a bijection is not the same as
not having a bijection.
Showing a not bijection proves
different sizes of sets.
I'll call that a Mückenheim.set.
Mückenheim.sets are finiteⁿᵒᵗᐧᵂᴹ sets.
No. The set  {1, 2, 3, 4, 5, ..., ω} is infinite, although there is a
last element.
Huh? If it has a last element its NOT infinite...
Under f(x) = 2x we get the image {2, 4, 6, 8, 10, ..., 2ω]. The mapping
restricted to the natural numbers shows less evens than naturals.
Post by Jim Burns
Post by WM
But we can assume that
very simple mappings like f(x) = x are true
even for dark elements.
Therefore
between the rational numbers and the natural numbers
f(n) = n/1 can be accepted,
also f(n) = 1/n,
but not f(n) = 2n.
Where are darkᵂᴹ numbers introduced?
They are the places where Bob rests when the mapping is finished.
Regards, WM
Chris M. Thomasson
2024-03-29 01:00:28 UTC
Reply
Permalink
Post by Chris M. Thomasson
Post by Jim Burns
Post by WM
Post by FromTheRafters
Not showing a bijection is not the same as
not having a bijection.
Showing a not bijection proves
different sizes of sets.
I'll call that a Mückenheim.set.
Mückenheim.sets are finiteⁿᵒᵗᐧᵂᴹ sets.
No. The set  {1, 2, 3, 4, 5, ..., ω} is infinite, although there is a
last element.
Huh? If it has a last element its NOT infinite...
Oh I missed something, sorry. Don't try to tell me that ω is your
infamous largest natural number... ?

Think of a line with points p0 and p1. Draw a line from p0 to p1. We
just covered an infinite number of points in between...
Post by Chris M. Thomasson
Under f(x) = 2x we get the image {2, 4, 6, 8, 10, ..., 2ω]. The
mapping restricted to the natural numbers shows less evens than naturals.
There are an infinite number of even naturals. There are an infinite
number of naturals. They both have an infinite number of items, so to speak.
Post by Chris M. Thomasson
Post by Jim Burns
Post by WM
But we can assume that
very simple mappings like f(x) = x are true
even for dark elements.
Therefore
between the rational numbers and the natural numbers
f(n) = n/1 can be accepted,
also f(n) = 1/n,
but not f(n) = 2n.
Where are darkᵂᴹ numbers introduced?
They are the places where Bob rests when the mapping is finished.
Regards, WM
Jim Burns
2024-03-29 09:46:58 UTC
Reply
Permalink
Post by Chris M. Thomasson
Post by Jim Burns
Post by WM
Post by FromTheRafters
Not showing a bijection is not the same as
not having a bijection.
Showing a not bijection proves
different sizes of sets.
I'll call that a Mückenheim.set.
Mückenheim.sets are finiteⁿᵒᵗᐧᵂᴹ sets.
No. The set {1, 2, 3, 4, 5, ..., ω} is infinite,
although there is a last element.
Huh?
If it has a last element its NOT infinite...
No,
the same set has transitive.trichotomous.orders
1 < 2 < 3 < 4 < ... < ω
and
ω <′ 1 <′ 2 <′ 3 <′ 4 <′ ...
and
... <″ 4 <″ 2 <″ ω <″ 1 <″ 3 <″ ...

The same set is infiniteⁿᵒᵗᐧᵂᴹ in each order.

What < <′ <″ have in common is
nonempty subsets without two ends.
1 < 2 < 3 < 4 < ...
2 < 3 < 4 < 5 < ...
3 < 4 < 5 < 6 < ...

ω <′ 1 <′ 2 <′ 3 <′ 4 <′ ...
1 <′ 2 <′ 3 <′ 4 <′ 5 <′ ...
2 <′ 3 <′ 4 <′ 5 <′ 6 <′ ...

... <″ 4 <″ 2 <″ ω <″ 1 <″ 3 <″ ...
ω <″ 1 <″ 3 <″ 5 <″ 7 <″ ...
... <″ 8 <″ 6 <″ 4 <″ 2 <″ ω

For an infiniteⁿᵒᵗᐧᵂᴹ set
_each transitive.trichotomous.order_
has nonempty non.two.ended subsets.

For a finiteⁿᵒᵗᐧᵂᴹ set
_each transitive.trichotomous.order_
does not have nonempty non.two.ended subsets.

No set exists
between finiteⁿᵒᵗᐧᵂᴹ and infiniteⁿᵒᵗᐧᵂᴹ with
some orders with non.two.ended subsets and
some orders without non.two.ended subsets.

For each set,
all orders are all.sub.two.ended or
no orders are all.sub.two.ended.
Chris M. Thomasson
2024-03-29 22:53:43 UTC
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Post by Jim Burns
Post by Chris M. Thomasson
Post by Jim Burns
Post by WM
Post by FromTheRafters
Not showing a bijection is not the same as
not having a bijection.
Showing a not bijection proves
different sizes of sets.
I'll call that a Mückenheim.set.
Mückenheim.sets are finiteⁿᵒᵗᐧᵂᴹ sets.
No. The set {1, 2, 3, 4, 5, ..., ω} is infinite,
although there is a last element.
Huh?
If it has a last element its NOT infinite...
No,
the same set has transitive.trichotomous.orders
[...]

Yeah. Think of two points p0, p1, draw a line, we just covered infinite
points even though it has a defined start and end.
Tom Bola
2024-03-29 14:13:34 UTC
Reply
Permalink
Post by Chris M. Thomasson
Post by Jim Burns
Post by WM
Post by FromTheRafters
Not showing a bijection is not the same as
not having a bijection.
Showing a not bijection proves
different sizes of sets.
I'll call that a Mückenheim.set.
Mückenheim.sets are finiteⁿᵒᵗᐧᵂᴹ sets.
No. The set  {1, 2, 3, 4, 5, ..., ω} is infinite, although there is a
last element.
Huh? If it has a last element its NOT infinite...
Oops, you might also write that this way: { ω, 1, 2, 3, 4, 5, ... }.
Chris M. Thomasson
2024-03-29 22:52:00 UTC
Reply
Permalink
Post by Tom Bola
Post by Chris M. Thomasson
Post by Jim Burns
Post by WM
Post by FromTheRafters
Not showing a bijection is not the same as
not having a bijection.
Showing a not bijection proves
different sizes of sets.
I'll call that a Mückenheim.set.
Mückenheim.sets are finiteⁿᵒᵗᐧᵂᴹ sets.
No. The set  {1, 2, 3, 4, 5, ..., ω} is infinite, although there is a
last element.
Huh? If it has a last element its NOT infinite...
Oops, you might also write that this way: { ω, 1, 2, 3, 4, 5, ... }.
I got confused. I thought that ω was WM's largest natural number, its
last element, so to speak.
Tom Bola
2024-03-29 23:29:44 UTC
Reply
Permalink
Post by Chris M. Thomasson
Post by Tom Bola
Post by Chris M. Thomasson
Post by Jim Burns
Post by WM
Post by FromTheRafters
Not showing a bijection is not the same as
not having a bijection.
Showing a not bijection proves
different sizes of sets.
I'll call that a Mückenheim.set.
Mückenheim.sets are finiteⁿᵒᵗᐧᵂᴹ sets.
No. The set  {1, 2, 3, 4, 5, ..., ω} is infinite, although there is a
last element.
Huh? If it has a last element its NOT infinite...
Oops, you might also write that this way: { ω, 1, 2, 3, 4, 5, ... }.
I got confused. I thought that ω was WM's largest natural number, its
last element, so to speak.
Sure...
Chris M. Thomasson
2024-03-29 23:31:52 UTC
Reply
Permalink
Post by Tom Bola
Post by Chris M. Thomasson
Post by Tom Bola
Post by Chris M. Thomasson
Post by Jim Burns
Post by WM
Post by FromTheRafters
Not showing a bijection is not the same as
not having a bijection.
Showing a not bijection proves
different sizes of sets.
I'll call that a Mückenheim.set.
Mückenheim.sets are finiteⁿᵒᵗᐧᵂᴹ sets.
No. The set  {1, 2, 3, 4, 5, ..., ω} is infinite, although there is a
last element.
Huh? If it has a last element its NOT infinite...
Oops, you might also write that this way: { ω, 1, 2, 3, 4, 5, ... }.
I got confused. I thought that ω was WM's largest natural number, its
last element, so to speak.
Sure...
I know for sure that WM thinks there is a largest natural number. That
must be his last element. Still, I do not know why he thinks that way.
Oh well. Shit happens.
Jim Burns
2024-03-30 13:56:51 UTC
Reply
Permalink
Post by Chris M. Thomasson
Post by Chris M. Thomasson
Post by WM
Post by Jim Burns
Post by WM
Post by FromTheRafters
Not showing a bijection is not the same as
not having a bijection.
Showing a not bijection proves
different sizes of sets.
I'll call that a Mückenheim.set.
Mückenheim.sets are finiteⁿᵒᵗᐧᵂᴹ sets.
No.
The set {1, 2, 3, 4, 5, ..., ω} is infinite,
although there is a last element.
Huh? If it has a last element its NOT infinite...
I know for sure that WM thinks
there is a largest natural number.
That must be his last element.
I think that, in WM's telling,
ω is _immediately after_ the largest natural number,
and
proofs that
"immediately before ω" does not describe
anything non.self.contradictory
are proofs that
darkᵂᴹ numbers exist.
In WM's telling.

I think that WM thinks that, in our telling,
ω is a natural number one gets to
by going out out out to "infinity"
and the word "infinite" is in the same family
as "humongous" and "ginormous" except it uses
special characters WM won't or can't read,
which which are what makes "infinite" mathematics.

For the record,
matheologians (AKA not.WM) do not say that.
That which we call ω
by any other word would smell as sweet.

Finite and infinite are as different as
rational and irrational are.
Post by Chris M. Thomasson
Still, I do not know why he thinks that way.
Oh well. Shit happens.
Why he thinks that
is a question for psychologists.

I speculate that
Wolfgang Mückenheim has been seduced by
decades of students at Hochschule Augsburg
memorizing whatever fool thing pops into his head
into thinking that
it is impossible for him to be wrong.
Saying otherwise, or worse, showing otherwise
is just _rude_
like a student interrupting a lecture.
Possibly.
WM
2024-03-30 16:17:37 UTC
Reply
Permalink
Post by Jim Burns
I think that, in WM's telling,
ω is _immediately after_ the largest natural number,
like the smallest unit fraction is existing.
Post by Jim Burns
I think that WM thinks that, in our telling,
ω is a natural number
No.
Post by Jim Burns
Why he thinks that
is a question for psychologists.
It is a simple result of the permanent existence of all numbers and
point.For ever n, there is 2n, is potential infinity. You first must give
an n. If however all 2n are there, then none can be created.

Regards, WM
Jim Burns
2024-03-30 19:04:41 UTC
Reply
Permalink
Post by WM
Post by Jim Burns
I think that, in WM's telling,
ω is _immediately after_ the largest natural number,
like the smallest unit fraction is existing.
Post by Jim Burns
I think that WM thinks that, in our telling,
ω is a natural number
No.
Hold back just a moment on assigning labels.
Is this a fair description of
what you think we think?

I think that WM thinks that, in our telling,
one gets to ω
by going out out out to "infinity"
and the word "infinite" is in the same family
as "humongous" and "ginormous" except it uses
special characters, which which are
what makes "infinite" mathematics.

WM
2024-03-30 16:09:20 UTC
Reply
Permalink
Post by Chris M. Thomasson
I know for sure that WM thinks there is a largest natural number. That
must be his last element. Still, I do not know why he thinks that way.
If there are all points on the real line permanently existing, then there
is a point next to zero.

Regards, WM
Tom Bola
2024-03-29 14:09:37 UTC
Reply
Permalink
Post by WM
The set {1, 2, 3, 4, 5, ..., ω} is infinite, although there is a
last element.
One may construct such a set "N-with-omega".
Post by WM
Under f(x) = 2x we get the image {2, 4, 6, 8, 10, ..., 2ω}.
One may also define such a function, which domain contains both sort of ordinals,
i.e. natural numbers and also a limit ordinal like ω.
Post by WM
The mapping restricted to the natural numbers shows less evens than naturals.
The above sentence is very idiotic nonsense, because this set would contain
the union of W1 = {2, 4, 6, 8, 10, ...} of card ω and W2 = {2ω} of card 1.

You are way too dense for even simplest thinking and math...
Tom Bola
2024-03-29 16:31:42 UTC
Reply
Permalink
Post by Tom Bola
Post by WM
The set {1, 2, 3, 4, 5, ..., ω} is infinite, although there is a
last element.
One may construct such a set "N-with-omega".
Sets are not ordered, so there cannot be a "last element" in any set, of course.
WM
2024-03-30 16:24:04 UTC
Reply
Permalink
Post by Tom Bola
Post by WM
The set {1, 2, 3, 4, 5, ..., ω} is infinite, although there is a
last element.
One may construct such a set "N-with-omega".
Post by WM
Under f(x) = 2x we get the image {2, 4, 6, 8, 10, ..., 2ω}.
One may also define such a function, which domain contains both sort of ordinals,
i.e. natural numbers and also a limit ordinal like ω.
Post by WM
The mapping restricted to the natural numbers shows less evens than naturals.
this set would contain
the union of W1 = {2, 4, 6, 8, 10, ...} of card ω and W2 = {2ω} of card 1.
No. Cantor accepts ω, ω + 2, ω + 4, ... These numbers can be realized
by repeated addition. Mutiplication is a shorthand of addition.

Regards, WM
Tom Bola
2024-03-30 16:37:25 UTC
Reply
Permalink
Post by WM
Post by Tom Bola
Post by WM
The set {1, 2, 3, 4, 5, ..., ω} is infinite, although there is a
last element.
One may construct such a set "N-with-omega".
Post by WM
Under f(x) = 2x we get the image {2, 4, 6, 8, 10, ..., 2ω}.
One may also define such a function, which domain contains both sort of ordinals,
i.e. natural numbers and also a limit ordinal like ω.
Post by WM
The mapping restricted to the natural numbers shows less evens than naturals.
this set would contain
the union of W1 = {2, 4, 6, 8, 10, ...} of card ω and W2 = {2ω} of card 1.
No. Cantor accepts ω, ω + 2, ω + 4, ... These numbers can be realized
by repeated addition. Mutiplication is a shorthand of addition.
You complete moron are unable to get that your set {1, 2, 3, 4, 5, ..., ω}
does not belong to the math of natural numbers because ω is the limiting
ordinal of IN = {1, 2, 3, 4, 5, ...}.

Piss off already, blithering retard...
FromTheRafters
2024-03-24 21:03:47 UTC
Reply
Permalink
Post by WM
Does ℕ = {1, 2, 3, ...} contain all natural numbers such that none cn be
added?
Yes, it is the set of natural numbers.
Post by WM
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.
Dieter Heidorn
2024-03-24 21:09:31 UTC
Reply
Permalink
Post by FromTheRafters
Post by WM
Does ℕ = {1, 2, 3, ...} contain all natural numbers such that none cn
be added?
Yes, it is the set of natural numbers.
Post by WM
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
FromTheRafters
2024-03-24 21:18:24 UTC
Reply
Permalink
Post by Dieter Heidorn
Post by FromTheRafters
Post by WM
Does ℕ = {1, 2, 3, ...} contain all natural numbers such that none cn be
added?
Yes, it is the set of natural numbers.
Post by WM
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.
"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.
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