Re: Incompleteness of Cantor's enumeration of the rational numbers (extra-ordinary)

Am Sat, 11 Jan 2025 12:44:59 +0100 schrieb WM:

On 11.01.2025 10:41, joes wrote:

Am Sat, 11 Jan 2025 09:50:02 +0100 schrieb WM:

On 10.01.2025 22:51, joes wrote:

Am Fri, 10 Jan 2025 22:38:51 +0100 schrieb WM:

On 10.01.2025 21:06, joes wrote:

Am Fri, 10 Jan 2025 20:17:37 +0100 schrieb WM:

On 10.01.2025 19:28, joes wrote:

Am Fri, 10 Jan 2025 18:21:43 +0100 schrieb WM:

>

I have no expectations about cardinality. I know that for every
finite initial segment the even numbers are about half of the
natural numbers.
This does not change anywhere. It is true up to every natural
number.

You wrongly expect this to hold in the infinite.

No, I expect it is true for all natural numbers, none of which is
infinite.

But it is true for every natural

Of course. Otherwise you would have to find a counterexample.

Good. It is not true for the infinite sets.

The natural numbers are an infinite set. For all of them it is true,

But not for omega, which is not a natural.

Therefore it is irrelevant. No bijection from ℕ contains it.

Then don't claim that some sentence held for omega.

(if you formalise it correctly)!

Irrelevant.

Mathematics is all about formalising.

No, that is only a habit of the last century.

Informal reasoning gets you nowhere, see the centuries before that.

There mathematics has flourished. Now mainly nonsense is produced.

Crawl back into your cave and marvel about infinity.

∀n ∈ ℕ: |{1, 2, 3, ..., 2n}|/|{2, 4, 6, ..., 2n}| = 2.

Those are not N and E.

Find an element of N or E that is not covered by the equation.

Not what I said. Every natural is finite, and so are the starting
segments of N and E.

And which are not?

See:

The whole sets (which can be seen as the limits) are not finite.

My claim holds for all numbers only. That is mathematics. [?]

Your claim does not hold for the sets.

That doesn't make it true for N and G.

I am not interested in these letters but only in all natural
numbers.
All natural numbers are twice as many as all even natural numbers.
If your N and G denote all natural numbers and all even numbers,
then 2 is true also for them.

No. For n->oo,

Every n is finite.

The *set* of all of them isn't.

Irrelevant. My claim holds for all natnumbers only.

That's what I'm saying.

E is both the set {2, 4, ..., 2n} and {2, 4, ..., 42n};
indeed, {2, 4, ..., 2kn} for every k e N.

And all of them can be denoted by n.

All what?

All natnumbers which Cantor uses in bijections: "such that every element
of the set stands at a definite position of this sequence". If this has
been accomplished, and then more numbers are created, the bijection
fails. This must not happen.

No numbers are "created" (I guess you mean the image is a subset of the
domain?).

"thus we get the epitome (ω) of all real algebraic numbers [...] and
with respect to this order we can talk about the th algebraic number
where not a single one of this epitome () has been forgotten." [E.
Zermelo: "Georg Cantor – Gesammelte Abhandlungen mathematischen und
philosophischen Inhalts", Springer, Berlin (1932) p. 116]
Afterwards no extension by 42 is allowed.

There is no "after" an infinity.

Cantor maps all natural numbers to a set. Afterwards these natural
numbers can be multiplied by 2. Not all remain those which Cantor has
applied.

Cantor bijectively maps the naturals to the algebraic numbers. You can
also biject the naturals and the evens. Call the first one f and the
second g; then you can compose g(f(n)): E->(N->)A, a bijection from
the even to the algebraic numbers. The function {(k, f(k) for all k e N}
is also bijective, but A->N: {(f(k), k) for all k e E} of course isn't,
if I parsed your last sentence correctly.

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Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.