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From: joes <noreply@example.org>
Newsgroups: sci.math
Subject: Re: Incompleteness of Cantor's enumeration of the rational numbers
 (extra-ordinary)
Date: Sat, 11 Jan 2025 13:58:28 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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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.

-- 
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.