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From: WM <wolfgang.mueckenheim@tha.de>
Newsgroups: sci.math
Subject: Re: Incompleteness of Cantor's enumeration of the rational numbers
 (extra-ordinary)
Date: Sat, 11 Jan 2025 12:44:59 +0100
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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.
> 
>>>>> (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.
> 
>>>> ∀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?

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

My claim holds for all numbers only. That is mathematics.
> 
>>>>> 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.
> 
>>> G 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.

>> "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.

Regards, WM