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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: Fri, 10 Jan 2025 09:03:03 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Am Thu, 09 Jan 2025 23:27:21 +0100 schrieb WM:
> On 09.01.2025 22:03, joes wrote:
>> Am Thu, 09 Jan 2025 11:54:27 +0100 schrieb WM:
> 
>>> If none is empty, then other numbers must be inside. Contradiction.
>> No number is an element of all segments.
> But what numbers are inside if all natural numbers are outside?
Inside what? Inside each segment are infinitely many naturals. Mind
your quantifiers: It is not required that the intersection be nonempty.

>>> Infinitely many endsegments need infinitely many indices. Therefore no
>>> natural number must remain as content in the sequence of endsegments.
>> Only insofar as every number eventually "leaves" the endsegments.
> But what remains?
In what? The intersection is empty.

>> This however does not imply and empty endsegment,
> What remains?
Nothing "remains". There is no end, only a limit.

>> since there inf. many of both naturals and therefore endsegments.
> Infinitely many numbers leave. All elements of ℕ leave.
Yes, in the limit.

>>>>> Unless you claim that the general law does not hold for ∀k ∈ ℕ.
>>>> It does not hold for the infinite intersection.
>>> Why not?
>> Sorry, I lost track. Which law?
> The law that the intersection gets empty but only by one element per
> term.
Well, in the limit (sigh) infinitely many numbers have been "lost".

>>>>> It is trivially true that only one element can vanish with each
>>>>> endsegment.
>>>> Which noone contradicted.
>>> Then the empty intersection is preceded by finite intersections.
>> Does not follow.
> It follows from the law that only one element per term can leave.
Care to write out that deduction?

>>>>> Jim "proved" that when exchanging two elements O and X, one of them
>>>>> can disappear. His "proofs" violate logic which says that lossless
>>>>> exchange will never suffer losses.
>>>> Wrong. The limit of the harmonic series is zero, even though none of
>>>> the terms are.
>>> There is no exchange involved.
>> Same reason: the limit may have different properties than the terms.
> There is no limit involved when counting the fractions.
It is an infinite "process".

>>>>> That's a simple fact. The sequence of natural numbers 1, 2, 3, ...,
>>>>> n, n+1, ...
>>>>> cannot be cut into two actually infinite sequences, namely indices
>>>>> and contents.
>>>> Why should it?
>>> Because an infinite sequence of indices followed by an infinite
>>> sequence of contentent would require two infinite sequences.
>> But this never happens.
> An infinite set of infinite endsegments happens according to matheology.
> It requires two infinite sequences.
No, it doesn't. There is no "infinitieth" segment.

>>> The claim of infinitely many infinite endsegments is false.
>> There are infinitely many naturals though.
> But all must be in the set of indices, if there are infinitely many
> endsegments. But infinitely many must be in the set of contents, if all
> endsegments are infinite.
Those are the same set N.

-- 
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.