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

Am Sat, 11 Jan 2025 10:03:01 +0100 schrieb WM:

On 10.01.2025 23:42, joes wrote:

Am Fri, 10 Jan 2025 22:44:31 +0100 schrieb WM:

Yes they can, because there are an infinity of them.

That is wrong. Infinitely many of them can only exist when no natural
natural number is missing an an index.

The naturals *are* the indices of the sequence. And it is infinite.

Not if any natural is missing because it remains in the content.

The limit is empty, no natural "remains" in every endsegment, the
sequence of naturals diverges beyond any (finite) bound. Furthermore,
every one of the inf.many naturals has a corresponding endsegment,
none is "missing"; the sequence is infinitely long.

Therefore none can remain in the content. Therefore your argument is
fools crap.

The limit is indeed empty.

There is no limit! All indices are required for bijections. Bijections
concern all elements, not limits. In particular a sequence of infinite
sets has no empty "limit".

Yes it does, the (infinite) sequence of (infinite) end segments
converges to the empty set, because no element can be in every
segment. Partial mappings of only a finite starting segment are
not bijections but converge to it as well.

A union of FISONs which stay below a certain threshold can surpass that
threshold.

Only a finite threshold. Every FISON is (surprise) finite, and every
finite union of consecutive ones is again finite and equal to the
largest one. Every infinite union of (not necessarily consecutive)
FISONs has no largest natural.

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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.