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

Am Thu, 12 Dec 2024 15:33:20 +0100 schrieb WM:

On 12.12.2024 15:23, joes wrote:

Am Thu, 12 Dec 2024 10:12:26 +0100 schrieb WM:

 

The end of the sequence is defined by ∀k ∈ ℕ : E(k+1) = E(k) \ {k}.

The sequence is endless, has no end, is infinite.

If a bijection with ℕ is possible, the sequence can be exhausted so that
no natural numbers remains in an endsegment.

For you: no, an infinite set cannot be exhausted in finite steps.

None of which are an infinite sets, so trying to take a "limit" of
combining them is just improper.

Most endsegments are infinite. But if Cantor can apply all natural
numbers as indices for his sequences, then all must leave the sequence
of endsegments. Then the sequence (E(k)) must end up empty. And there
must be a continuous staircase from E(k) to the empty set.

It makes no sense not being able to „apply” numbers. Clearly Cantor
does.

He claims it. That means no numbers remain unpaired in endsegments.

As you haven’t disproven.

The sequence IS continuous. It’s just that you misconceive of the limit
as reachable.

Cantor does. If the limit is not reachable, then complete bijections
cannot be established.

The limit is only reachable… at infinity.

"If we think the numbers p/q in such an order [...] then every number
p/q comes at an absolutely fixed position of a simple infinite sequence"
[E. Zermelo: "Georg Cantor – Gesammelte Abhandlungen mathematischen und
philosophischen Inhalts", Springer, Berlin (1932) p. 126]
"The infinite sequence thus defined has the peculiar property to contain
the positive rational numbers completely, and each of them only once at
a determined place." [G. Cantor, letter to R. Lipschitz (19 Nov 1883)]

If you accept these claims, then no number must remain in an endsegment.

True: no number is in all endsegments.
False: there is a segment with no numbers.

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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.