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

Am Thu, 28 Nov 2024 11:39:05 +0100 schrieb WM:

On 28.11.2024 09:34, Jim Burns wrote:
 

Consider the sequence of claims.
⎛⎛ [∀∃] for each end.segment ⎜⎜ there is an infinite set such that ⎜⎝
the infinite set subsets the end.segment

and its predecessors!

Naturally.

If each endsegment is infinite, then this is valid for each endsegment
with no exception. because all are predecessors of an infinite
endsegment. That means it is valid for all endsegments.

That is what "every" means.

The trick here is that the infinite set has no specified natural number
(because all fall out at some endsegment) but it is infinite without any
other specification.

Yes. You can call it omega or N.

⎜⎛ [∃∀] there is an infinite set such that ⎜⎜ for each end.segment ⎝⎝
the infinite set subsets the end.segment
We cannot SEE,
just by looking at the claims,
that, after [∀∃], [∃∀] is not.first.false.

 
I have proved above that [∃∀] is true for all infinite endsegments.

Uh, that is wrong.

A simpler arguments is this: All endsegments are in a decreasing
sequence.

There is no decrease, they are all infinite.

Before the decrease has reached finite endsegments, all are
infinite and share an infinite contents from E(1) = ℕ on. They have not
yet had the chance to reduce their infinite subset below infinity.

All segments are infinite. Nothing can come "afterwards".

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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.