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

Am Sat, 07 Dec 2024 12:09:19 +0100 schrieb WM:

On 06.12.2024 19:17, Jim Burns wrote:

On 12/6/2024 3:19 AM, WM wrote:

 

⎜ With {} NOT as an end.segment,

all endsegments hold content.

But no common.to.all finite.cardinals.

Show two endsegments which do not hold common content.

Not two, but infinitely many.

⎜ there STILL are ⎜ more.than.finite.many end.segments,

Not actually infinitely many however.

More.than.finitely.many are enough to break the rules we devise for
finitely.many.

More than finitely many are finitely many,

No. Larger than any finite number = infinite.

For each finite.cardinal,
up.to.that.cardinal are finitely.many.
A rule for finitely.many holds.
All.the.finite.cardinals are more.than.finitely.many.
A rule for more.than.finitely.many holds.

All the fite cardinals are actually infinitely many. That is impossible
as long as an upper bound rests in the contentents of endsegments.

This right here is you cardinal (heh) mistake. Infinite means there
is no upper bound.

If all endsegments have content, then not all natnumbers are indices,

That seems to be based on the idea that no finite.cardinal is both
index and content.

By an unfortunate definition (made by myself) there is always one
cardinal content and index: E(2) = {2, 3, 4, ...}. But that is not
really a problem.

And since the "content" is infinite, there are inf.many indices.

Elsewhere, considering one set, that's true.
No element is both index(minimum) and content(non.minimum).
However,
here, we're considering all the end segments.
Each content is index in a later set.
Each non.zero index is content in an earlier set.

"All at once" is the seductive attempt of tricksters. All that happens
in a sequence can be investigated at every desired step.

There are no steps. Something that holds for every finite part
may not hold for the infinite whole.

Each content is index in a later set.

Only if all content is lost. That is not possible for visible
endsegments.

It is possible in the limit.

They all are infinite and therefore are finitely many.

Ridiculous. The "contents" "become" indices.

⎜ And therefore,
⎜ the intersection of all ⎝ STILL holds no finite cardinal.

No definable finite cardinal.

Wasn't there a time when you (WM)
thought 'undefinable finite.cardinal'
was contradictory?

Yes, until about six years ago.
Although I was a strong opponent of Cantor's
actual infinity, an internet discussion in 2018 [1] has changed my mind
in that without actual infinity the real axis would have gaps.

[citation needed]

Round up the usual suspects and label them 'definable'.

∀k ∈ ℕ : E(k+1) = E(k) \ {k} cannot come down to the empty set in
definable numbers. No other way however is accessible.

It doesn't "come down", only in the infinite.

The intersection of all non.empty.end.segments of the definable
finite.cardinals,
which are each infinite non.empty.end.segments,
is empty.

A clear selfcontradiction because of inclusion monotony.

No, there are infinitely many.

Generalizing,
the intersection of all non.empty end.segments is empty.
It is an argument considering finites,
of which there are more.than.finitely.many.

It is violating mathematics and logic. Like Bob.

It's just not finite.

--
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.