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

On 11/29/2024 10:53 AM, Jim Burns wrote:

On 11/28/2024 5:39 AM, WM wrote:

On 28.11.2024 09:34, Jim Burns wrote:

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

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and its predecessors!

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For each end.segment of finite.cardinals,
that end.segment and its predecessors
are not
each end.segment.
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In particular,
for each end.segment of finite.cardinals,
there is a successor.end.segment which is
not one of
that end.segment and its predecessors.
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If each endsegment is infinite,
then this is valid
for each endsegment with no exception

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Yes,
if each end.segment is infinite
then each end.segment is infinite.
>

because all are
predecessors of an infinite endsegment.

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Each end.segment of finite.cardinals
  is staeckel.infinite
because
each finite.cardinal is countable.past
each finite.cardinal is not its second end
each end.segment has a non.empty subset (itself)
which is not.two.ended.
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That means it is valid for all endsegments.
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The trick here is that
the infinite set has no specified natural number (because all fall out
at some endsegment)

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all fall out == empty
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but it is infinite

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infinite and empty
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without any other specification.

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A finite.cardinal is specified to be a cardinal and finite.
>
There is no other specification for a finite cardinal.
>
>

Ah, it appears the slide there, a bit, the back-slide.
It reminds of the Matrix movie, "what was that?",
"well I just saw a cat walk past, and then it
did again", and it was like "that's bad, usually
it means something got changed".
You appear to have added another law of restriction
of comprehension - "no other specification", I don't
believe you've established criteria sufficient "uniqueness".
So what are these cardinoids again you say don't exist?