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

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

On 05.12.2024 23:20, Jim Burns wrote:

Depending upon how 'end.segment' is defined,
{} either is or isn't an end.segment.
Consider the options.
⎛ With {} as an end.segment,
⎜ there are more.than.finite.many end.segments,
⎜ too many for any finite.cardinal to be
⎜ in common with all end segments.
⎜ And therefore,
⎜ the intersection of all holds no finite cardinal.

>
Yes.
>

⎜ With {} NOT as an end.segment,

>
all endsegments hold content.

But no common.to.all finite.cardinals.

⎜ 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.
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.

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.
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.

then the indices have an upper bound.
>

⎜ too many for any finite.cardinal to be
⎜ in common with all end segments.

>
Too many for all definable natnumbers.
But by assumption of content
not all natnumbers have become indices.

No.
Each content is index in a later set.

⎜ 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?
Round up the usual suspects
and label them 'definable'.
The intersection of all non.empty.end.segments
of the definable finite.cardinals,
which are each infinite non.empty.end.segments,
is empty.

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

>
That is a wrong conclusion because
inclusion monotony prevents
an empty intersection of non-empty endsegments:

inclusion monotony: pairwise subset.or.superset
The non.empty end.segments of
the definable finite.cardinals
⎛ are pairwise subset.or.superset
⎜ have inclusion monotony
⎜ have an empty intersection --
⎝ which hasn't been prevented by inclusion.monotony.

Because,
⎛ for each finite.cardinal,
⎜ there are fewer finite.cardinals before it
⎝ than there are finite.cardinals after it.

>
That is true only for definable or accessible cardinals.

What does it mean to be
an undefinable or inaccessible finite.cardinal?

Not because
( an end.segment is empty.

>
This argument is wrong if
infinite bijections are assumed to exist.

Each finite.cardinal k is in
(finite) k+1.many end.segments ⟦j,ℵ₀⦆ such that j ∈ ⟦0,k⟧
There are at least
(finite) k+2.many end.segments ⟦j,ℵ₀⦆ such that j ∈ ⟦0,k+1⟧
which are enough for us to know that
⎛ k is not common to all of those finitely.many end.segments.
⎜ k is not common to all end.segments.
⎝ k is not in the intersection of all end segments.
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.