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

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Sujet : Re: Incompleteness of Cantor's enumeration of the rational numbers (extra-ordinary)
De : james.g.burns (at) *nospam* att.net (Jim Burns)
Groupes : sci.math
Date : 03. Dec 2024, 01:32:26
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <9bcc128b-dea8-4397-9963-45c93d1c14c7@att.net>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27
User-Agent : Mozilla Thunderbird
On 12/2/2024 9:28 AM, WM wrote:
On 02.12.2024 12:53, FromTheRafters wrote:

[...]
>
Infinite endsegments contain an infinite set each,
infinitely many elements of which
 are in the intersection.
Yes to:
⎛ regarding finite.cardinals,
⎜ for  each  end.segment   E(k)
⎜ there is a subset S such that
⎝ for each finite cardinal j, j < |S| ≤ |E(k)|
No to:
⛔⎛ regarding finite.cardinals,
⛔⎜ ⮣ there is a subset S such that ⮧
⛔⎜ ⮤ for  each  end.segment   E(k) ⮠
⛔⎝ for each finite cardinal j, j < |S| ≤ |E(k)|
A quantifier shift tells you (WM) what you (WM) _expect_
  but a quantifier shift is untrustworthy.

An empty intersection cannot come before
an empty endsegment has been produced by
losing one element at every step.
No.
Because see below.

E(1), E(2), E(3), ...
and
E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
are identical for every n and in the limit
because
E(1)∩E(2)∩...∩E(n) = E(n).
They are
identical COUNTER.EXAMPLES to what you expect.
----
An empty intersection cannot come before
an empty endsegment has been produced by
losing one element at every step.
No.
For the set of finite cardinals,
EVEN IF NO END.SEGMENT IS EMPTY,
  the intersection of all end segments is empty.
⎛ The set of finite.cardinals holds
⎜  only finite.cardinals.

⎜ Each finite.cardinal is finite.

⎜ For each finite.cardinal,
⎜  only finitely.many finite.cardinals are ≤ it.

⎜ For each finite.cardinal,
⎜  only end.segments which start ≤ it
⎜ hold it.

⎜ For each finite.cardinal,
⎝  only finitely.many end.segments hold it.
⎛ The set of finite cardinals holds
⎜  all finite.cardinals.

⎜ Each finite.cardinal is followed by
⎜  another finite.cardinal.

⎜ No finite.cardinal is last.

⎜ The set of finite cardinals has
⎜  a subset (itself) which is not two.ended.

⎜ The set of finite cardinals is infinite.

⎜ Each finite.cardinal starts an end.segment.

⎝ There are infinitely.many end.segments.
⎛ For each finite.cardinal,
⎜  only finitely.many end.segments hold it.

⎜ There are infinitely.many end.segments.

⎜ For each finite.cardinal,
⎜  not all end.segments hold it.

⎜ For each finite.cardinal,
⎜  the intersection doesn't hold it

⎜ EVEN IF NO END.SEGMENT IS EMPTY,
⎜⎛ For each finite.cardinal,
⎜⎜  only finitely.many end.segments hold it.
⎜⎜ For each finite.cardinal,
⎜⎜  only finitely.many end.segments hold it.
⎜⎜ For each finite.cardinal,
⎜⎝  the intersection doesn't hold it

⎜ EVEN IF NO END.SEGMENT IS EMPTY,
⎝  the intersection of all end segments is empty.

Date Sujet#  Auteur
23 Dec 24 o 

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