Re: Incompleteness of Cantor's enumeration of the rational numbers

On 12/1/2024 5:02 AM, WM wrote:

On 01.12.2024 00:34, Jim Burns wrote:

On 11/30/2024 4:56 PM, WM wrote:

Inclusion monotony prevents
an empty set of common finite cardinals
 without an empty endsegment.

>
No

>
It does.
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).

Identical sequences without an empty end.segment.
Identical empty set of common finite.cardinals.
Inclusion monotony does not prevent
an empty set of common finite.cardinals
  without an empty end.segment.
ENDS is the set of end.segments of finite.cardinals.
ENDS⁺ = ENDS\{{}}  is
  the set of end.segments of finite.cardinals
  without the empty end.segment
Even if {} is in ENDS, {} isn't in ENDS⁺
⋂ENDS⁺ = {}
For each finite.cardinal k
|ENDS⁺| > k
For each finite.cardinal k
⎛ |{fore.segments k is not.in}| < |ENDS⁺|
⎜ |{end.segments k is in}| < |ENDS⁺|
⎜ |{in ENDS⁺, end.segments k is in}| < |ENDS⁺|
⎜ k is not in common with each end segment in ENDS⁺
⎝ k is not in ⋂ENDS⁺
Each finite.cardinal k is not.in ⋂ENDS⁺
⋂ENDS⁺ = {}

inclusion.monotony does not prevent
an empty set of common finite.ordinals
without an empty end segment.

>
Stupid or impudent. No reason to continue.