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.mathDate : 29. Nov 2024, 19:53:29
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <9e4bb6a7-9e7a-41a3-bd3b-4dbb7d2452f7@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
User-Agent : Mozilla Thunderbird
On 11/28/2024 5:39 AM, WM wrote:
On 28.11.2024 09:34, Jim Burns wrote:
Consider the sequence of claims.
⎛⎛ [∀∃] for each end.segment
⎜⎜ there is an infinite set such that
⎜⎝ the infinite set subsets the end.segment
>
and its predecessors!
For each end.segment of finite.cardinals,
that end.segment and its predecessors
are not
each end.segment.
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.
If each endsegment is infinite,
then this is valid
for each endsegment with no exception
Yes,
if each end.segment is infinite
then each end.segment is infinite.
because all are
predecessors of an infinite endsegment.
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.
That means it is valid for all endsegments.
>
The trick here is that
the infinite set has no specified natural number (because all fall out at some endsegment)
all fall out == empty
but it is infinite
infinite and empty
without any other specification.
A finite.cardinal is specified to be a cardinal and finite.
There is no other specification for a finite cardinal.