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 : 08. Dec 2024, 00:38:44
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <10ebeeea-6712-4544-870b-92803ee1e398@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
User-Agent : Mozilla Thunderbird
On 12/7/2024 4:37 PM, WM wrote:
On 07.12.2024 20:59, Jim Burns wrote:
On 12/7/2024 6:09 AM, WM wrote:
On 06.12.2024 19:17, Jim Burns wrote:
On 12/6/2024 3:19 AM, WM wrote:
On 05.12.2024 23:20, Jim Burns wrote:
⎜ With {} NOT as an end.segment,
>
all endsegments hold content.
>
But no common.to.all finite.cardinals.
>
Show two endsegments which
do not hold common content.
>
I will, after you
show me a more.than.finitely.many two.
>
There are no more than finitely many
natural numbers which can be shown.
All which can be shown have common content.
2 is in 3 end.segments: ⟦0,ℵ₀⦆...⟦2,ℵ₀⦆
There are more.than.3 end.segments.
Therefore, 2 is not in common with all end.segments.
For each end.segment ⟦j,ℵ₀⦆
there is a subset ⟦j,j+2⟧ holding more.than.2
2 < |⟦j,j+2⟧| ≤ ⟦j,ℵ₀⦆|
Therefore, each end.segment holds more.than.2
For each finite cardinal k
⎛ k is in k+1 end.segments: ⟦0,ℵ₀⦆...⟦k,ℵ₀⦆
⎜ There are more.than.k+1 end.segments.
⎝ Therefore, k is not in common with all end.segments.
and
⎛ for each end.segment ⟦j,ℵ₀⦆
⎜ there is a subset ⟦j,j+k⟧ holding more.than.k
⎜ k < ⟦j,j+k⟧ ≤ ⟦j,ℵ₀⦆
⎝ Therefore, each end.segment holds more.than.k
Each end.segment is more.than.finite and
the intersection of the end.segments is empty.
⎜ 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.
>
More than finitely many are finitely many,
>
You (WM) define it that way,
which turns your arguments into gibberish.
>
This is not gibberish but mathematics:
∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n).
True.
Every counter argument has to violate this.
False.
For example, see above.
That is inacceptable.
More than finitely many are finitely many,
unless they are actually infinitely many.
Therefore they are not enough.
>
Call it 'potential'.
>
It is.
Each finite.cardinal is not in common with
more.than.finitely.many end.segments.
Each end.segment has, for each finite.cardinal,
a subset larger than that cardinal.
No empty intersection without an empty endsegment.
Only end.segments larger than any finite cardinal
which hold no common.to.all finite.cardinals.
…