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

On 12/4/2024 8:31 AM, WM wrote:

On 04.12.2024 11:33, FromTheRafters wrote:

WM formulated the question :

On 03.12.2024 21:34, Jim Burns wrote:

On 12/3/2024 8:02 AM, WM wrote:

E(1)∩E(2)∩...∩E(n) = E(n).
Sequences which are identical in every term
have identical limits.

>
An empty intersection does not require
an empty end.segment.

>
A set of non-empty endsegments has
a non-empty intersection.
The reason is inclusion-monotony.

>
Conclusion not supported by facts.

@WM
Below, you (WM) are headed toward proving
⎛ for the intersection of set S which is FINITE
⎜ holding only non.empty sets
⎜⎛ if the intersection of S is empty,
⎝⎝ then S is not pairwise subset.or.superset
I accept that.
And I accept that
the end.segments of the finite.cardinals
are pairwise subset.or.superset.
However,
that doesn't conflict with
⎛ An empty intersection does not require
⎝ an empty end.segment.
Below: two is finite.

In two sets A and B which
are non-empty both
but have an empty intersection,
there must be at least
two elements a and b which are
in one endsegment but not in the other:
a ∈ A but a ∉ B and b ∉ A but b ∈ B.
>
Same with a set of endsegments.
It can be divided into two sets
for both of which the same is required.

No.
Not all sets of end.segments
can be subdivided into two FINITE sets.
----
⎛ An empty intersection does not require
⎝ an empty end.segment.
ℕ is the set of finite.cardinals.
E(k) is an end.segment of the finite.cardinals.
E(k) = {i∈ℕ:k≤i}
ENDS is the set of end.segments.
G(k) is a greater.segment of the finite.cardinals.
G(k) = {i∈ℕ:k<i}
G(k) is non.empty.  [!]
GREATERS is the set of greater.segments.
GREATERS is inclusion.monotonic and {}.free.
G(k) = E(k+1)
ENDS = GREATERS∪{ℕ}
∀k ∈ ℕ:
⎛ G(k) ≠ {}   [!]
⎜ k ∉ G(k) ∈ GREATERS
⎝ k ∉ ⋂GREATERS
⋂GREATERS = {}
{} ∉ GREATERS
⋂ENDS = (⋂GREATERS)∩ℕ = {}∩ℕ = {}
{} ∉ ENDS
⎛ An empty intersection does not require
⎝ an empty end.segment.