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

On 11.01.2025 10:26, joes wrote:

Am Sat, 11 Jan 2025 10:03:01 +0100 schrieb WM:

On 10.01.2025 23:42, joes wrote:

Am Fri, 10 Jan 2025 22:44:31 +0100 schrieb WM:

 

Yes they can, because there are an infinity of them.

That is wrong. Infinitely many of them can only exist when no natural
natural number is missing an an index.

The naturals *are* the indices of the sequence. And it is infinite.

Not if any natural is missing because it remains in the content.

The limit is empty,

There is no limit in bijections between sets. There is only the unlimited application of
∀k ∈ ℕ : E(k+1) = E(k) \ {k+1}
such that no number has been forgotten. Every number leaves the content. No number remains, let alone infinitely many.

no natural "remains" in every endsegment, the
sequence of naturals diverges beyond any (finite) bound. Furthermore,
every one of the inf.many naturals has a corresponding endsegment,
none is "missing"; the sequence is infinitely long

only if all numbers are indices and therefore no number remains in the content, let alone infinitely many.

 

Therefore none can remain in the content. Therefore your argument is
fools crap.

The limit is indeed empty.

There is no limit! All indices are required for bijections. Bijections
concern all elements, not limits. In particular a sequence of infinite
sets has no empty "limit".

Yes it does, the (infinite) sequence of (infinite) end segments
converges to the empty set,

Contradicitio in adjecto. There can only be one infinite sequence when 1, 2, 3, ... is cut into indices and content. It is always the second part unless this is empty.

because no element can be in every
segment.

All natural numbers and all their predecessors leave. If all endsegments remain infinite, then they must contain numbers which have not left and therefore are not predecessors of natural numbers.

A union of FISONs which stay below a certain threshold can surpass that
threshold.

Only a finite threshold. Every FISON is (surprise) finite, and every
finite union of consecutive ones is again finite and equal to the
largest one. Every infinite union of (not necessarily consecutive)
FISONs has no largest natural.

That is true. Nevertheless a union of FISONs which stay below 1 % of |ℕ| cannot surpass that threshold.
Regards, WM