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

Am Fri, 13 Dec 2024 11:55:16 +0100 schrieb WM:

On 13.12.2024 03:23, Richard Damon wrote:

On 12/12/24 9:25 AM, WM wrote:

if Cantor can apply all natural numbers as indices for his bijections,
then all must leave the sequence of endsegments. Then the sequence
(E(k)) must end up empty.
And there must be a continuous staircase from E(k) to the empty set.

But a segment that is infinite in length is, by definiton, missing at
least on end.

That means that the premise "if Cantor can apply all natural numbers as
indices for his bijections" is false.

Nah. Just imagine all of the inf.many steps as a whole - y’know, ACTUAL
infinity.

So, which bijection from Cantor are you talking about? Of are you
working on a straw man that Cantor never talked about?

 
There are many. The mapping from natumbers to the rationals, for
instance, needs all natural numbers. That means all must leave the
endsegments. Another example is Cantor's list "proving" uncountable
sets. If not every natural number has left the endsegment and is applied
as an index of a line of the list, the list is useless.

Then the list were finite. It isn’t, though.

But if every natural number has left the endsegments, then the
intersection of all endsegments is empty.

Yes.

Then the infinite sequence of
endegments has a last term (and many finite predecessors, because of ∀k
∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}).

No. It is literally „without an end”, and yet can be „completed”, if
only you were able to conceive of infinity.

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Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.