Re: Incompleteness of Cantor's enumeration of the rational numbers

Am Mon, 25 Nov 2024 13:11:26 +0100 schrieb WM:

On 24.11.2024 19:16, Jim Burns wrote:

On 11/23/2024 3:45 PM, WM wrote:

 

for every interval (0,n]
the relative covering is 1/10,
independent of how the hats are shifted. This cannot be remedied in
the infinite limit because outside of all finite intervals (0, n]
there are no further hats available.

 
All the hats for which
  if there are G.many then there are G^G^G.many
are enough to "remedy" the 1/10.relative.covering

 
Not at all!

If each G can match G^G^G

That is a false assumption.
If there are enough hats to cover G then there are enough hats to cover
G^G^G. G does not cover G^G^G.

Matching a proper subset is the sort of behavior which permits Bob to
disappear with enough room swapping _inside_ the Hotel.
You (WM) treat that behavior as proof that we are wrong and you (WM)
are right.

Exchange of two elements never leads to loss of one of them.

That's still true in the infinite case.

What it is  is proof that not all sets behave like finite sets.

Nonsense. Logic is also prevailing in infinite sets.

Infinite sets are not finite.

If there are enough hats for G natural numbers,
then there are also enough for G^G^G natural numbers.
The number G^G^G is not.first for which there are NOT enough hats.

There is no such number because the set of definable hats is potentially
infinite.

Right!

Therefore,
the number G^G^G is not.first for which there are NOT enough hats.

We do not disagree. Therefore you need not prove a difference for G and
G^G^G.

Nor for G and 10G.

A similar argument can be made for each natural number.

No, it can be made for each definable natural number, i.e., for a number
belonging to a tiny finite initial segment which is followed bay almost
all numbers.

Yes, like he said: for every natural.

Consider the set of natural numbers for which there are NOT enough
hats.

It is dark.

It is not dark what we mean by 'natural number'.
A natural number is countable to from 0.

That is a definable number.

Yes, the naturals are all "definable". That is all mathematicians talk
about, so you can just imagine they always mean your "N_def" when they
say N.

The natural numbers "fail" at being finitely many.
It is nothing more than that.

If they are infinitely many but complete, then they and their number
don't vary. |ℕ| - 1 < |ℕ| < |ℕ| + 1.

What would it even mean for the naturals to not be "complete"? Either
you have the whole set or you don't.

--
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.