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

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.

I accept logic. Exchange of two elements never leads to loss of one of them. You do not.

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

Nonsense. Logic is also prevailing in infinite sets.

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.

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.

 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.

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.

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.
Regards, WM