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

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

If there are enough hats for G natnumbers,
then there are also enough for G^G^G natnumbers.

On 11/24/2024 6:06 AM, WM wrote:

On 24.11.2024 06:37, Jim Burns wrote:

If there are enough hats for G natural numbers,
then there are also enough for G^G^G natural numbers.

>
So it is.

Thank you.
That is enough to see that
the hats behave like they are
  more than unreasonably.large.but.finite.

But that does not negate the fact that
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
If each G can match G^G^G
then
all of those numbers G match
a proper subset G^G^G of all of those numbers 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.
What it is  is proof that
not all sets behave like finite sets.
By 'finite', I mean our 'finite', with
its fundamentally different behavior,
not your (WM's) 'finite', which merely means
more of the same, even though
the 'more' is 'super.colossally more'.

If there are NOT enough for G^G^G natural numbers,
then there are also NOT enough for G natural numbers.
>
G precedes G^G^G.
If, for both G and G^G^G, there are NOT enough hats,
G^G^G is not first for which there are not enough.
>
That generalizes to
each natural number is not.first for which
there are NOT enough hats.

>
It seems so but
the sequence 1/10, 1/10, 1/10, ... has limit 1/10
with no doubt.

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.
⎛ Assume otherwise.
⎜ Assume G^G^G is first for which
⎜ there are NOT enough hats.
⎜
⎜ G < G^G^G
⎜ There ARE enough hats for G.
⎜ If there are enough hats for G natural numbers,
⎜ then there are also enough for G^G^G natural numbers.
⎜ There ARE enough hats for G^G^G
⎜
⎜ However,
⎜ there are NOT enough hats for G^G^G
⎝ Contradiction.
Therefore,
the number G^G^G is not.first for which
there are NOT enough hats.
A similar argument can be made for
  each natural number.
Therefore,
each natural number is not.first for which
  there are NOT enough hats.

This dilemma is the reason
why dark numbers are required.
>

----
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 knowledge narrows the possibilities to two.

Since it is a set of natural numbers,
there are two possibilities:
-- It could be the empty set.
-- It could be non.empty and hold a first number.

>
Both attempts fail.

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