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

On 24.11.2024 06:37, Jim Burns wrote:

On 11/23/2024 5:01 PM, WM wrote:

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

So it is. 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.

 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. 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.

 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. That is the reason why dark numbers are required.

 Its first number, if it existed, would be
the first natural number for which
there are NOT enough hats.
 However,
the FIRST natural number for which
there are NOT enough hats
does not exist.

That however does not negate the fact that for all intervals (0, n] and also for all intervals (10n, 10(n+1)] with no exception from which another load of hats could be acquired there are not enough hats to cover all n.

Therefore,
for each natural number,
there are enough hats.

Then mathematics fails. I don't accept that a constant sequence has another limit than this constant.
Regards, WM