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

On 31.12.2024 21:26, Jim Burns wrote:

On 12/31/2024 1:20 PM, WM wrote:

Every union of FISONs which
stay below a certain threshold
stays below that threshold.
Note: Every FISON stays below 1 % of ℕ.

 For each finite ⟦0,j⦆, there are
more.than.#⟦0,j⦆.many finite ⟦0,k⦆

So it is for all definable natural numbers. But all natural numbers are finite, and nevertheless there are natural numbers which have only few successors because when they are removed nothing remains
ℕ \ {1, 2, 3, ...} = { } .

The set {⟦0,i⦆:#⟦0,i⦆<#⟦0,i+1⦆} of
all the finite ⟦0,k⦆
is not a finite set.

It is not a set but a potentially infinite collection.

Bob is placed within
the dark parts of the matrix.

 Then not all swaps are swapped

If all X_n1 from the first column
XOOO...
XOOO...
XOOO...
XOOO...
...
were transferred into the second column X_n2 or the first row X_1n: Would you then believe that not all swaps had been swapped?

Bob comes to rest within the dark part.

 Then not all swaps are swapped

For people not blinded by set theory it is clear that the X cannot cover the whole matrix. Therefore the limit in
XOOO... XXOO... XXOO... XXXO... ... XXXX...
XOOO... OOOO... XOOO... XOOO... ... XXXX...
XOOO... XOOO... OOOO... OOOO... ... XXXX...
XOOO... XOOO... XOOO... OOOO... ... XXXX...
...............................................
can only represent a small part of the matrix.

Every exchange happens between two sites.

 Yes.
Always between two visibleᵂᴹ sites.
And no single swap disappears Bob.

In no single swap *that is observed*!

Infinitely.many is different from finitely.many.
For each finitely.many, there are more.

But the first column is exactly as long as the first row.

 

After all swaps,
⎛ Bob has never swapped into any darkᵂᴹ room,
⎝ Bob has swapped out of any visibleᵂᴹ room,
Bob is nowhere.

>
But Bob is never out of the matrix
because
there is no exchange at all between outside and inside.

 Yes, Bob never exits the matrix.
However, after all swaps, Bob isn't in the matrix.
Therefore, infinite is different.

No, set theorists have been blinded.
Regards, WM