Re: The non-existence of "dark numbers"

On 18.03.2025 17:34, Jim Burns wrote:

On 3/18/2025 5:00 AM, WM wrote:

Bob can only disappear in a set of finite size
because all exchanges appear at finite sizes.

 Bob disappears from (is swapped from)
each finite set A.
Bob disappears to (is swapped to)
a different finite set

Yes. But he never leaves the matrix!

After all the swaps,
without ever disappearing into
anywhere other than a finite set,
Bob disappears out of all finite sets.

No, that is impossible. If there is an "after all swaps", then all O have settled within the matrix. Lossless exchanges with losses are not allowed.

 The swaps are ⟨n⇄n+1⟩ for all n,
in order by n,
in the emptiest inductive set.

Yes, for all n reachable by induction.

 

The set of all finite.sizes is same.sized as
the set of all finite sizes and Bob.

>
Give the index where Bob is lost.

 Before all the swaps,
Bob is not lost from all finite indices.

And all his infinitely many colleagues are also not lost from all finite indices.

 After all the swaps,
Bob is lost from all finite indices.

No, that is impossible. Here lies your mistake. The matrix has no drain! All X destinated to be exchanged with O are placed inside the matrix at the beginning.

 There is no finite index,
either visible or dark,
from which Bob is last.lost,
or to which Bob is last.lost.

That means he cannot get lost. All motions happen at finite indices. He remains in the matrix, but only the X's are visible after all:
XOOO... XXOO... XXOO... XXXO... ... XXXX...
XOOO... OOOO... XOOO... XOOO... ... XXXX...
XOOO... XOOO... OOOO... OOOO... ... XXXX...
XOOO... XOOO... XOOO... OOOO... ... XXXX...
..............................................................................
Regards, WM
Regards, WM