Re: The non-existence of "dark numbers"

On 3/18/2025 12:58 PM, WM wrote:

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!

WM.logic claims that
sizes of
⎛ sets of places and swaps
⎜ such that,
⎜⎛ if Bob is in.matrix before.swaps
⎝⎝ then Bob is in.matrix after.swaps.
are
the only sizes, the WM.sizes.
If the set of WM.sizes has a WM.size,
then it has a subset larger than it, that set.
Matheologians
reject subsets larger than their sets, and
accept sizes contrary to WM.logic,
  sizes larger than each WM.size.
There are sets of places and swaps
having those larger, contra.WM.logical sizes
such that
⎛ Bob is in.matrix before.swaps, but
⎝ Bob is NOT in.matrix after.swaps.

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

>
No, that is impossible.

Then there are subsets larger than their sets.

If there is an "after all swaps",
then all O have settled within the matrix.

That matrix,
with rows and columns indexed by
the emptiest inductive set,
after all the swaps,
that matrix doesn't hold any O.
⎛ Assume otherwise.
⎜ Assume that it's after all swaps,
⎜ and O is in cell <k,1>
⎜
⎜ However,
⎜ it isn't after swap ⟨⟨k,1⟩⇄⟨k(k+1)/2,1⟩⟩
⎜ which swaps an X to ⟨k,1⟩
⎝ Contradiction.

Lossless exchanges with losses
are not allowed.

Set.level swaps between a set and
a same.sized proper.subset
preserve size.
If same.sized proper.subsets
are not allowed,
then
the set of all allowed sizes
has subsets larger than it, that set.

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.

All n in the emptiest inductive set
are reachable by
proving a subset is inductive,
because
the emptiest inductive set
is its.own.only.inductive.subset.

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

>
No, that is impossible.
Here lies your mistake.
The matrix has no drain!

If the set of
all set.sizes needing a drain
is a set of a size needing a drain,
then
that set has subsets larger than it, the set.