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

On 12/29/2024 5:27 PM, WM wrote:

On 29.12.2024 21:09, Jim Burns wrote:

Bob can disappear within a larger.enough set,
because
no finite is last of the finites,
because
swapping two values of a one.to.one map
leaves another one.to.one map.

>
No.
Exchange of two entities does never result in
only one of them.

Elsethread.
<WM>

Every FISON is less than 1 % of ℕ
because
by expanding it by a factor 100
the situations remains the same - forever.

</WM>
Date: Tue, 31 Dec 2024 10:39:12 +0100
Message-ID: <vl0e3v$25vs8$1@dont-email.me>
For each ordinal k, there is
an ordinal k+1 which is fuller.by.one.
⎛ k = ⟦0,k⦆
⎝ k+1 = ⟦0,k⦆∪⦃k⦄
For each finite.ordinal k, there is
a finite.ordinal k+1 which is larger.by.one.
⎛ finite k
⎜ #⟦0,k⦆ < #⟦0,k+1⦆
⎜ ¬∃f one.to.one: ⟦0,k+1⦆ ⇉ ⟦0,k⦆
⎜ ¬∃g one.to.one: ⟦0,k+2⦆ ⇉ ⟦0,k+1⦆
⎜ #⟦0,k+1⦆ < #⟦0,k+2⦆
⎝ finite k+1
----
⎛ not (finite k and infinite k+1)
⎜ no first.infinite k+1
⎜ well.order
⎜ no infinite k+1
⎝ '+1' closed in the finites
⎛ define j+(k+1) = (j+k)+1
⎜ not (finite j,k+1,j+k, infinite (j+k)+1)
⎜ no infinite (j+k)+1
⎜ no first.infinite j+(k+1)
⎜ well.order
⎜ no infinite j+(k+1)
⎝ '+' closed in the finites
⎛ define j×(k+1) = (j×k)+j
⎜ not (finite j,k+1,j×k, infinite (j×k)+j)
⎜ no infinite (j×k)+j
⎜ no first.infinite j+(k+1)
⎜ well.order
⎜ no infinite j×(k+1)
⎝ '×' closed in the finites
⎛ define j^(k+1) = (j^k)×j
⎜ not (finite j,k+1,j^k, infinite (j^k)×j)
⎜ no infinite (j^k)×j
⎜ no first.infinite j^(k+1)
⎜ well.order
⎜ no infinite j^(k+1)
⎝ '^' closed in the finites
⎛ define j^^(k+1) = j^(j^^k)
⎝ '^^' closed in the finites
⎛ define j^^^(k+1) = j^^(j^^^k)
⎝ '^^^' closed in the finites
...
----
For each finite.ordinal k, there is
room k
For each room k, there is
swap k⇄k+1
Consider all
finite⇄larger.by.one swaps k⇄k+1,
swapped in order by k
Before any swap,
Bob is in room 0
After swaps 0⇄1 1⇄2 ... k-1⇄k and
before swap k⇄k+1
Bob is in room k
⎛ If Bob is in room k
⎜ them k⇄k+1 is not swapped
⎜
⎜ If Bob is in any room
⎜ then not all swaps are swapped
⎜
⎜ For any claims P and Q
⎜ P⇒¬Q and Q⇒¬P are equally.true and equally.false
⎜
⎜ If all swaps are swapped
⎝ them Bob is not in any room.

Exchange of two entities does never result in
only one of them.

Two is finite.
How.many there are of
all swaps, all rooms, or all finite.ordinals
is more than any finite.ordinal,
is more.than.finite.
After all swaps,
⎛ Bob has never swapped into any darkᵂᴹ room,
⎝ Bob has swapped out of any visibleᵂᴹ room,
Bob is nowhere.