Re: How many different unit fractions are lessorequal than all unit fractions? (infinitary)

On 10/24/2024 6:58 AM, WM wrote:

On 23.10.2024 20:39, Jim Burns wrote:

On 10/22/2024 2:34 PM, WM wrote:

On 22.10.2024 19:38, Jim Burns wrote:

The description of each element in ℕ
requires its double to also be in ℕ

>
All that is in ℕ, according to your opinion,
is accepted.

>
In the ℕ which is our ℕ,
n ∈ ℕ  ⇔  ∃⟨0,1,...,n-1,n⟩
>
In the ℕ which is our ℕ,
for each j ∈ ℕ, ∃k ∈ ℕ: k = j+1
for each j ∈ ℕ, j=0  ∨  ∃i ∈ ℕ: i+1=j
for each S s ℕ, S = {}  ∨  ∃m ∈ S: m=min.S

>
All that is accepted and doubled.

The subset of those with a double not.in ℕ
holds no minimum.
We see above that
the only subset which holds no minimum is {}
The subset of those with a double not.in ℕ
is {}
The subset of those with a double in ℕ
is ℕ

Bob is in room 0 of our ℕ.Hotel
Swap guests in 0⇄1, 1⇄2, 2⇄3, 3⇄4, ...
After all swaps,
there is no first room Bob is in.
There is no ▒▒▒▒▒ room Bob is in.
'Bye, Bob.

>
Obviously he has occupied a dark room.

ℕ\{0} holds the rooms Bob is swapped into.
For each j ∈ ℕ\{0},
there is a swap j-1⇄j into room j
Are any dark? It doesn't matter.
ℕ holds the rooms Bob is swapped out of.
For each j ∈ ℕ,
there is a swap j⇄j+1 out of room j
There is no room which Bob is swapped into
and which, after all swaps, he isn't swapped out of.
If any of those rooms are dark,
he isn't in those, either.

If you find that the set is complete,
then it is doubled.

>
In a WM.complete ℕ.Hotel,
dark rooms are added for Bob to disappear to
when he isn't in the visible rooms,
repairing 'bye.Bob,
  leaving can't.see.Bob.
>
However,
none of these swaps
move Bob to a dark room.

>
No definable swap.

Definable or not.definable,
I refer to Swaps: 0⇄1, 1⇄2, 2⇄3, 3⇄4, ...
⎛
⎜ for each j⇄j+1 ∈ Swaps, ∃k⇄k+1 ∈ Swaps: k = j+1
⎜ for each j⇄j+1 ∈ Swaps, j=0 ∨ ∃i⇄i+1 ∈ Swaps: i+1=j
⎜ for each S ⊆ Swaps, S = {} ∨ ∃m⇄m+1 ∈ S: m⇄m+1=min.S
⎝
None of them moves Bob into a room which
no other swap moves Bob out of.
After all swaps,
Bob is not in any room he was ever in.
'Bye, Bob.
Note well:
After all _infinitely many_ swaps,
Bob is not in any of _infinitely many_ rooms he was ever in.
Infinite is different from finite.

But in case of completeness of definable rooms,
Bob could pass all rooms
(because in case of completeness all rooms exist) and occupy the last room
(because in case of all rooms
there is a last room necessary to establish completeness).
This can only be prevented by dark rooms.

You are using 'complete' to say 'finite'.
So be it: completeᵂᴹ == finiteⁿᵒᵗᐧᵂᴹ
If any set B is incompleteᵂᴹ (Bob can disappear from B)
then any superset A ⊇ B is incompleteᵂᴹ
Bob disappearing from B disappears Bob from A
Adding dark rooms to an incompleteᵂᴹ ℕ.Hotel
produces a superset of ℕ.Hotel rooms
which are still incompleteᵂᴹ
from which Bob can still disappear.

Bob disappeared
without going to a dark room.

 Impossible with an indestructible Bob.

...in finiteⁿᵒᵗᐧᵂᴹ completeᵂᴹ sets.
{j,j+1} is finiteⁿᵒᵗᐧᵂᴹ
ℕ is infiniteⁿᵒᵗᐧᵂᴹ

Adding dark rooms without Bob in them
does not repair 'bye.Bob.
>
There is no
'bye.Bob.repairing WM.complete Hotel.

>
There is no chance to repair mathematics
when Bob disappears.

The problem with Bob disappearing is that
Bob disappearing is not a problem.
It's not repairable because it's not broken.