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From: WM <wolfgang.mueckenheim@tha.de>
Newsgroups: sci.math
Subject: Re: How many different unit fractions are lessorequal than all unit
 fractions? (infinitary)
Date: Thu, 24 Oct 2024 12:58:05 +0200
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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.
> 
> 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.
> 
>> 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 0⇄1, 1⇄2, 2⇄3, 3⇄4, ...
> move Bob to a dark room.

No definable swap. 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.

> Bob disappeared
> without going to a dark room.

Impossible with an indestructible Bob.
> 
> 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.

Regards, WM