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A.

Hilberts Hotel

Hilbert’s Hotel is full. Guest X arrives. Every guest n moves intro room n+1 synchronously. Guest X moves into room 1.

X and all other guests are in a room.

If A is true, then B is true.

B.

It does not matter if an infinite amount of moves taken simultaneously or an infinite amount of moves taken one after another.

Hilberts Hotel is full. Guest X arrives. X moves intro room 1 and guest 1 moves out of room 1 simultaneously. Guest 1 moves into room 2 and guest 2 to moves out of room 2 simultaneously… After an infinite amount of moves taken, ever guest n is in room n+1.

X and all other guests are in a room.

If B is true, then C is true.

C.

It does not matter if guests moving in a chain or X takes all the moves by himself.

Hilberts Hotel is full. Guest X arrives. X moves into room 1 and guest 1 moves out of room 1 simultaneously. Then they switch back, guests 1 moves back into room 1 and X moves out of room 1 simultaneously. X moves into room 2 and guest 2 moves out of room 2 simultaneously. Then they switch back, guests 2 moves back into room 2 and X moves out of room 1 simultaneously… After an infinite amount of moves taken, every guest n is in room n. X is in a room, too.

X and all other guests are in a room.

If C is true, then D is true.

D.

Example C with X and no other guests

Hilberts Hotel is empty. Guest X arrives. X moves into room 1. X moves into room 2… After an infinite amount of moves taken, X is in a room and all rooms n are empty.

X is in a room. All rooms n are empty in Hilberts Hotel.

Edit:

So some of you point out, that C is false.

It seems like to matter, if I can point out the moving guest. Am I getting it right?

Let’s try again.

If A is true, then E is true.

E.

E. is like B. but with a red hut.

Hilberts Hotel is full. Guest X arrives. Guest X has a red hat and all guests in the hotel have no hat. X moves intro room 1 and guest 1 moves out of room 1 simultaneously. Guest 1 moves into room 2 and guest 2 to moves out of room 2 simultaneously… After an infinite amount of moves taken, ever guest n is in room n+1.

X with his red hat is in a room and all other guests are in a room.

Conclusion: If A is true, then E is true.

F.

F. is like E. but the red hut is moving.

Hilberts Hotel is full. Guest X arrives. Guest X has a red hat and all guests in the hotel have no hat. X moves intro room 1 and guest 1 moves out of room 1 simultaneously. X gives guest 1 the red hat. Guest 1 moves into room 2 and guest 2 to moves out of room 2 simultaneously. Guest 1 gives guest 2 the red hat… After an infinite amount of moves taken, ever guest n is in room n+1.

X is in a room and all other guests are in a room. There is no room with a guest, which has the red hat.

Is then the Conclusion: If E. is true, F. don’t have to be true?

Every guest in E. is moving exactly in the same room as in F. But in F. there is always a guest with a red hat moving, who can never find a room?

Then it seems like, if you can track the last guest somehow, then Hilberts Hotel is not working but if you don’t track the last guy it is working?

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    $\begingroup$ C is false. $ $ $\endgroup$ Commented Oct 7, 2020 at 22:34
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    $\begingroup$ In C: "After an infinite amount of moves taken, every guest n is in room n. X is in a room, too." Really? Why is that? What room exactly is X in? $\endgroup$ Commented Oct 7, 2020 at 23:10
  • $\begingroup$ In C you can not name the room for X, because there is no number for infinity. But infinity exits anyway like in B. If B is true, but when giving a red hut to this example B, then B is wrong? When in B, X would give a red hut to the next guest and this guest would give the hut to next and so on, then B is false? Because the guest with the red hut never finds a room? Then you can can say, name me the guest which has the red hut? $\endgroup$
    – dakiprae
    Commented Oct 8, 2020 at 9:17

1 Answer 1

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There is an essential difference between $B$ and $C$.

In $B$, every guest gets a precise order: "vacate your room and move to room $n$", with $n$ a specific natural number. No one has any problem.

In $C$, the guest $X$ is never done. Whatever room they are in, they can't stay there. If there was a room they were in, it would have a room number, and that room number would be a natural number (since all room numbers are) but every single room with a natural number is already taken by a guest.

This shows that transfinite numbers are not something we should use our intuition for; rather we should rely on formal definitions and formal logic derivations.

The difference between $B$ and $C$ shows that $1 + \omega = \omega \neq \omega + 1$.

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    $\begingroup$ To clarify for the OP, in the last line of this answer ordinal, rather than cardinal, arithmetic is being employed. Of course in cardinal arithmetic we have $1+\aleph_0=\aleph_0=\aleph_0+1$, that's the (initial) point of Hilbert's Hotel. Ordinal arithmetic is finer-grained than cardinal arithmetic, though, and is often a better model for analyzing "infinitary processes" like this one. $\endgroup$ Commented Oct 7, 2020 at 23:16
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    $\begingroup$ (Actually I don't think the last line really fits at all - it's a bit of a stretch to connect that ordinal arithmetic to the actual question, and may confuse the OP at this point in their studies.) $\endgroup$ Commented Oct 7, 2020 at 23:19
  • $\begingroup$ That is fair; although when I was a student, these concepts were taught back-to-back, IIRC (it's been a few decades, so I might misremember). $\endgroup$ Commented Oct 7, 2020 at 23:41
  • $\begingroup$ I edit my question with an example, guest giving each other a red hat. Here has every guest a precise order: "vacate your room and move to room 𝑛" with 𝑛 a specific natural number. No one has any problem. But there in no room can be a guest with a red hat. $\endgroup$
    – dakiprae
    Commented Oct 8, 2020 at 19:43

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