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In Hilbert's hotel if we get each guest to his next room (guest in the room $n$ will get in room $(n + 1)$) then the first room will be empty. About this method Wikipedia says:

"By repeating this procedure, it is possible to make room for any finite number of new guests."

But why only finite? Why can't we add countably infinite number of new guests this way?

Hanul Jeon
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    Because you don't want to move any guest an infinite number of places or they won't know when to stop at their new room. – RobertTheTutor Apr 08 '21 at 23:23
  • What room number would you tell each of the current guests to relocate to? –  Apr 08 '21 at 23:28
  • I would do like that: for each guest first I would apply the procedure and then get him to the room 1. –  Apr 08 '21 at 23:29
  • You're going to ask every other guest to move to the "next room over" a countably infinite number of times?! –  Apr 08 '21 at 23:31
  • Got it. Thank you all for the help. –  Apr 08 '21 at 23:33

2 Answers2

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Recall that infinite hotels do not actually exist; the whole story is just a metaphor for constructing functions between various sets. It would be futile to attempt to motivate the conditions in the story by thinking about how infinite hotels "ought to" work if they did exist; at the end of the day which kind of solutions the story should allow is measured purely by whether or not they can be interpreted as valid mathematics.

When one guest arrives, what the story tries to illuminate is that we can construct an bijective function, for example $f: (\mathbb N \cup \{\frac12\})\to\mathbb N$ by defining $$ f(\frac12) = 0,\ f(0) = 1, \ f(1) = 2, \ f(2) = 3, \ \ldots $$

We can run this a finite number of times to get for example $g: (\mathbb N\cup\{\frac12,\frac13,\frac14\})\to\mathbb N$: $$ g(\tfrac14)=0,\ g(\tfrac13)=1,\ g(\tfrac12) = 2,\ g(0) = 3, \ g(1) = 4, \ g(2) = 5, \ \ldots $$

But if we try to make space for an infinite number of new elements by simply repeating the process, trying to make a $h:(\mathbb N\cup\{\frac12,\frac13,\frac14,\frac15,\ldots\}) \to\mathbb N$, then it doesn't work. Nothing ends up mapping to anything in particular, and we don't have any function to look at in the end.

Back in the hotel metaphor we can interpret this problem by demanding that every guest must end up in some room where they can go to sleep undisturbed. Requiring every guest to keep moving for an infinity of times does not achieve that.

Troposphere
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  • Do you think we should tell OP about Banach-Tarski...? –  Apr 08 '21 at 23:46
  • @boojum: Surely you mean So-and-so's theorem of immeasurable measure. – Troposphere Apr 08 '21 at 23:52
  • (heh) Why pure mathematicians and physicists can't communicate... I wonder if the Scottish Cafe had a continuous orange on hand for just such demonstration purposes...? –  Apr 09 '21 at 00:06
  • +1 especially for the first paragraph, which I think gets at the root of many people's confusion about the "Hilbert's hotel" metaphor – Atticus Stonestrom Apr 09 '21 at 00:11
  • But the disjoint union $\Bbb{N} \sqcup \Bbb{N}$ is equipollent with $\Bbb{N}$, so you don't need to make any guest move infinitely many times to make room for a countable infinity of new guests. – Rob Arthan Apr 09 '21 at 00:16
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    @RobArthan: Right. However, the point that confused the OP was why the particular proposed strategy of "just repeat the procedure that worked for one guest, ad infinitum" doesn't achieve that. – Troposphere Apr 09 '21 at 00:18
  • OK, if that's your interpretation of the question. I read it as saying "why can't we use a different process to add countably infinite [new guests]". – Rob Arthan Apr 09 '21 at 00:24
  • @RobArthan I agree that OP's question is ambiguous as written, but see their comment on paul garrett's post, which corroborates Troposphere's interpretation of the question. (in particular, they write "do you agree that the (n -> n+1) method works for a countably infinite number of new guests?") – Atticus Stonestrom Apr 09 '21 at 00:24
  • In particular the words "this way" at the end of the question body are essential to my interpretation of it. – Troposphere Apr 09 '21 at 00:30
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:) Indeed. This can be accomplished all at once, too: for example, by sending guests from room $n$ to room $2n$, all the odd-numbered rooms will be emptied. :)

paul garrett
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  • I know about that method too, but do you agree that (n -> n + 1) method works for countably infinite number of new guests? –  Apr 08 '21 at 23:27
  • @AlexJ: no, see the comment on your question. No specific existing guest would have a specific assigned room number in that case. – abiessu Apr 08 '21 at 23:29
  • I would be a little more uneasy about a proposed procedure to repeat the move-$n$-to-$n+1$ countably-infinitely many times, I think. Somehow asking "infinitely-many individuals" to do just one thing each is (?!?!) less obviously fallacious? And, specifically, as @abiessu comments, doing a literal shift-by-one would seem to put all the guest out of the hotel entirely. So, at the very least, a different description of an iterative process would be necessary. Something that would leave all guests with a (well-defined) room number. It is a fun+productive thing to think about. :) – paul garrett Apr 08 '21 at 23:29
  • I believe what Alex J proposes is an example of a "super-task", but would the hotel guests all want to be involved in it? –  Apr 08 '21 at 23:35