In Hilbert's Hotel scenario where an infinite number of new guests arrive, why can't we just kick everyone out and accommodate them again?

Since there is already an infinite number of new guests waiting, kicking an infinite number of guests out of the hotel should leave us with an infinite number of empty rooms and an infinite number of guests waiting for a room. From here we just accommodate everyone again.

Is there a flaw in this logic? I couldn't find an answer to this problem other than the one with moving everyone to a room 2n where n is their current room number.

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    $\begingroup$ Roughly speaking, yes. And since you are asking "why can't", does this mean that you've found references that says you can't? Can you provide a link? $\endgroup$
    – Trebor
    Mar 22, 2022 at 16:17
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    $\begingroup$ The goal is to accommodate the newcomers without kicking anyone out. With finite sets, this is not possible. If you have $20$ occupied rooms and $20$ new quests show up, somebody will go without a room. $\endgroup$
    – lulu
    Mar 22, 2022 at 16:17
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    $\begingroup$ Keep in mind that HH is just a metaphor for a precise mathematical result, or rather sequence of results developing a single theme. So rather than think too much about the story, just think about the results themselves. E.g. the one relevant here is "There is a bijection between $\mathbb{N}$ and $\mathbb{N}\times\{0,1\}$." $\endgroup$ Mar 22, 2022 at 16:18
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    $\begingroup$ That said, re: specifics: your statement "Since there is already an infinite number of new guests waiting, kicking an infinite number of guests out of the hotel should leave us with an infinite number of empty rooms and an infinite number of guests waiting for a room." assumes that "infinity + infinity = infinity" in ... precisely the sense we're investigating! How do you know that the two infinite crowds of hotel-goers don't combine into a "super-infinite" crowd? $\endgroup$ Mar 22, 2022 at 16:19
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    $\begingroup$ You need not kick out anyone. You move every guest to the doubled room number and then the (infinite many) odd-number rooms are empty. Of course we cannot imagine this process since we can neither have infinite many guests nor infinite many rooms. $\endgroup$
    – Peter
    Mar 22, 2022 at 16:37

1 Answer 1


Here's how I think about this. Let's say you kick out all your current guests. You now have two infinite groups of guests, new arrivals and the ones you just kicked out. Just saying you put both of those groups back into rooms isn't rigorous, you still need to define how you order them, or mix them back together.

One way of doing such ordering/mixing is the common answer you have heard. Assign the first room to a kicked-out guest, the second room to a new guest, etc. After repeating this infinitely, all of your guests will have a room.

However, there are orderings/mixings that will not work. Imagine first giving all of the kicked-out guests rooms and then giving the new guests rooms. Even after an infinite number of steps, you still will be giving kicked-out guests rooms and you will have given no rooms to new guests.

Also note there are many other valid orderings (in fact, infinitely many). An infinite family of such orderings are all the orderings where you first give $n \in \mathbb{N}$ rooms to the kicked-out guests and then start giving every other room to a new guest. The reason you have only heard of the alternating assignment ordering is just that it is simple.

TL;DR: Just saying "just accommodate everyone again" is not enough.

The way mathematicians talk about these ideas is in terms of "bijective functions". Bijective functions are mappings between two sets that satisfy an important property. The mapping must map every member of the second set to one and only one member of the first set. It turns out when you have a bijection between two sets, they are the same size. This is the fundamental idea that HH is built to illustrate.

In this case, the alternating room ordering is a bijection between your guests and the hotel rooms, but the other mapping I gave is not a bijection, in fact, it is not even a function, because the new guests never get mapped to a room.

Bijections can be used to show a lot of crazy results that are even more un-intuitive than HH. Two of the most famous examples are showing the integers and the rationals are the same size and Cantor's diagonal argument which shows there is no bijection between the integers and the reals!

Welcome to infinite set theory!

  • $\begingroup$ (+1) for pointing out that the way of assigning the rooms is important. $\endgroup$
    – Peter
    Mar 22, 2022 at 16:39
  • $\begingroup$ Why do we have to assign rooms to those groups separately? Shouldn't those who left just become one group with the ones that arrived? We can't have 2*∞, right? $\endgroup$ Mar 22, 2022 at 16:58
  • $\begingroup$ @CantGetEnough mixing infinite sets is not a trivial task. $\endgroup$ Mar 22, 2022 at 17:02
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    $\begingroup$ @CantGetEnough my point is that showing that $\mathbb{N} \times \{0, 1 \}$ is the same as $\mathbb{N}$ still requires you to create a bijection between the two. You are right that they are the same, but WHY are you right? $\endgroup$ Mar 22, 2022 at 17:07
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    $\begingroup$ @CantGetEnough This is important because there are cases where it is not true that the union of two infinite sets will just be able to fit back into the hotel. For example, imagine you have a continuous bus, where you have $|\mathbb{R}|$ many passengers. Kicking out your $|\mathbb{N}|$ many guests and mixing them with the passengers results in an infinite set that won't fit back into the hotel. See this. $\endgroup$ Mar 22, 2022 at 17:14

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