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Many of you have probably heard about Hilbert's Hotel problem. Mr Hilbert owns a hotel with countably infinite amount of one-bed rooms. All the rooms are, of course, taken.

A (finite or infinite) group of k people walks in and wishes for accommodation. However, here comes the tricky part. The current guests are quite tired and Mr Hilbert does not wish to make them move from one room to another. How does he achieve accommodating the new batch of people without moving the already accommodated ones?

This problem, although it's quite known and documented is nowhere to be found here, or anywhere else on the Internet (which is kind of strange).

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  • $\begingroup$ Sending the k people to the hotel in front of the street...without moving the current guests and being all the rooms taken I don't think the newly arrived guests-to-be can be accommodated. $\endgroup$
    – Timbuc
    Commented May 5, 2015 at 16:46
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    $\begingroup$ build a new Hotel? $\endgroup$
    – mandata
    Commented May 5, 2015 at 16:49
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    $\begingroup$ I read through the wiki page a bit and it seems you don't want to assume no guests can be moved. Otherwise all rooms are full and there's nowhere to put a new guest. $\endgroup$ Commented May 5, 2015 at 16:50
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    $\begingroup$ That's a peculiar use of "of course". $\endgroup$ Commented May 5, 2015 at 16:52
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    $\begingroup$ But the difference is, the hotel will never actually be full using that strategy. $\endgroup$ Commented May 5, 2015 at 16:59

2 Answers 2

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My understanding of the Hilbert Hotel is that the hotel has $n$ rooms. Since the number of rooms is infinite a new guest will get the room $n_{+1}$. So if $k$ people arrive the will stay in the rooms $n_{+1},...,n_{+k}$, and there is no need for the other guests to move.

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The usual specification is that the hotel is already full when the new guests arrive. If that is true, you cannot accommodate them without moving the guests that are already there. If we just have to have an infinite number of guests in the hotel and accommodate a number more without moving the existing ones, just put the existing guests in the odd numbered rooms and put the new guests in the even numbered rooms. The question is not clearly stated.

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