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I understand how in the infinite hotel 'paradox' moving every person in room $n$ to room $n+1$, and then putting the new quests in room $1$, generates a new space in the countable, but infinite, set.

What I don't understand is why the new guests can't be moved straight to $n+1$, where $n$ is the final room. Or have the current guests at $n$ move to $n+1$ and the new guests move to $n$.

Is it due to $n$ being inaccessible directly, so that you need to build to it iteratively? Is it a genuine requirement?

If it is, then why not build a function $f$ such that:

$$f : X \to X', f(n) = \left\{ \begin{array} {ll} n, \ \text{if room} \ n \ne 0 \\ \text{new family, otherwise} \end{array} \right.$$ (terminating after the second condition is met, and iterating over $X$ in order beforehand)?

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    $\begingroup$ The hotel is supposed to be infinite. $\endgroup$ – WillO Sep 8 '15 at 1:57
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    $\begingroup$ If there are infinitely many rooms, then which room is room $n$? If the entire hotel is full, then how is room $n+1$ empty? $\endgroup$ – Omnomnomnom Sep 8 '15 at 1:58
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    $\begingroup$ If we put the guest into Room $n$ say, and move the current occupant to $n+1$, then we will have to put the current occupant of $n+1$ somewhere, say $n+2$, and then the occupant of $n+2$ will have to be moved, and so on. Remember, the hotel has infinitely many rooms, Room 1, Room 2, Room 3, and so on, and they are all occupied. $\endgroup$ – André Nicolas Sep 8 '15 at 2:00
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    $\begingroup$ @liqiudilk you still have to clarify what $n$ is supposed to be here. There is certainly no integer value of $n$ that makes sense here. $\endgroup$ – Omnomnomnom Sep 8 '15 at 2:02
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    $\begingroup$ There is no final room, but you could start from room n+1 instead of room 1. I.e. Leave everybody in the first n rooms where they are and move n+1 to n+2 etc so creating a vacancy in n+1. $\endgroup$ – Tom Collinge Sep 8 '15 at 11:16
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You're assuming that Hilbert's hotel has an empty "$(n+1)$th" room. Note, however, that the entire hotel is full. So, there are no empty rooms.

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  • $\begingroup$ Answered also in discussion in the comment section above. $\endgroup$ – liqiudilk Sep 8 '15 at 2:18

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