md65536 Posted September 1, 2011 Share Posted September 1, 2011 (edited) This puzzle deals with a variation of Hilbert's infinite hotel... http://en.wikipedia.org/wiki/Hilbert_hotel. In the simplest example, the hotel can be full but you can make room for another guest by moving every guest one room over. By induction, you can repeat this infinitely many times, and accommodate an infinite number of new guests. You can also accommodate a countably infinite number of guests all at once by moving every guest in room n to room 2n, freeing up all the infinitely many odd numbered rooms. All the above stuff should be clear to anyone familiar with the paradox. Variation: A hotel with an infinite number of rooms can take up a finite volume of space. For example, if the first room takes up 1 m^3, and each room n+1 takes up half the volume of room n, then the space of all the infinite hotel rooms is 2 m^3. Suppose that each guest is exactly as large as their room. In this case, the hotel is full and it doesn't matter if you can find an empty room or not; the hotel spatially is not infinite and cannot fit any additional guests with non-zero volume. UNLESS... the guests can be compressed! Say each guest in room n can be squished to half their size and fit into room n+1. Then you can free up a room. So here is my paradox. Suppose each guest can be compressed but not to an infinitesimal size. Suppose for any guest in room n, they can be compressed only enough to fit into room 2n. Then, if the hotel starts full but uncompressed, you can free up an infinite number of rooms and accommodate an infinite number of new guests all at once by moving every guest in room n to room 2n. However, if you had instead freed up rooms one at a time by moving every guest in every room n to room n+1, it is clear you could not do this an infinite number of times. So how is it that you can accommodate an infinite number of guests arriving all at once, but not if they arrive one at a time? Is there a flaw in my logic? How do you resolve the paradox? Edited September 1, 2011 by md65536 Link to comment Share on other sites More sharing options...

John Cuthber Posted September 1, 2011 Share Posted September 1, 2011 "Suppose each guest can be compressed but not to an infinitesimal size." Then you can't fill the last room because no guest would fit in it, so there's always a guest left over. With that stipulation the hotel goes out of business even before you try to add more guests (though I'm not sure if that bankruptcy takes zero time, infinite time or finite time) Link to comment Share on other sites More sharing options...

md65536 Posted September 1, 2011 Author Share Posted September 1, 2011 "Suppose each guest can be compressed but not to an infinitesimal size." Then you can't fill the last room because no guest would fit in it, so there's always a guest left over. With that stipulation the hotel goes out of business even before you try to add more guests (though I'm not sure if that bankruptcy takes zero time, infinite time or finite time) Well, there is no last room, but I'd better fix my wording. To fill each of an infinite number of rooms that take up in total a finite space, it would require that the guests can have infinitesimal size. I should have simply said "Suppose each guest can be compressed but that there are limits to how much a particular guest can be compressed." Note that as it's set up (with room n+1 half as big as room n), and allowing guest n to be compressed to fit in room 2n, we must have that an arbitrarily high value of n will allow for an arbitrarily high compression ratio. So we must allow for compression to infinitesimal size. Please consider the problem with the correction that "but not to an infinitesimal size" is removed. The paradox still stands. Not all the guests can keep being compressed indefinitely. Yes, assumptions must be made in any of the Hilbert hotel variations in order to allow them to be possible, but the impossibility of the paradox isn't based on an assumption. You can assume the best case for every assumption. There is no "hidden factor" like money or time. You can assume that an infinite number of people can be checked in in a finite time, for example. Link to comment Share on other sites More sharing options...

John Cuthber Posted September 1, 2011 Share Posted September 1, 2011 That looks like a contradiction " not to an infinitesimal size is removed." "Not all the guests can keep being compressed indefinitely." Why not? Link to comment Share on other sites More sharing options...

md65536 Posted September 1, 2011 Author Share Posted September 1, 2011 That looks like a contradiction " not to an infinitesimal size is removed." "Not all the guests can keep being compressed indefinitely." Why not? Cuz I says so! Literally, I have chosen (and specified adequately I think) the particular details of the example to arrive at the paradox. If the guests can't be compressed at all and they entirely fill the finite space of such a hotel, there is no more spatial room for any new guests*. If they can be compressed indefinitely, then their spatial aspects don't matter at all, and the variation of the hotel is equivalent to the original hotel. In between is a hotel that can take an infinite number of additional guests one way, but not another way (though a simple change would allow both ways). I guess it should be noted that these guests by definition are not all the same. Each of the rooms is a different size, and each of the guests fills their starting room so each guest is a different starting size. * ... of non-zero size. I suppose I should also specify that each guest has a finite size. Note that a spatially full hotel is full, no matter how small a potential new guest is. If the rooms are sizes 1, 1/2, 1/4, 1/8 etc, and so are the guests, then every guest in the hotel has a non-zero size or volume. The hotel is 2 m^3 and the guests take up exactly 2 m^3 and if you add any non-zero positive fraction to two you get more than two. Link to comment Share on other sites More sharing options...

uncool Posted September 29, 2011 Share Posted September 29, 2011 You can accomodate one at a time: For the first guest, move everyone one down and put the guest in room 1. And for the nth guest, move everyone in the 2n-1th room or further down 1, and place the guest in the 2n-1th room. =Uncool- Link to comment Share on other sites More sharing options...

md65536 Posted September 29, 2011 Author Share Posted September 29, 2011 (edited) You can accomodate one at a time: Yes. That's the answer I was looking for. The key to accommodating guests is the assumption that the new guests will be specifically small enough to fit in the rooms (which approach a volume of 0 ie. infinitesimal as the room number, or number of guests, approaches infinity). The original bad algorithm of trying to put each new guest in room 1 means that incoming smaller guests are put in a large room with a lot of wasted space, and the larger guests that we try to move can't fit into the smaller higher-room-number rooms. If we allow the new incoming guests to be similarly compressed as the "original" guests, such that they can be compressed to fit into room 2n, where n is the room they're first put into, then it should also be possible to accommodate an infinite number of infinitely large groups (that is: large in terms of count, not volume), and after each check-in have all of the rooms occupied and spatially full. I think... Edited September 29, 2011 by md65536 Link to comment Share on other sites More sharing options...

## Recommended Posts

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account## Sign in

Already have an account? Sign in here.

Sign In Now