Hilbert's Grand Hotel: Difference between revisions
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Hilbert's Grand Hotel is |
Hilbert's Grand Hotel is a famous analogy and paradox used to explain the notion of [[countability]]. One starts off by imagining a hotel, with an infinite amount of rooms, and each is occupied. One's intuition says that it's not possible to fit any more people - however, due to the way infinite [[Bijection|bijections]] work and the fact that they go against common sense, it is possible to still fit many more people. |
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Firstly, if there is a single new guest who wants a room, it is possible to accommodate them like so - |
Firstly, if there is a single new guest who wants a room, it is possible to accommodate them like so - the hotel night manager can tell everybody to move up one room, so the person checked into room zero moves to room one, the person checked into room one moves to room two, and so on. Because the set of rooms is never-ending, we don't run out of rooms and everybody who was checked in still has a room. Yet room number zero is now empty - the new guest can check in there. This is analogous to the proof that [[Omega|\(\omega\)]] and \(\omega+1\) are equinumerous. Similarly, if someone in room \(n\) checks out of the hotel, then everybody in room \(m\) for \(m > n\) can move one room to the left, and all the rooms will be filled again. |
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One can also accommodate countably infinitely many new guests, by requiring that every current guest in room \(n\) goes to room \(2n\) and the \(n\)th of the new guests goes to room \(2n+1\). This frees up all the odd-numbered rooms, which the new guests can fill up. Therefore, \(\omega 2\) is equinumerous with \(\omega\). |
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Similarly, if someone in room \(n\) checks out of the hotel, then everybody in room \(m\) for \(m > n\) can move one room to the left, and all the rooms will be filled again. |
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In fact, it's even possible to accomodate a countably infinite collection of countably infinitely many sets of new guests! One can assign the current guest in room \(n\) to room \(2^n\), the \(n\)th guest in the first collection of new guests to room \(3^n\), the \(n\)th guest in the next collection of new guests to room \(5^n\), then \(7^n\), \(11^n\), and so on. Because there are infinitely many prime numbers, and powers of primes never overlap, everybody can be accomodated - even with many rooms now empty, such as room 6, which isn't a power of any prime number! |
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Revision as of 19:03, 24 September 2023
Hilbert's Grand Hotel is a famous analogy and paradox used to explain the notion of countability. One starts off by imagining a hotel, with an infinite amount of rooms, and each is occupied. One's intuition says that it's not possible to fit any more people - however, due to the way infinite bijections work and the fact that they go against common sense, it is possible to still fit many more people.
Firstly, if there is a single new guest who wants a room, it is possible to accommodate them like so - the hotel night manager can tell everybody to move up one room, so the person checked into room zero moves to room one, the person checked into room one moves to room two, and so on. Because the set of rooms is never-ending, we don't run out of rooms and everybody who was checked in still has a room. Yet room number zero is now empty - the new guest can check in there. This is analogous to the proof that \(\omega\) and \(\omega+1\) are equinumerous. Similarly, if someone in room \(n\) checks out of the hotel, then everybody in room \(m\) for \(m > n\) can move one room to the left, and all the rooms will be filled again.
One can also accommodate countably infinitely many new guests, by requiring that every current guest in room \(n\) goes to room \(2n\) and the \(n\)th of the new guests goes to room \(2n+1\). This frees up all the odd-numbered rooms, which the new guests can fill up. Therefore, \(\omega 2\) is equinumerous with \(\omega\).
In fact, it's even possible to accomodate a countably infinite collection of countably infinitely many sets of new guests! One can assign the current guest in room \(n\) to room \(2^n\), the \(n\)th guest in the first collection of new guests to room \(3^n\), the \(n\)th guest in the next collection of new guests to room \(5^n\), then \(7^n\), \(11^n\), and so on. Because there are infinitely many prime numbers, and powers of primes never overlap, everybody can be accomodated - even with many rooms now empty, such as room 6, which isn't a power of any prime number!