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I am a layman in this field so my understanding of the problem of "Hilbert's Hotel" is limited to the popular version presented to the public.

We know that Hilbert's Hotel can accommodate any finite number of guests; if one more guest arrives, we simply move the guest in room $1$ to room $2$, the guest in room $2$ to room $3$, and so on. If an infinite number of guests come, we move the guest in room $1$ to room $2$, the guest in room $2$ to room $4$, the guest in room $3$ to room $6$, and so on.

However, what happens if we have an infinite number of people leaving the hotel? Is every room still occupied? One way I thought about it is to work backwards from what we do if we add an infinite number of guests to the hotel. If the guest in room $3$ leaves, we just move the guest from room $6$ to room $3$, but.....oooops! The guest in room $6$ is already gone!

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    $\begingroup$ Which guests are leaving? $\endgroup$
    – Chris Culter
    Mar 3, 2016 at 7:50
  • $\begingroup$ @ChrisCulter Can I say that all the guests whose room number is an element of the natural numbers are leaving? Would that make the hotel empty? $\endgroup$
    – Ovi
    Mar 3, 2016 at 7:51
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    $\begingroup$ Oh, all of the current guests are leaving? Then yes, you're right, the hotel becomes empty: there is nobody remaining to put in any room. $\endgroup$
    – Chris Culter
    Mar 3, 2016 at 7:57
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    $\begingroup$ Let the hotel is initially full. If all the guests leave, the hotel becomes empty. If say guests from only odd or only even rooms leave, the hotel is still full. (even though in both cases infinite, countable, amount of guests leave the hotel) $\endgroup$
    – Andrew
    Mar 3, 2016 at 8:06

1 Answer 1

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As mentioned in the comments, it depends on which guests leave. Depending on which guests leave, the number of guests remaining can be any finite number, or infinity. In other words, "$\infty - \infty$" isn't well-defined.

This is related to the following puzzle. Initially you have some dollars. The Devil comes to you and tells you that for every positive integer $n$, when the clock strikes $\frac{1}{2^n}$ minutes before midnight, he will give you two dollars and take away a dollar from you. How much money do you have at midnight?

The curious answer is that while the amount of money you have is increasing up until midnight, at midnight it can be infinite or any finite amount, including zero, depending on which dollars the Devil took from you. In particular it's possible for the Devil to take all of your money.

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    $\begingroup$ Can the devil example be represented as a series of infinately many terms? I onow that you can have the difference of 2 divergent series equalling a finite value, but im interested in how different series ending in different ammounts of money at midnight would look. $\endgroup$
    – Ovi
    Mar 3, 2016 at 20:36
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    $\begingroup$ @Ovi: it can be represented as a series of infinitely many terms, but the terms are functions (called indicator functions), not numbers. This is related to the monotone convergence theorem. The idea for different amounts of money at midnight is that e.g. each time the Devil could give you 2 dollars and take one of the dollars he just gave you, which gives you infinitely many dollars at midnight. But the Devil could also give you 2 dollars and take the oldest dollar you have, which gives you zero dollars at midnight, because eventually any particular dollar you have is taken. $\endgroup$
    – Qiaochu Yuan
    Mar 3, 2016 at 22:30

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