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Mark44 said:There is no next-to-last room, any more than there is a last room. "At infinity" doesn't apply here. In a similar vein, there is also no last '9' digit in the number 0.999...

Fair enough. Although I just try to give an abstract idea - I have already mentioned in post #27

QuantumQuest said:As has already been pointed out, there is no next to infinity as there is also no previous to it; in other words you can't add or subtract or do any other math operations at infinity.

I think that I did it in a rather violating way regarding math lingo (referring also to post #28 by @jbriggs444). So, I apologize for it. What I essentially mean is that our counting goes till very close to infinity (not "at infinity" as I said).Now, let me give it in a more formal manner as a continuation of the first paragraph of post #27.

We use the bijection ##f(n) = n + 1## in order to relocate all guests. This holds for the simple variation of one guest arrives each time. For the other variants of the problem we can also create an appropriate bijection. The whole idea is that we can put a set having infinitely many elements into one-to-one correspondence with (any) one of its proper subsets.