4 ms·
I find this hard to follow. The "first" guest (who was in room 1, in the article) moves over one room. The "last" guest (showed as being in 'room n') has no roo
by username91 5y ago
I find this hard to follow. The "first" guest (who was in room 1, in the article) moves over one room. The "last" guest (showed as being in 'room n') has no room to go to; the hotel is full.
- snakeboy 5y agoWell the hotel has [countably] infinite-many rooms, so the nth guest moves to room n+1.
- username91 5y agoRoom n+1's guest, similarly, has nowhere to go; the hotel is full. It has countably infinitely many rooms, and all of them are occupied. I can only see a shift of guests being possible if you momentarily ignore the initial constraints.
- beervirus 5y agoThere is no last guest. There are infinitely many guests.
- username91 5y agoThanks for chiming in. I have trouble with that, though - if there's a "first guest", then I don't see there can't be a "last guest".. "room 1" and "room n" (as the article labels them) are equally arbitrary if there are infinitely many.
- beervirus 5y agoThat’s the difference between all the integers (from negative infinity to positive infinity) vs just the natural numbers (from 0 to positive infinity). You can have infinitely many guests starting from a particular first guest.
- username91 5y agoAh, cool! I can see that, thanks. :) In that case I'd probably rephrase my original objection to say the "next guest" has nowhere to go, except when the they depend on the "next next" (etc.) guest finding a room in a "full" hotel - but I can see how that situation "never" comes up since they can pass the burden along infinitely. I just feel like someday at the end of time, this poor unassuming new guest is gonna get cheated out of a room..
- beervirus 5y agoYep, this is what makes Hilbert's hotel (and infinity in general) counterintuitive. When we try to apply reasoning that's correct for finite sets to infinite sets, very often it becomes incorrect.