Actually, the hotel manager could relocate guests from each Nth room to the (2×N)th (every even-numbered room), as there are infinity rooms. This way, there’ll be as infinity free odd-numbered rooms as there are infinity booked guests. Sisyphus can then choose any odd room for himself and another for his boulder.
Wouldn’t you just move the ground floor people in the hotel up one floor and that floors people up one etc to infinity to clear space for the rock?
The ship might be repairable, but that’s not so true of the crew
Actually, the hotel manager could relocate guests from each Nth room to the (2×N)th (every even-numbered room), as there are infinity rooms. This way, there’ll be as infinity free odd-numbered rooms as there are infinity booked guests. Sisyphus can then choose any odd room for himself and another for his boulder.
Person in room 4.3*10^53 when asked to move 4.3*10^53 rooms be like