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.
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.