When infinity appears, making room stops being a question of gaps and becomes a question of order. This problem has the rare joy of ideas that seem impossible and yet fit entirely on one line.
Hilbert II's Hotel
Riddle statement
A hotel has infinitely many rooms numbered 1, 2, 3, …, and they are all occupied.
An infinite queue of new guests now arrives, numbered \(g_1, g_2, g_3, \dots\).
How can you accommodate them all?
Show solution
Solution
Answer: Simply move each current guest from room \(n\) to room \(2n\). Then the new guests occupy the odd ones: \(g_1\) goes to 1, \(g_2\) to 3, \(g_3\) to 5, and so on.
Explanation:
The hotel is full, but that does not prevent it from being rearranged. If you send each old guest from room \(n\) to room \(2n\), everyone still has a room and there are no collisions: two different numbers produce different even rooms.
You are then left with all the odd ones free. Since they also form an infinite and countable collection, you can assign them one by one to new guests: \(g_k\) goes to room \(2k - 1\).