A hotel has infinite 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?
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Hilbert II's Hotel
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.
Hints
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- Don't try to find a free room: make it by displacing current guests.
- A good strategy is to send all former guests to an infinite subset of rooms.
- If the old ones occupy all the even ones, the odd ones are free for the new ones.
Solution
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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 doesn't stop us from reorganizing it. - The paired rooms are infinite.
- The odd rooms are also infinite. If you send each old guest from \(n\) to \(2n\), everyone still has a room and there are no collisions, because two different numbers produce different even rooms. Then all the odd ones are free. And since they also form a countable infinite collection, you can assign them one by one to new guests. The central idea is that an infinite set can be redistributed so that an infinite part remains free.
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