Big Maths Ideas · Beta

Hilbert’s hotel and infinity

Move guests to make room in a hotel with infinitely many occupied rooms.

Imagine rooms 1, 2, 3, …, all occupied. Give each old guest a unique new room.

n↦n+1n\mapsto n+1
Imagine rooms 1, 2, 3, …, all occupied. Give each old guest a unique new room.Old guests (first eight shown)1234567812345678910111213141516…Blue: old guests. Red: rooms for new guests.

Every old guest gets a distinct room. The rule continues beyond the visible rooms.

New guests
1
New guest m goes to
Room m, for 1 ≤ m ≤ 1
The idea behind the experiment

Hilbert’s hotel is a thought experiment about countable infinity. It has rooms 1, 2, 3, … with no last room. The finite picture shows only the beginning of an infinite one-to-one mapping. A real finite hotel cannot use this rule to create space.

Harvey Mudd on one-to-one pairings
Predict, test and explain

Step 1 of 3

Predict

If every old guest n moves to room 2n, which rooms are free for new guests?

InvestigateHow can every occupied room move to make room for one more guest?

Explore and compare

Choose the investigation above. Predict what will change before you move a control.

Try it in the simulation. Change one thing at a time.

Notes and saved values stay in this tab. They disappear when you reload.

Hints, self-check and connections
Try a self-check

Fixed values, separate from the diagram controls.

If every old guest n moves to room 2n, which rooms are free for new guests?

n↦2nn\mapsto2n

Hint

Write a rule that gives each old guest a distinct room.

Reveal answer

All old guests occupy even-numbered rooms; every odd-numbered room becomes free.

odd rooms free\text{odd rooms free}

Build an explanation

Hint 1

Write a rule that gives each old guest a distinct room.

Hint 2

The hotel has no last room.

Hint 3

The drawing shows only the beginning of an infinite mapping.

Worked example

This example uses fixed values, separate from the diagram controls.

Make room for countably infinitely many new guests.

  1. Move old guest n to room 2n.
  2. Send new guest m to room 2m − 1.
  3. Even and odd rooms never overlap.

Answer: Every old and new guest has a distinct room.

Connect this idea

Watch out: The old guests use even rooms, so odd rooms remain for new guests.