I'm trying to expand on Hilbert's paradox. The original version states that:

Suppose there is a hotel with a countable infinity of rooms (eg. $\mathbb N$), all of which are occupied.

  • It can accomodate a countably infinite number of new guests (say, $\mathbb N$).
  • It can also accomodate a countably infinite number of coaches with countably infinite guests ($\mathbb N \times\mathbb N$), and so on for every finite number of "layers".

Now, I'm trying to expand on this by imagining a hotel with an uncountable infinity of rooms: say, $\mathbb R$. I successfully managed to "fit" $\mathbb R$ and $\mathbb R^2$ many guests, and in general $\mathbb R^n$ many guests ($n\in\mathbb N$); however, I'm not sure that my process is correct, partly because of the counterintuitive implications (you can map a 3-dimensional space $\mathbb R^3$ to a 1-dimensional segment $[0, 1)$).
This is my reasoning.

Accomodating $\mathbb R$ guests

To accomodate $\mathbb R$ guests, we "map" them to a segment. Albeit I don't know the expression of such a map, I know that it can be done by the following geometrical reasoning:

Let $[0, +\infty)$ be "the first half of $\mathbb R$", so to speak. Draw it as a ray $r$, starting in A.
Draw a segment AB perpendicular to this ray.
Put a point O on the other side of the segment (if the ray is to the left, O is to the right), at the same height of B.
For each point $P\in r$, a segment PO exists. PO crosses AB in a point P'. P' is the projection of P on AB. Each point of $r$ is mapped to another point on the segment PO.
Repeat the above process for the other half of $\mathbb R$.
Take the two segments thus obtained, and put them together.

This segment is equivalent to, say, $(0, 1)$. Therefore, we can just move all positive real numbers by $1$, and then put the new guests in the freed space.

Accomodating $\mathbb R^2$ guests

This one is the one I'm the most unsure about.
To accomodate $\mathbb R^2$ guests, we assign each of them a real, unique identifier (i.e. we map $\mathbb R^2$ to $\mathbb R$, if such a thing is possible); then, we treat them as if we had to accomodate $\mathbb R$ guests, following the proceeding above.

Suppose $\mathbb R$ coaches arrive, each with $\mathbb R$ guests.
Let $C$ be the coach number: $C = C_\infty\:...\:C_3\:C_2\:C_1,\:C_{-1}\:C_{-2}\:C_{-3}\:...\:C_{-\infty}$, where $C_n$ is the $n$-th digit, and $,$ is the decimal comma.
Let $O$ be the guest number: $O = O_\infty\:...\:O_3\:O_2\:O_1,\:O_{-1}\:O_{-2}\:O_{-3}\:...\:O_{-\infty}$.
We'll assign each guest a unique identifier number $I$: $I = C_\infty\:O_\infty\:...\:C_3\:O_3\:C_2\:O_2\:C_1\:O_1,\:C_{-1}\:O_{-1}\:C_{-2}\:O_{-2}\:C_{-3}\:O_{-3}\:...\:C_{-\infty}\:O_{-\infty}$

My question is: are the above reasonings correct? If yes, does it mean we can map each point of eg. a square to a segment?

  • $\begingroup$ We have to exercise some care, since some numbers have $2$ decimal expansions. As it stands your "interleaving decimal expansions" argument is not quite right. It can be fixed, but that takes some work. $\endgroup$ Sep 26, 2014 at 18:42
  • $\begingroup$ @AndréNicolas: can you provide a specific example of such numbers? $\endgroup$ Sep 26, 2014 at 18:47
  • $\begingroup$ $1.230000\dots$ and $1.229999\dots$. That's the only type of example. $\endgroup$ Sep 26, 2014 at 18:48
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    $\begingroup$ Also, you have not really dealt with pairs that may involve a positive and a negative. But that can be first dealt with by mapping to $(0,1)$ in the case of $\mathbb{R}$, and the open square in the case of $\mathbb{R}^2$. $\endgroup$ Sep 26, 2014 at 18:51

1 Answer 1


Yes, you can map each point of a square to a segment. $\mathbb R^n$ and $\mathbb R$ both have the same cardinality.


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