# Hilbert's hotel with uncountably infinite rooms: can you fit $\mathbb R^2$ guests?

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?

• 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. Sep 26, 2014 at 18:42
• @AndréNicolas: can you provide a specific example of such numbers? Sep 26, 2014 at 18:47
• $1.230000\dots$ and $1.229999\dots$. That's the only type of example. Sep 26, 2014 at 18:48
• 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 $\mathbb{R}^2$. Sep 26, 2014 at 18:51

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