# [Points in space] Pairing function for reals (Set cardinality problem) [closed]

This might be a simple question, but are there "more" points on a line in (3D) space than on a plane? Or in more mathematical terms: Do $\mathbb {R}$ and $\mathbb {R}^2$ have equal cardinalities (which $\mathbb {N}$ and $\mathbb {N}^2$ have).

If this is true, what could a bijection/pairing function ($\pi \colon\ \, \mathbb {R}^2 \to \mathbb {R}$) for reals look like?

-

## closed as too localized by Reid Barton, Chris Schommer-Pries, Scott Morrison♦Feb 19 '10 at 0:00

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Yes. In fact, assuming the axiom of choice, any infinite set A has the property that A^2 = A. See in particular en.wikipedia.org/wiki/… – Qiaochu Yuan Feb 18 '10 at 18:20
Probably not the right place for this question. The answer: yes, R and R^2 have the same cardinal. But it is interesting that Cantor reports thinking "I see it, but I don't believe it" when he first proved it. But for R the Axiom of Choice is not required. – Gerald Edgar Feb 18 '10 at 18:23
@darios: Chris has deleted his answer, so you may not have seen my response to your question about it. Any real number, transcendental or not, has a binary expansion which is unique if we require that it does not end in a string of 1s. In particular, the number of binary expansions is uncountable. The problem with Chris' strategy (interweaving digits) is that the slight non-uniqueness of binary expansion is trickier to handle than it seems at first glance, which is why I think it's easier just to argue by Cantor-Bernstein-Schroeder. – Qiaochu Yuan Feb 18 '10 at 18:35
It may be possible to fix the answer I gave by dealing with the non-uniqueness more carefully, but I decided to delete my answer because I don't think this question is exactly the right level for MO. Also Qiaochu Yuan is right that it is easier to just use a more general argument, anyway. – Chris Schommer-Pries Feb 18 '10 at 18:43
Closed, see Gerald's comment above. (Also, there's a satisfactory answer by now.) – Scott Morrison Feb 19 '10 at 0:00

You might want to look into space filling curves, which were first described by Peano and Hilbert in the late 1800's. These are continuous surjections from $[0,1]$ onto $[0,1]^2$ (and higher powers) but they are not bijections. However, they are visualizable to a certain extent. A quick Google search gave a lot of hits, in particular this one at Cut The Knot which has an illustrative java applet.
As for the existence of a bijection, you can derive it from the fact that $\aleph_0\cdot2 = \aleph_0$ and the usual exponent rules: $$(2^{\aleph_0})^2 = 2^{\aleph_0\cdot2} = 2^{\aleph_0}$$ It is also easy to write an explicit bijection between Cantor space $\{0,1\}^{\mathbb{N}}$ (the space of infinite binary sequences) and its square by splitting the even and odd coordinates. This, together with a bijection between $\mathbb{R}$ and $\{0,1\}^{\mathbb{N}}$, gives what you want. Note that it is this last bijection which is harder to visualize. The reason is that $\mathbb{R}$ is connected while $\{0,1\}^{\mathbb{N}}$ is totally disconnected (with the product topology).