[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?

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

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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).