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Is $\mathbb{R}$ equipotent to $\mathbb{R}^2$? thanks for answering.

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up vote 6 down vote accepted

Answer : Yes

Hint : Since $\mathbb{R}$ is equipotent to $]0,1[$, you just have to prove $]0,1[$ is equipotent to $]0,1[^2$. Now you can use decimal expansion and the same kind of trickery you would use to show $\mathbb{N}$ and $\mathbb{N}^2$ are equipotent.

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Personally I detest decimal expansions. I would be interested in other proofs than the standard one (which is in Joel's reply) that avoid this.

(By the way, to show that $|\mathbb{N}\times \mathbb{N}|=|\mathbb{N}|$ I prefer the argument that $(m,n)\mapsto 2^m3^n$ is injective by the fundamental theorem of arithmetic.)

For example, we could use that for an infinite field $K$, its algebraic closure has the same cardinality. (This follows because $|K^n|=|K|$ and then $|K[x]|=|\cup_{n\in\mathbb{N}}K^n|=|\mathbb{N}||K|=|K|$.) Hence $|\mathbb{R}|=|\mathbb{C}|=|\mathbb{R}^2|$.

Who knows a (edit: non-circular) argument?

A nice (general) argument is based on some elementary point-set topology, as described at wikipedia: any compact Hausdorff space with more than two point, and all of whose singletons are non-open, must be uncountable.

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But... isn't this circular? – t.b. May 8 '11 at 19:55
@Theo: yes, $|K^n|=|K|$ is of course what we want to prove. I didn't pay attention. I now recall a better argument, see my edit. – wildildildlife May 8 '11 at 20:04
Well, since you ask for variants: Geometrically much more appealing (for the case of $\mathbb{R}$) than decimal expansion is to use a Peano curve. – t.b. May 8 '11 at 20:06
Cantor-Bernstein-Schroeder is generally the way to go in situations like this where it's awkward to construct a bijection but not too bad to construct surjections or injections. – Qiaochu Yuan May 8 '11 at 20:13
Just because a compact Hausdorff space satisfying properties is uncountable does NOT mean that you know its cardinality. Uncountable means that the cardinality is larger than the smallest infinite cardinal. There are quite a large number of cardinals which satisfy that condition. – Aaron May 8 '11 at 21:17

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