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Is $\mathbb{R}$ equipotent to $\mathbb{R}^2$?

user26857
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fdqsrt
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2 Answers2

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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.

Joel Cohen
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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.

wildildildlife
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  • But... isn't this circular? – t.b. May 08 '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 08 '11 at 20:04
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    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 08 '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 08 '11 at 20:13
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    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 08 '11 at 21:17
  • OK the proof that a perfect set is uncountable actually shows there is a copy of the Cantor set in it. But if you reject writing in binary, now you need to show that the Cantor set has power at least c in some other way. – GEdgar May 09 '11 at 00:11
  • @Aaron: yes, I was not very sharp yesterday. I was at the same time thinking about two things: showing that $\mathbb{R}^2$ and $\mathbb{R}$ are equipotent, and showing that $\mathbb{R}$ is uncountable, both without using decimal expansions. – wildildildlife May 09 '11 at 16:36