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Let $a,b\in (0,1)\subset\mathbb{R}$,

$a=0,a_1 a_2 ...$; $\;b=0,b_1 b_2 ...$

Why is

$\pi:\mathbb{R}²\rightarrow\mathbb{R}: (a,b)\mapsto 0,a_1 b_1 a_2 b_2...$

not bijective, if the constraint $\forall N\in\mathbb{N}\exists i>N:a_i\neq9$ is applied?

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    @GEdga$r$ indeed. But I gue$s$s a de$f$inition can be $f$ound that allows for bijective mappin$g$?!2012-11-04

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It is true that $\mathbb R$ and $\mathbb R^2$ are isomorphic as sets, as groups, or as vector spaces over $\mathbb Q$.

They are not isomorphic as topological spaces, as rings, or as vector spaces over $\mathbb R$ though.

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    @Seirios: No. This result requires the axiom of choice.2012-11-06