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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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Isomorphic as what? –  lhf Nov 3 '12 at 13:55
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I think your constraint is not phrased exactly right. You probably mean that there does not exist N such that $a_i = 9$ for all $i > N$. With your constraint there's no way to write $.09090909\ldots$. –  Noah Snyder Nov 3 '12 at 13:58
    
Wrong term, sorry, I meant, why arent they of the same Cardinality? There is a term for that, I dont know the correct word in english terminology however. –  CBenni Nov 3 '12 at 13:58
    
@NoahSnyder indeed. Sorry. I was correcting it by the time you posted. –  CBenni Nov 3 '12 at 14:00
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CBenni: and you did it incompletely. Your map is not surjective. There is no pair $(a,b)$ that maps to $0.123\overline{49}$ –  GEdgar Nov 4 '12 at 0:05
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1 Answer 1

up vote 12 down vote accepted

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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Interesting! That means $\mathbb{R}$ is not isomorphic to $\mathbb{C}$ as algebraic stuctures with both multiplication and addition, but isomorphic as sets or as groups with + or $\cdot$? I didnt know that, I always assumed $|\mathbb{R}|<|\mathbb{R}²|$ –  CBenni Nov 3 '12 at 14:04
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@CBenni: That is true, as fields they are not isomorphic but simply as sets it follows easily from cardinal arithmetics: $$|\mathbb R|=2^{\aleph_0}=2^{\aleph_0\cdot 2}=(2^{\aleph_0})^2=|\mathbb R^2|$$ –  Asaf Karagila Nov 3 '12 at 14:29
    
Do you know an explicit isomorphism of groups between them? –  Seirios Nov 6 '12 at 16:00
    
@Seirios: No. This result requires the axiom of choice. –  Asaf Karagila Nov 6 '12 at 16:09
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