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Question: How to show that the additive groups $\mathbb{R}^n$ for $n\geq 1$ are all isomorphic (as abstract groups).

I've read this statement in the comments of this mathoverflow question. We can certainly find set theoretic bijections $\mathbb{R}\to\mathbb{R}^n$ because both sets have the same cardinality, but how do we ensure that it respects addition?

Community
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Paul
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1 Answers1

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You can do this with the axiom of choice.

As a $\mathbb{Q}$-vector space, $\mathbb{R}^n$ has the same dimension as $\mathbb{R}$. Therefore you can find a $\mathbb{Q}$-linear bijection between $\mathbb{R}$ and $\mathbb{R}^n$, and this map gives an isomorphism of abelian groups.

spin
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  • Are you saying that $\dim_{\mathbb{Q}}\mathbb{R}^n=\operatorname{card}(\mathbb{R})$ for all $n\geq 1$? – Paul Sep 02 '16 at 13:23
  • @Paul: Yes. You can show $\dim_\mathbb{Q} \mathbb{R} = \operatorname{card}(\mathbb{R})$, and then $\dim_\mathbb{Q} \mathbb{R}^n = \operatorname{card}(\mathbb{R}) \times\operatorname{card}(\mathbb{R})\times \cdots \times \operatorname{card}(\mathbb{R})$ ( $n$ times) $= \operatorname{card}(\mathbb{R})$ – spin Sep 02 '16 at 13:28
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    How do you multiply cardinalities? Do you rather mean $\dim_{\mathbb{Q}}\mathbb{R}^n=\operatorname{card}(\mathbb{R}^n)=\operatorname{card}(\mathbb{R})$? – Paul Sep 02 '16 at 13:47
  • Product of cardinals is defined in the obvious way: $\alpha\beta$ is the cardinal of the cartesian product, which, as soon as at least one is infinite (and the other nonzero), equals $\max(\alpha,\beta)$. – YCor Sep 04 '16 at 23:09