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Consider $\mathbb{R}^n$ as measurable space with the Borel algebra. If $\mathbb{R}^n$ and $\mathbb{R}^m$ are isomorphic (in the category of measurable spaces, i.e. there are measurable maps in both directions, which are inverse to each other), can we conclude $n=m$? Note that this statement is stronger than the invariance of dimension in topology and I doubt that it is true. Can you give a counterexample?

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All uncountable standard Borel spaces are Borel isomorphic (see A. S. Kechris, Classical Descriptive Set Theory, page 90, Theorem 15.6).

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If you use $2^\mathbb{N}$ instead of $\mathbb{R}$, then isomorphism of $n$ and $m$-fold products is easy. And since the Cantor set $2^\mathbb{N}$ measurably embeds in $\mathbb{R}$, the construction in the Schroder-Bernstein theorem lets you soup this up to the genuine $\mathbb{R}$ without need of Axiom of Choice.

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  • $\begingroup$ pretty elementary :). vote $\endgroup$ – Martin Brandenburg Mar 9 '10 at 13:13

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