most recent 30 from http://mathoverflow.net 2013-05-25T07:44:43Z http://mathoverflow.net/feeds/question/17472 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/17472/dimension-of-the-measurable-space-mathbbrn Martin Brandenburg 2010-03-08T13:49:03Z 2010-03-08T16:45:23Z

<p>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?</p>

http://mathoverflow.net/questions/17472/dimension-of-the-measurable-space-mathbbrn/17475#17475 Pandelis Dodos 2010-03-08T13:58:48Z 2010-03-08T13:58:48Z

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

http://mathoverflow.net/questions/17472/dimension-of-the-measurable-space-mathbbrn/17493#17493 Gerald Edgar 2010-03-08T16:41:18Z 2010-03-08T16:41:18Z

<p>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.</p>