Dimension of the measurable space \$\mathbb{R}^n\$ - MathOverflow 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 Dimension of the measurable space \$\mathbb{R}^n\$ 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 Answer by Pandelis Dodos for Dimension of the measurable space \$\mathbb{R}^n\$ 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 Answer by Gerald Edgar for Dimension of the measurable space \$\mathbb{R}^n\$ 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>