Product of Borel sigma algebras - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T17:38:06Z http://mathoverflow.net/feeds/question/39882 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/39882/product-of-borel-sigma-algebras Product of Borel sigma algebras Bill Johnson 2010-09-24T18:11:10Z 2012-11-30T21:15:52Z <p>If $X$ and $Y$ are separable metric spaces, then the Borel $\sigma$-algebra $B(X \times Y)$ of the product is the the $\sigma$-algebra generated by $B(X)\times B(Y)$. I am embarrassed to admit that I don't know the answers to: </p> <p>Question 1. What is a counterexample when $X$ and $Y$ are non separable?</p> <p>Question 2. If $X$ is an uncountable discrete metric space, does $B(X)\times B(X)$ generated the Borel $\sigma$-algebra on $X \times X$?</p> <p>Question 3. If $X$ and $Y$ are metric spaces with $X$ separable, does $B(X)\times B(Y)$ generated the Borel $\sigma$-algebra on $X \times Y$?</p> http://mathoverflow.net/questions/39882/product-of-borel-sigma-algebras/39883#39883 Answer by Gerald Edgar for Product of Borel sigma algebras Gerald Edgar 2010-09-24T18:20:10Z 2010-09-24T18:20:10Z <p>Q1. Discrete spaces with cardinal > c ... then the diagonal is a Borel set, but not in the product sigma-algebra.</p> <p>This also answers Q2 (no)</p> <p>but not Q3.</p> http://mathoverflow.net/questions/39882/product-of-borel-sigma-algebras/39886#39886 Answer by Byron Schmuland for Product of Borel sigma algebras Byron Schmuland 2010-09-24T18:38:34Z 2010-09-24T18:38:34Z <p>The answer to question 3 is yes. At least according to Lemma 6.4.2 of the second volume of Bogachev's book "Measure Theory".</p> <p>He requires both spaces to be Hausdorff and one of them to have a countable base. They need not be metric spaces. </p> http://mathoverflow.net/questions/39882/product-of-borel-sigma-algebras/81491#81491 Answer by Michael Greinecker for Product of Borel sigma algebras Michael Greinecker 2011-11-21T10:37:43Z 2011-11-21T10:37:43Z <p>To close a gap: From the answer of Gerald Edgar, we know that the answer to the second question is no if the spaces involved have cardinality larger than $\mathfrak{c}$. This leaves open what happens when they <em>do</em> have cardinality $\mathfrak{c}$. The answer is yes under the continuum hypothesis, and in general it holds that $2^{\omega_1}\otimes 2^{\omega_1}=2^{\omega_1\times\omega_1}$. This was shown in </p> <p>B. V. Rao, <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.bams/1183530570" rel="nofollow">On discrete Borel spaces and projective sets</a> Bull. Amer. Math. Soc. Volume 75, Number 3 (1969), 614-617. </p> <p>In Bogachev's remarkable book, it can be found as Proposition 3.10.2. </p> http://mathoverflow.net/questions/39882/product-of-borel-sigma-algebras/115024#115024 Answer by Martin for Product of Borel sigma algebras Martin 2012-11-30T21:10:38Z 2012-11-30T21:15:52Z <p>This is should probably rather be a comment to Michael Greinecker's answer, but I do not have the necessary privileges.</p> <p>Michael Greinecker's answer leaves open what happens with a continuum-sized discrete space when one does <em>not</em> assume the continuum hypothesis.</p> <p>Arnold W. Miller showed in section 4 of <a href="http://www.math.wisc.edu/~miller/res/hier.pdf" rel="nofollow">On the length of Borel hierarchies</a> that it is consistent relative ZFC that no universal analytic set $U \subset [0,1] \times [0,1]$ belongs to the product $\sigma$-algebra $\mathcal{P}[0,1] \otimes \mathcal{P}[0,1]$. Combined with Rao's result mentioned by Michael Greinecker, this shows that $2^{\mathfrak{c \times c}} = 2^{\mathfrak{c}} \otimes 2^\mathfrak{c}$ is independent of ZFC.</p> <p>See my answer to <a href="http://math.stackexchange.com/q/177416" rel="nofollow">Universally measurable sets of $\mathbb{R}^2$</a> on math.stackexchange.com for related results and more details and references.</p>