Borel sets preserved under open maps? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T13:23:38Z http://mathoverflow.net/feeds/question/106933 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/106933/borel-sets-preserved-under-open-maps Borel sets preserved under open maps? Zhang Jing 2012-09-11T17:35:10Z 2012-09-11T21:58:09Z <p>Given open map f: $R^n$ to $R^n$ such that each open set $U\in R^n$, $f(U)$ is also open. Are Borel sets in $R^n$ preserved under f?</p> <p>Motivation: Pre-image of Borel sets under continuous map is a Borel set in $R^n$.</p> <p>The problem of the analogous statement above is when $U$ and $V$ are open, $f(U\cap V) \subset f(U)\cap f(V)$, but they may not be equal. Is it possible to construct a concrete counter-example to the statement?</p> http://mathoverflow.net/questions/106933/borel-sets-preserved-under-open-maps/106936#106936 Answer by Joel David Hamkins for Borel sets preserved under open maps? Joel David Hamkins 2012-09-11T18:20:44Z 2012-09-11T20:05:02Z <p>Every <a href="http://en.wikipedia.org/wiki/Analytic_set" rel="nofollow">analytic set</a> ($\Sigma^1_1$ set) of reals is the projection of a Borel subset of $\mathbb{R}\times\mathbb{R}$, and the projection map $p(x,y)\mapsto x$ is an open map. So the standard examples of non-Borel $\Sigma^1_1$ sets are also examples where Borel sets are not preserved by an open map $\mathbb{R}^2\to\mathbb{R}$.</p> <p>But you asked for an open map $\mathbb{R}^n\to\mathbb{R}^n$, with the same domain and codomain, and the reasoning above concerned only an open map $\mathbb{R}^2\to\mathbb{R}$. Here is one way to fix the issue and make an open map $\mathbb{R}^3\to\mathbb{R}^3$ having the image of a Borel set being non-Borel. Let $h:\mathbb{R}\to\mathbb{R}^2$ be any function whose restriction to every open interval is onto. One can make such a function by using Cantor's interleaving digits trick, combined with the idea of <a href="http://en.wikipedia.org/wiki/Conway_base_13_function" rel="nofollow">Conway's base 13 function</a>. This function is an open map, since every nonempty open set maps onto the whole space. Now, define $f(x,y,z)=(x,z_0,z_1)$, where $h(z)=(z_0,z_1)$. It is easy to see that the function $f$ is an open map. Meanwhile, every analytic set $A$ has the form $x\in A\iff \exists y B(x,y)$, where $B\subset\mathbb{R}^2$ is a Borel set. Let $C=B\times\mathbb{R}$, which is Borel. Consider the image set $f[C]$, and note that $(x,0,0)\in f[C]$ if and only if there is some $y$ such that $(x,y)\in B$, since in this case we will find a $z$ with $h(z)=(0,0)$; hence, $(x,0,0)\in f[C]$ if and only if $x\in A$, and so the intersection of $f[C]$ with the $x$-axis is $A$, a non-Borel set. So $f[C]$ cannot be Borel if $A$ is not. So this is a case where we have an open map $f:\mathbb{R}^3\to\mathbb{R}^3$ taking a Borel set to a non-Borel set.</p> http://mathoverflow.net/questions/106933/borel-sets-preserved-under-open-maps/106937#106937 Answer by Gerald Edgar for Borel sets preserved under open maps? Gerald Edgar 2012-09-11T18:20:50Z 2012-09-11T21:58:09Z <p>No. Projection from $\mathbb R^2$ onto $\mathbb R$ is open, but the image of a $G_\delta$ set can be any analytic set. So there are non-Borels among them. See <a href="http://en.wikipedia.org/wiki/Analytic_set" rel="nofollow">http://en.wikipedia.org/wiki/Analytic_set</a> </p> <p><strong>edit</strong> Not a counterexample from a space to itself as required. </p> <p>So a counterexample has to be an open map $\mathbb R^n$ to itself, but <em>not</em> at-most-countable-to-one, since those maps do preserve Borel, as I recall.</p>