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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?

Motivation: Pre-image of Borel sets under continuous map is a Borel set in $R^n$.

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?

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  • $\begingroup$ You need injectivity for the mapping f. $\endgroup$
    – Changyu Guo
    Commented Oct 12, 2012 at 6:26

2 Answers 2

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Every analytic set ($\Sigma^1_1$ 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 $\mathbb{R}^2\to\mathbb{R}$.

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 Conway's base 13 function. 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.

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  • $\begingroup$ But I admit, this doesn't quite get a counterexample with an open map $\mathbb{R}^n\to\mathbb{R}^n$ as requested... $\endgroup$
    – Joel David Hamkins
    Commented Sep 11, 2012 at 18:38
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    $\begingroup$ If one uses Baire space or Cantor space, which are homeomorphic to their squares, rather than $\mathbb{R}$, then one can easily fold in another dimension to transfer this example to have the same domain and codomain. $\endgroup$
    – Joel David Hamkins
    Commented Sep 11, 2012 at 18:56
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    $\begingroup$ I have realized how to do it with $\mathbb{R}^n\to\mathbb{R}^n$ itself, and edited the answer. $\endgroup$
    – Joel David Hamkins
    Commented Sep 11, 2012 at 20:00
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    $\begingroup$ Here's a cheap trick that gives an open $f \colon \mathbb{R} \to \mathbb{R}$. Let $E$ be Vitali equivalence on $\mathbb{R}$, and let $g \colon \mathbb{R}^2 \to \mathbb{R}$ be any Borel function sending distinct points to $E$-unrelated points. Then define $f \colon \mathbb{R} \to \mathbb{R}$ by $f(x) = y$ if $\exists z\ (x \mathrel{E} g(y,z))$, and say $f(x) = 0$ if no such $(y,z)$ exists. This function is open (since the image of any open set is $\mathbb{R}$ by density of $E$-classes). [cont.] $\endgroup$
    – Clinton Conley
    Commented Sep 11, 2012 at 20:39
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    $\begingroup$ Now if $B \subseteq \mathbb{R}^2$ is any Borel set with non-Borel projection, then $g[B]$ is again Borel (as $g$ is injective), but $f[g[B]]$ is the projection of $B$, which was non-Borel. $\endgroup$
    – Clinton Conley
    Commented Sep 11, 2012 at 20:39
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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 http://en.wikipedia.org/wiki/Analytic_set

edit Not a counterexample from a space to itself as required.

So a counterexample has to be an open map $\mathbb R^n$ to itself, but not at-most-countable-to-one, since those maps do preserve Borel, as I recall.

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    $\begingroup$ The open map was to be from $\mathbb{R}^n$ to itself. See Joel's second comment below his answer. $\endgroup$
    – Todd Trimble
    Commented Sep 11, 2012 at 19:38

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