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Aug 19 at 0:44 comment added 183orbco3 There is even a continuous $f$, using a continuous open map $h:\mathbb{R}^3\rightarrow\mathbb{R}^4$
Sep 13, 2012 at 16:11 vote accept Jing Zhang
Sep 11, 2012 at 20:39 comment added Clinton Conley 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.
Sep 11, 2012 at 20:39 comment added Clinton Conley 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.]
Sep 11, 2012 at 20:05 history edited Joel David Hamkins CC BY-SA 3.0
added 14 characters in body
Sep 11, 2012 at 20:00 comment added Joel David Hamkins I have realized how to do it with $\mathbb{R}^n\to\mathbb{R}^n$ itself, and edited the answer.
Sep 11, 2012 at 19:59 history edited Joel David Hamkins CC BY-SA 3.0
Making domain and codomain match
Sep 11, 2012 at 18:56 comment added Joel David Hamkins 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.
Sep 11, 2012 at 18:38 comment added Joel David Hamkins But I admit, this doesn't quite get a counterexample with an open map $\mathbb{R}^n\to\mathbb{R}^n$ as requested...
Sep 11, 2012 at 18:20 history answered Joel David Hamkins CC BY-SA 3.0