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Consider a Polish space $(X,d)$ and any metric space $(Y,e)$. If we have a continuous surjection $f:X\to Y$ then is the image $f(U)$ of any open subset $U\subset X$ a Borel set in $Y$?

I know that this is true if we allow $X$ to be compact, since every open subset of a metric space is $F_{\sigma}$ and closed subsets of a compact metric space are compact, and moreover continuous images of compact sets are compact. So in this case, the image of an open set would be $F_{\sigma}$ in $Y$.

If this is not the case for the given $f$, would it do any difference if we allowed $Y$ to be in addition Polish? Or compact? Or should $f$ have more properties? How about $X$? Or is this all just a triviality that I have overlooked?

Thanks for all the input in advance.

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As noted, this is not true in general. It might be worth pointing out that the answer becomes "yes" if $f(U)$ is disjoint from $f(X\setminus U)$ by an application of Suslin's theorem that Borel subsets of an analytic space are precisely the analytic/coanalytic sets. – user83827 Jun 18 '12 at 12:20
A slight extension of your observation is sometimes useful: if $U$ is a countable union of compact sets then $f(U)$ is Borel. In particular, if $X$ is locally compact and Polish then images of open sets are Borel. – t.b. Jun 18 '12 at 12:49
@t.b. True, that would do the job. Thanks. – T. Eskin Jun 18 '12 at 13:03
up vote 10 down vote accepted

The answer is no. Endow both $\mathbb{R}^2$ and $\mathbb{R}$ with the usual Borel $\sigma$-algebra and let $\pi_1:\mathbb{R}^2\to\mathbb{R}$ be the projection onto the first coordinate. Now it is a well known fact that there exists Borel sets $B\subseteq\mathbb{R}^2$ such that $\pi_1(B)$ is not Borel. Now $B$ will be in general not open, but we can make it so. We just apply the following theorem (Lemma 4.58 in Alipranits & Border 2006):

Let $\mathcal{C}$ be a countable family of Borel subsets of a Polish space $(X, \tau)$. Then there is a Polish topology $\tau'\supseteq\tau$ on $X$ with the same Borel $\sigma$-algebra for which each set in $\mathcal{C}$ is both open and closed.

In particular, we can make $B$ open. Then $B$ is open and $\pi_1$ is a continuous surjection, but $\pi_1(B)$ not Borel. Since one can take $B$ in the original space to be bounded, the example works even when $Y$ is compact.

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Wouldn't it be easier to appeal to the definition given on Wikipedia that analytic sets are precisely the continuous images of the Baire space $\mathbb{N^N}$ (which is open in itself) and using the fact that not all analytic sets are Borel? – t.b. Jun 18 '12 at 12:45
Uhm...yes.$~~~~~$ – Michael Greinecker Jun 18 '12 at 12:49
@MichaelGreinecker: Thanks! – T. Eskin Jun 18 '12 at 13:04
Oh, now I see: I missed the hypothesis that $f$ be onto $Y$, then you need indeed a bit more than what I said. Sorry! – t.b. Jun 18 '12 at 13:10
@t.b.: An easy modification of that direct argument works, though. Fix two non-Borel analytic sets $X_1$ and $X_2$ whose union is $X$ (this is easy to do, just make sure that each contains some designated Borel half of the space). Then surject a copy of Baire space onto $X_1$, $X_2$ respectively, and use the fact that Baire space is homeomorphic to two copies of itself to get a surjection to $X$ such that the image of each of the clopen subcopies of Baire space is properly analytic. – user83827 Jun 18 '12 at 15:51

I see that Michael Greinecker has preempted the idea on which I was working; I post this only because it contains a little more detail about the retopologizing, since I didn’t know the general lemma that Michael quotes and had to make my own way at that point.

Let $\mathscr{N}=\omega^\omega$ with the product topology $\tau$. Let $d$ be any complete metric on $\omega^\omega$ that generates $\tau$ and is bounded by $1$. Let $F$ be a closed subset of $\mathscr{N}$, and define a new metric $\rho$ on $\omega^\omega$ as follows:

$$\rho(x,y)=\begin{cases} d(x,y),&\text{if }x,y\in F\text{ or }x,y\in\mathscr{N}\setminus F\\ 2,&\text{otherwise}\;. \end{cases}$$

It’s easy to check that $\rho$ is a metric. Let $\tau_\rho$ be the topology on $\omega^\omega$ generated by $\rho$, and let $X$ be $\omega^\omega$ with the topology $\tau_\rho$. Clearly $F$ is clopen in $X$. Any $\rho$-Cauchy sequence must be eventually in $F$ or in $X\setminus F$ and must therefore be $d$-Cauchy, so $\langle X,\rho\rangle$ is complete. Moreover, $\tau_\rho$ is just the join of $\tau$ and $\{\varnothing,F,\mathscr{N}\setminus F,\mathscr{N}\}$, so it’s second countable, and $X$ is therefore a Polish space in which $F$ is a clopen set.

Now let $A\subseteq\mathscr{N}$ be analytic but not Borel. Because $A$ is analytic, there is a closed $H\subseteq\mathscr{N}\times\mathscr{N}$ such that $A=\pi_0[H]$, where $\pi_0:\mathscr{N}\times\mathscr{N}\to\mathscr{N}:\langle x,y\rangle\mapsto x$. Let $h:\mathscr{N}\to\mathscr{N}\times\mathscr{N}$ be a surjective homeomorphism, and let $F=h^{-1}[H]$. Apply the construction of the first paragraph to this $F$. Then $f\triangleq\pi_0\circ h:X\to\mathscr{N}$ is continuous, and $f[F]=A$, so $A$ is a non-Borel image of the clopen set $F$ in the Polish space $X$.

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Thanks Brian, this answer is also very enlightening. – T. Eskin Jun 18 '12 at 18:20

On the other hand if $Y$ is Polish and $f$ is a countable to one Borel map, then the image of every Borel set is Borel.

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