image of homeomorphism Borel?

Question

Let $X,Y$ be a Polish space, $A\subset X$ a Borel subset and $f:A \to B\subset Y$ a homeomorphism. Is $B$ then still Borel in $Y$ ?

Answer 1

Short answer: yes.

Long answer:

Recall $(X,\Sigma)$ is Standard Borel if $\Sigma$ is the $\sigma$-algebra coming from the borel sets of a polish topology (analogous to a polish topology being the open sets coming from some separable complete metric).

If $X$ and $Y$ are Standard Borel spaces, $f : X \to Y$ is Borel, and $f \upharpoonright A$ is injective, then $f[A]$ is Borel.

In your case $X$ and $Y$ are Polish, in particular standard borel, so if $f \upharpoonright A$ is injective, $f[x] = B$ is borel. Thankfully, $f$ is a homeomorphism, so it is (among other things) injective and borel.

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Longer Answer:

This theorem is true because of a fact in descriptive set theory, namely $B$ is Borel if and only if it is the continuous, injective image of (a closed subset of) Baire Space.

To prove the above theorem, then, we write $A = g[C]$ for $C$ a closed subset of Baire Space and $g$ injective. Now $B = f[A] = fg[C]$ is also an injective image of a closed subset of Baire Space.

Proving that borel sets are exactly continuous injective images of closed subsets of Baire Space is somewhat hard, but it can be found in any good descriptive set theory reference.

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Because I can't help myself:

The theorem would be true even if $f$ were just injective and borel! Homeomorphism-ness is very strong, so it makes sense that $B$ should be borel in that setting, but it seems surprising we should be able to get away with so little. The solution is one of my favorite tricks in mathematics, so I feel obligated to mention it here. Often we fix topologies, and try to find functions which behave nicely. Rarely do we fix a function and try to find topologies under which it behaves nicely! However here we have the opportunity to do exactly that!

If $X$ and $Y$ are polish, and $f : X \to Y$ is borel, then we can find a new polish topology on $X$ (with the same borel sets!) rendering $f$ continuous. We then use the characterization of borel sets above to conclude. If you are interested in the actual construction of the new topology (which is quite easy to comprehend), I recommend this article, which does quite a good job explaining it.