The projection of a Borel set of $\mathbb{R}^n$ needn't be Borel, although the projection of a closed set is a countable union of compact sets, hence Borel.
My question is if $A,B \subset \mathbb{R}^n$ are Borel such that $\pi_k(A), \pi_k(B)$ are known to be Borel, is it true $\pi_k(A \cap B)$ is Borel?
I haven't found an easy way to prove this, and can't find an answer, but it feels possible because intersecting seems limited in how much it can sabotage the projections. I proved a geometric theorem about such sets/projections, but lack that the projections are Borel.