If $A\subseteq{}^\omega2$ is a Borel set, is $\{x\in{}^\omega2\!:(\exists y\in A)\ x\le y\}$ also a Borel set?

Question

For $x,y\in{}^\omega2$, let us write $x\le y$ iff $(\forall n\in\omega)\ x(n)\le y(n)$. Let $A\subseteq{}^\omega2$ be a Borel set. Does it follow that $B=\{x\in{}^\omega2\!:(\exists y\in A)\ x\le y\}$ is a Borel set?

Answer 1

It doesn't. Consider the $[\mathbb{N}]^\omega$, the Polish space of infinite subsets of $\mathbb{N}$, with the relative topology from $^\omega 2$ (where I'm identifying subsets with characteristic functions). By Exercise (27.16) in Kechris' Classical Descriptive Set Theory, there exists a closed $F\subset [\mathbb{N}]^\omega$ such that $$ F^* := \{H\in[\mathbb{N}]^\omega : \exists H'\in F (H'\subseteq H) \} $$ is $\Sigma^1_1$-complete (hence not Borel). Therefore neither $F^*$ is Borel regarded as a subset of $^\omega 2$. The image of $F^*$ by the homeomorphism $H\mapsto \omega\setminus H$ is $$ \{G\in {}^\omega 2 : \exists H'\in F (H'\subseteq \omega\setminus G) \} $$ which equals $\{x\in{}^\omega2:(\exists y\in A)\ x\le y\}$ when taking $A:=\{\omega\setminus H' \in {}^\omega 2 : H'\in F \}$.

So the implication is false already for $G_\delta$ subsets $A$.