Operation over the Baire space $\omega^\omega$ that preserves Borel sets

Question

We have the following operation on sequences of natural numbers (elements of $\omega^\omega$): $$\begin{align}*:\omega^\omega\times\omega^\omega &\longrightarrow \omega^\omega\\ (x,y) &\longmapsto x*y\end{align}$$ with $$x*y(n) = \begin{cases} x(k) & \text{if }n = 2k \\ y(k) & \text{if }n = 2k+1 \end{cases}$$ So $*$ merges two sequences. My question is whether this operation preserves Borel sets, i.e. if given $A,B \in \mathcal{B}(\omega^\omega)$ then $A*B = \{x*y \mid x \in A, y \in B\} \in \mathcal{B}(\omega^\omega)$

Kechris in his Classical Descriptive Set Theory, when talking about Wadge's Game, indirectly mentions this fact without proving it. It is clear that $*$ is continuous, but besides this I'm not able to prove the statement. Any help?

Thanks!

Answer 1

$*$ is a homeomorphism, and if $A$ and $B$ are Borel subsets of $\omega^\omega$, then $A\times B\subseteq \omega^\omega\times \omega^\omega$ is Borel, so $*$ preserves Borel sets.

Answer 2

The map $* : \omega^\omega \times \omega^\omega \to \omega^\omega$ is in fact a homeomorphism and therefore a Borel isomorphism. In particular, Borel sets of the form $A \times B$ for $A, B \in \mathcal B(\omega^\omega)$ are mapped to Borel sets.

Bijection is clear since you are just interlacing two sequences. For continuous inverse, note that if you want to determine the accuracy of the domain up to say $N$ bits in each coordinate, then you simply need $2N + 1$ bits of info in the range.