If $A\subset B$ are definable sets, and $A$ is open in $B$, then there exists a definable open $U$ with $U\cap B=A$

Question

This is Lemma 3.4, chapter 1 in the book "Tame Topology and O-minimal Structures." Consider an o-minimal structure $\mathcal{S}=(\mathcal{S}_{n})$ on a dense linearly ordered nonempty set without endpoints, $R$, which is equipped with the interval topology. Let $A\subset B\subset R^{m}$ are definable sets, and $A$ is open in $B$, prove that there is a definable open $U\subset R^{m}$ with $U\cap B=A$.

The book tells that we can take for $U$ the union of all boxes (a Cartesian product of intervals) in $R^{m}$ whose intersection with $B$ is contained in $A$. However, since $\mathcal{S}_{n}$ is a Boolean algebra instead of topological space, I don't know why the union of these boxes which might be arbitrary many is definable. Can someone explain to me a little bit?

Answer 1

I have an answer now:

note the following equivalence:

\begin{align} (x_1,\dots , x_m)\in U&\Leftrightarrow \exists y_1\dots\exists y_m\exists z_1\dots\exists z_m \\ [&(y_1<x_1<z_1)\wedge\dots\wedge (y_m<x_m<z_m)] \wedge\ \\ \big[\forall w_1\dots\forall w_m[&(y_1<w_1<z_1)\wedge\dots\wedge (y_m<w_m<z_m) \wedge (w_1,\dots, w_m)\in B] \\ \rightarrow[&(w_1,\dots, w_m)\in A]\big] \end{align}

Therefore, we have that $U$ is a definable open set.