The canonical open set which is equal to a set with the BP modulo meager sets is regular open (Kechris)

Question

A set $U$ in a topological space $X$ is called regular open iff $U=(\overline{U})^°$.

Exercise $(8.30)$ (Kechris, "Classical Descriptive Set Theory") Prove that $$U(A)=\bigcup \{U\,\text{open}\mid U\Vdash A\}$$ is regular open. Moreover, if $X$ is a Baire space and $A$ has the BP (Baire property), then $U(A)$ is the unique regular open set $U$ with $A=^*U$.

Actually, the first part of this question has already been asked, but the answer doesn't seem to work (at least for me): indeed, what the hint allows to prove is that $A$ is comeager in $(\overline{U(A)})^°$, but how does the emptyness of $(\overline{U(A)})^°\setminus U(A)$ follow?

For the second part, assume that $A=^*V$ for some open regular open set in $X$. I don't see how the assumption should help.

NOTE: I decided to ask for it once again because the user Brian M. Scott seems to be no more active on this site .

Answer 1

The aforementioned hint is that $U(A)\Vdash A$. As you indicate, this allows to show $(\overline{U(A)})^°\Vdash A$.

By the hint and its definition, $U(A)$ it is the largest open set $U$ such that $U\Vdash A$. On the other hand, $U(A)\subseteq (\overline{U(A)})^°$. Hence $U(A) = (\overline{U(A)})^°$ and therefore it is regular open. (I fail to see which emptiness should be relevant here.)

For the second part, assume $A$ the BP and $A=^*V$ with $V$ regular open. Then $V\Vdash A$ and hence $V\subseteq U(A)$.

From $A=^*V$ we also conclude $A\setminus V$ is meager, and hence $U(A)\setminus A \cup A\setminus V \supseteq U(A) \setminus V$ is meager and has empty interior. This shows that $ U(A)\setminus V \subseteq \overline{V}$, and therefore $U(A) \subseteq (\overline{V})^° = V$. We have both inclusions.