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 .


1 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.

  • $\begingroup$ Thank you for your plain answer. You are right, you fail to see the emptyness because the above expression is imprecise: I meant that $(\overline{U})^°\setminus U$ should be empty. I apologize for this. Thank you once again. $\endgroup$
    – LBJFS
    Apr 6, 2019 at 15:48
  • $\begingroup$ You're welcome! $\endgroup$ Apr 6, 2019 at 16:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.