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Suppose that $X$ is an extremally disconnected topological space (meaning that the closure of an open set is still open). Then $X$ has the following property: the family of all sets $S$ such that $S$ differs from a clopen (open and closed simultaneously) set by the set of first category is a $\sigma$-algebra which contains all Borel sets. I wonder whether the assumption of $X$ being extremally disconnected may be weakened—for example to such assumption which allows $X$ to be compact metrizable.

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  • $\begingroup$ So concretely you ask for a compact metrisable $X$ such that $S$ (all sets that differ from a clopen set by a first-category set) is a $\sigma$-algebra (maybe it always is, I'm not sure) and contains all Borel sets? $\endgroup$ Jan 24, 2016 at 6:51
  • $\begingroup$ Yes, but if such space does not exist still I'm interested in finding some examples wth this property which are not extremally disconnected. $\endgroup$
    – truebaran
    Jan 24, 2016 at 14:33

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I claim that if such a space is a Baire space then it is extremally disconnected. Suppose to the contrary that $X$ is a Baire space which is not extremally disconnected. Furthermore, I claim that if $U$ is an open set where $\overline{U}$ is not clopen, then $U\not\in S$.

Suppose that $U$ is open but $\overline{U}$ is not clopen. Let $C$ be a clopen set. If $U\not\subseteq C$, then $U\setminus C$ is a non-empty open set, so $U\setminus C$ is not meager. On the other hand, if $U\subseteq C$, then $\overline{U}\subseteq C$. Since $\overline{U}$ is not clopen, $\overline{U}\neq C$, so $C\setminus\overline{U}$ is a non-empty open set, so $C\setminus\overline{U}$ is not meager. Therefore, $C\setminus U$ is not meager either. We conclude that $U\not\in S$.

Of course, if $X$ is a countable union of nowhere dense subsets, then every set differs from a clopen set by a meager set since every set is meager.

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  • $\begingroup$ Why is $U\setminus C$ clopen? It is open and this seems to be enough. $\endgroup$ Jan 28, 2016 at 13:33
  • $\begingroup$ @Jochen Wengenroth. Yes. $U\setminus C$ is only open. Thanks for pointing that out. $\endgroup$ Jan 28, 2016 at 23:17
  • $\begingroup$ I also noted that but nevertheless the argument is fine. Thank you $\endgroup$
    – truebaran
    Jan 28, 2016 at 23:39

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