Kechris states the following result in "Classical Descriptive Set Theory", pp. $48$:

($\boldsymbol{8.26}$) Proposition Let $X$ be a topological space and suppose that $A\subseteq X$ has the BP (Baire Property). Then either $A$ is meager or there is a nonempty open set $U\subseteq X$ s.t. $A$ is comeager in $U$.

If $X$ is a Baire space, exactly one of these alternatives holds.

Well, I'm not able to prove the last assertion. I've worked on it for some time and maybe I can't something different from what I have already tried (I think the proof is very elementary).

For the convenience of the reader, I recall that

$A$ has the BP iff there is an open set $U\subseteq X$ s.t. the symmetric difference $A\triangle U$ is meager;

$X$ is a Baire space If it satisfies one of the following equivalent conditions: (i) every nonempty open set in $X$ is non-meager; (ii) every comeager set in $X$ is dense; (iii) the intersection of countably many dense open sets on $X$ is dense.

Thank you in advance for your help.


If you've already proved the first assertion, then the second assertion just says that a meager set $A$ can't be comeager in a nonempty open set $U$. Well, suppose $A$ is meager, $U$ is nonempty and open, and $U-A$ is meager. Then $U$, being included in the union of two meager sets $A$ and $U-A$, is meager, contrary to the assumption that $X$ is a Baire space.

  • $\begingroup$ I thought that it was simple, but this is a great blunder! I would like to thank you for this proof... I was in trouble at that moment :) $\endgroup$ – LBJFS Mar 29 '19 at 19:02

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