# Baire Property on Baire Spaces (Kechris' book)

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.

## 1 Answer

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.

• 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 :) – LBJFS Mar 29 '19 at 19:02