I'm referring to "Classical Descriptive Set Theory" by Kechris.
I found this question on the same theorem $(8.29)$ pp. $49$, but I need further explanations about both the proof and a consequence of it.
First of all, recall that $U\Vdash A$ means $U\setminus A$ meager in $U$ and that $A=^* U$ means the simmetric difference is meager.
(8.29) Theorem. Let $X$ be a topological space and $A \subseteq X$. Put
$$U(A) = \bigcup \{ U\text{ open} : U \Vdash A \}.$$
Then $U(A) \setminus A$ is meager, and if $A$ has the Baire Property, $A \setminus U(A)$, and thus $A \mathop{\Delta} U(A)$, is meager, so $A =^* U(A)$.
Question 1: the proof begins considering a maximal pairwise disjoint subfamily of $\{U\,\text{open}\mid U\Vdash A\}$.
(i) What does it mean? As I understand it, I have to consider subfamilies consisting of disjoint open sets with the property I require.
(ii) How can I guarantee the existence of a maximal element? The trivial idea is to use Zorn's Lemma, but to do this I have to define a partial order and it's not clear to me which is the right choice.
Question 2: the author says that the above theorem can be expressed in the following way:
For a topological space $X$ and a subset $A$ with the Baire property, we have:
$$x\in A\iff \exists U(x)\,\text{open nhbd}:U(x)\Vdash A.$$
I don't see how to prove $(\implies)$.
Best Answer
Part (1) is quite easy: consider the poset $\mathscr{P}$ of all pairwise disjoint families of open subsets of $U(A):=\{U: U \Vdash A\}$, ordered by inclusion (non-empty, as a singleton subset is trivially pairwise disjoint, so we just need one such $U$ to exist). It's quite trivial to see this is an inductive poset (unions of chains are upperbounds) and so Zorn applies to conclude there is a maximal element $\mathcal{U} \in \mathscr{P}$, which means that $\mathcal{U}$ is pairwise disjoint (so at most countable, as $X$ is separable), for all $U \in \mathcal{U}$ we have $U \Vdash A$ and if $U \notin \mathcal{U}$ and $U \Vdash A$ we cannot have (by maximality) that $\{U\} \cup \mathcal{U}$ is pairwise disjoint, so $U$ intersects some member of $\mathcal{U}$. The union of the $\mathcal{U}$ will then have good properties, like being dense in $U(A)$. It's a common technique to take such maximal p.d. families.