# Theorem $(8.29)$ (Kechris)

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 $$(\implies)$$.

• The second statement is only "for the generic $x \in X$", i.e. for all $x$ in a comeagre set of $X$, not all $x \in A$! – Henno Brandsma Apr 2 '19 at 8:48
• @Henno Ops, I missed your comment! I hope what I'm going to write is correct: let, by definition of BP, $W\ne\emptyset$ open s.t. $A=^*W$, i.e. the symmetric difference $A\triangle W$ is meager in $X$. If $x\in W$ we are done; if $x\in A\setminus W$ we are done once again because $x\in X$ is generic and $A\setminus W$ is meager. – LBJFS Apr 11 '19 at 14:02

## 1 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.