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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 $(\implies)$.

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  • $\begingroup$ The second statement is only "for the generic $x \in X$", i.e. for all $x$ in a comeagre set of $X$, not all $x \in A$! $\endgroup$ – Henno Brandsma Apr 2 '19 at 8:48
  • $\begingroup$ @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. $\endgroup$ – LBJFS Apr 11 '19 at 14:02
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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.

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