Exercise 3, § 5

Let $A$ be a subset of a topological space $X$, and let $D(A)$ denote the set of all $x \in X$ such that, for each neighbourhood $V$ of $x$, the set $V \cap A$ is not meagre. $D(A)$ is contained in $\overline{A}$.

$a$ If $B$ is a subset of $X$ such that $D(B) = \varnothing$, then

for every subset $A$ of $X$.

$b$ $D(A) = \varnothing$ if and only if $A$ is meagre. [Begin by showing that if $(U_i)$ is a family of pairwise disjoint open subsets of $X$ and if, for each $i$, $V_i$ is a nowhere dense set (relative to $X$) contained in $U_i$, then $\bigcup V_i$ is nowhere dense. Then consider a maximal set $\mathfrak{M}$ of open subsets of $X$, each pair of which are disjoint, and such that $A \cup U$ is meagre for each $U \in \mathfrak{M}$; the existence of such a maximal set is established by Zorn's lemma. Show that if $D(A) = \varnothing$, then $A \cap \mathbf{C}W$ is nowhere dense, where $W$ is the union of the sets of $\mathfrak{M}$, and hence show that $A$ is meagre.]

$c$ Show that $D(A)$ is closed and that $A \cap \mathbf{C}D(A)$ is meagre for all $A \subset X$ [show, with the help of $b$], that $A \cap \mathbf{C}D(A)$ is meagre relative to the open subspace $\mathbf{C}D(A)$].

$d$ Show that $D(A)$ is equal to the closure of its interior. [If $D'(A)$ is the closure of the interior of $D(A)$, show that $A \cap \mathbf{C}D'(A)$ is meagre, by observing that this set is the union of