Lemma 1

Let X be a topological space and let A be a subset of X. In order that A be meagre, it is necessary and sufficient that there exist an open covering $(U_i)_{i\in I}$ of X such that $A\cap U_i$ be meagre in $U_i$ for every $i\in I$.

Let $\mathscr{O}$ be the set of open subsets U of X such that $A\cap U$ be meagre. The set of subsets of $\mathscr{O}$ consisting of pairwise disjoint open subsets, ordered by inclusion, is inductive. Let $\mathfrak{U}$ be a maximal element; there exists one by E, III, p. 20, th. 2.

Let O be the union of the open subsets of X belonging to $\mathfrak{U}$. For every open subset $U\in \mathfrak{U}$, let $(B_{U,n})$ be a sequence of rare subsets of U whose union is $A\cap U$. For every integer $n$, the union $B_n$ of the subsets $B_{U,n}$, with U ranging over $\mathfrak{U}$, is rare relative to O (TG, IX, p. 52, prop. 1). By TG, IX, p. 53, prop. 2, $B_n$ is a rare subset of X, since O is open in X. Consequently, the set $A\cap O$, equal to the union of the $B_n$, is a meagre subset of X.

Let F be the complement of O. It is a closed subset of X; let us prove that it has empty interior. In the contrary case, let $x$ be an interior point of F. By assumption, there exists an open neighbourhood V of $x$, which may be supposed contained in F, such that $A\cap V$ be meagre. Then V is disjoint from the open subsets of X belonging to $\mathfrak{U}$ and $\mathfrak{U}\cup \{V\}$ is a set of pairwise disjoint open subsets belonging to $\mathscr{O}$, which contradicts the maximality of $\mathfrak{U}$. The relation

then implies that A is meagre, which was to be proved.

Let X be a topological space.

Recall (TG, IX, p. 69) that a subset A of X is said to be approachable if there exists an open subset U of X such that $U\cap \complement A$ and $A\cap \complement U$ be meagre in X. The set of approachable subsets of X is a tribe which contains the Borel tribe (TG, IX, p. 69, lemma 8 and its proof). A meagre subset is approachable.

For every subset A of X, let D(A) be the set of points $x\in X$ such that for every neighbourhood V of $x,A\cap V$ is not meagre. One also sets $D^*(A) = A\cup D(A)$.