Lemma 2

Let X be a topological space and let A be a subset of X.

a) The set D(A) is closed in X; its complement is the largest open subset U of X such that $A\cap U$ be a meagre set.

b) In order that A be meagre, it is necessary and sufficient that D(A) be empty.

c) The set $D^*(A)$ is approachable.

d) For every approachable set B of X containing A, $D^*(A)\cap$ $\complement B$ is meagre.

Let U be the union of the open subsets V of X such that $A\cap V$ be meagre. In order that a point $x$ belong to U, it is necessary and sufficient that it possess a neighbourhood V such that $A\cap V$ be meagre. Thus one has $U =\complement D(A)$ and therefore D(A) is a closed subset of X. By construction, the subspace U has a covering by open subsets whose intersection with A is meagre. By lemma 1, applied to the topological space U and to the subset $A\cap U$, $A\cap U$ is meagre in U, hence also in X. This proves assertion a), since, by construction, every open subset V of X such that $A\cap V$ be meagre is contained in U.

Assertion b) follows immediately from this.

c) One has $D^*(A) = A\cup D(A) = (A\cap U)\cup D(A)$. The set D(A) is closed, hence approachable (IV, p. 361); the meagre subset $A\cap U$ is so as well. Hence $D^*(A)$ is approachable (loc. cit.).

d) Let B be an approachable subset of X containing A. Its complement $\complement B$ is then an approachable subset of X (loc. cit.) and there therefore exists an open subset V of X such that $V\cap B$ and $\complement V\cap \complement B$ are meagre. Since $A\subset B$, $V\cap A$ is still meagre, whence $V\subset U$. Since $\complement U\cap \complement B$ is contained in $\complement V\cap \complement B$, it is also a meagre subset of X. Finally, the inclusions

show that $D^*(A)\cap \complement B$ is a meagre subset of X.