It is enough to show that the union of two nowhere dense sets $A, B$ is nowhere dense, and without loss of generality we may assume that $A$ and $B$ are closed. The proposition is then equivalent to saying that the intersection of two dense open sets $\overline{C}A, \overline{C}B$ is dense. Now if $U$ is a non-empty open set, then $U \cap \overline{C}A$ is open and non-empty, hence

is open and non-empty.

Let $Y$ be a subspace of a topological space $X$. A subspace $A$ of $Y$ is said to be nowhere dense relative to $Y$ if $A$ is nowhere dense when considered as a subset of the topological space $Y$.

Suppose that $A$ is nowhere dense relative to $Y$. If the closure $\overline{A}$ of $A$ in $X$ contains a non-empty open set $U$, then $U \cap A$ is not empty (by the definition of closure); hence $U \cap Y$ is a non-empty open set relative to $Y$, and is contained in the closure $\overline{A} \cap Y$ of $A$ with respect to $Y$, which is contrary to the hypothesis.

Now suppose that $Y$ is open in $X$ and that $A \subset Y$ is nowhere dense relative to $X$. If $U$ is open in $Y$ and is not empty, then $U$ is open in $X$ and therefore contains a non-empty set $V$ which is open in $X$ (and a fortiori in $Y$) and does not meet $A$; hence $A$ is nowhere dense relative to $Y$.

The second part of Proposition 2 is clearly not valid if $Y$ is not open in $X$; consider, e.g., the situation where $Y \neq \emptyset$ is nowhere dense in $X$, and $A = Y$.

This follows from (EB"), for every open (resp. meagre) set in $Y$ is open (resp. meagre) in $X$.

It follows from Proposition 3 that every point of a Baire space has a fundamental system of neighbourhoods, each of which is a Baire space. Conversely:

Let $A$ be a non-empty open set in $X$, let $x \in A$ and let $V$ be an open neighbourhood of $x$ which is a Baire space. If $A$ were meagre in $X$, then $V \cap A$ would be meagre in $V$ and open in $V$, which is contrary to hypothesis.

Let $A$ be a meagre set in $X$; then its complement $Y = \complement_A$ in $X$ is dense in $X$. Let $B$ be a meagre set relative to $Y$; $B$ is also meagre relative to $X$, hence $A \cup B$ is meagre relative to $X$. Hence the complement of $A \cup B$ in $X$, which is also the complement of $B$ in $Y$, is dense in $X$ and a fortiori dense in $Y$. Hence $Y$ is a Baire space.

We shall show that axiom (EB) is satisfied in each case. Let $(A_n)$ be a sequence of dense open sets in $X$, and let $G$ be any non-empty open set. We can then define inductively a sequence $(G_n)$ of non-empty open sets such that $G_1 = G$ and $\overline{G}_{n+1} \subset G_n \cap A_n$: for since by hypothesis $G_n$ is not empty, $G_n \cap A_n$ is a non-empty open set; and since $X$ is regular in both cases envisaged, there is a non-empty open set $G_{n+1}$ such that $\overline{G}_{n+1} \subset G_n \cap A_n$. Hence the set $G \cap \bigcap_{n=1}^\infty A_n$ contains the intersection of the sets $G_n$, which is equal to the intersection of the sets $\overline{G}_n$; hence it is enough to show that $\bigcap_{n=1}^\infty \overline{G}_n \neq \varnothing$. Now if $X$ is locally compact, we may assume that $\overline{G}_2$ is compact; in the compact space $\overline{G}_2$, the $\overline{G}_n (n \geq 2)$ form a decreasing sequence of non-empty closed sets, and their intersection is therefore not empty by axiom (C'') (cf. Chapter I, § 9, no. 1, Definition 1]. If $X$ is a complete metric space (with respect to a metric compatible with its topology) we may suppose that $\overline{G}_n$ has been chosen so that its diameter (with respect to this metric) tends to 0 as $n$ tends to $+\infty$; the $\overline{G}_n$ therefore form a Cauchy filter base which converges to a point $x$, and $x$ necessarily belongs to the intersection of the sets $\overline{G}_n$.

Q.E.D.

This theorem may also be stated in the form that the set of points in the neighbourhood of which the family $(f_\alpha)$ is uniformly bounded above is a dense open set.

Let $f = \sup_\alpha f_\alpha$ be the upper envelope of the family $(f_\alpha)$. The function $f$ is lower semi-continuous (Chapter IV, § 6, no. 2, Theorem 4) and finite at every point of $X$. It is therefore enough to carry through the proof in the case where the family $(f_\alpha)$ consists of a single function $f$. Let $A_n$ be the set of points $x \in X$ such that $f(x) \leq n$; $A_n$ is closed (Chapter IV, § 6, no. 2, Proposition 1), and the assumptions imply that $X$ is the union of the sets $A_n$; hence at least one of the $A_n$ has an interior point, and therefore there is a non-empty open set on which $f$ is bounded above (by an integer $n$). If we apply this result to any non-empty open subset of $X$ (this subspace is also a Baire space by Proposition 3 of no. 3), we have the theorem.

In applications of this theorem it is generally the case that the $f_\alpha$ are continuous on $X$.