Lemma 3

Let $X$ be a normal space and $(X_i)_{i \in I}$ a locally finite open covering of $X$. For all $(i, j) \in I \times I$ and all $x \in X_i \cap X_j$, let $V_{ij}(x)$ be a neighbourhood of $x$ contained in $X_i \cap X_j$. Then we can associate with every $x \in X$ a neighbourhood $V(x)$ of $x$ such that the following conditions are fulfilled: (a) the relation $x \in X_i \cap X_j$ implies $V(x) \subset V_{ij}(x)$; (b) if $V(x)$ and $V(y)$ meet, there exists $i \in I$ such that $V(x) \cup V(y) \subset X_i$. There exists an open covering $(X'_i)_{i \in I}$ of $X$ such that $\overline{X'_i} \subset X_i$ for all $i \in I$ (General Topology, Chapter IX, § 4, Theorem 3). Let $x \in X$. Let $V_1(x)$ be the intersection of the $V_{ij}(x)$ and the $X'_k$ which contain $x$; this is an open neighbourhood of $x$. Let $V_2(x)$ be a neighbourhood of $x$ contained in $V_1(x)$ and meeting only a finite number of $X_i$. Then $V_2(x)$ meets only a finite number of $\overline{X'_i}$ and hence the set

is a neighbourhood of $x$. If $x \in X_i \cap X_j$, then $V_1(x) \subset X_i \cap X_j$ and hence $V(x) \subset X_i \cap X_j$. Let $x, y$ be in $X$ and suppose that $V(x)$ and $V(y)$ meet. There exists $i \in I$ such that $x \in X'_i$. Then $V_1(x) \subset X'_i$, hence $V(x) \subset X'_i$ and hence $V(y) \cap \overline{X'_i} \neq \emptyset$. Then $y \in \overline{X'_i}$ by definition of $V(y)$, whence $y \in X_i$ and $V(y) \subset X_i$. Thus $X_i$ contains $V(x)$ and $V(y)$.

§ 2.1