Theorem 3

Let $(A_i)_{i \in I}$ be a locally finite open covering of a closed set $Y$ in a normal space $X$. Then there is an open covering $(B_i)_{i \in I}$ of $Y$ such that $\overline{B}_i \subset A_i$ for each $i \in I$.

Well-order the index set $I$ (Set Theory, Chapter III, § 2, no. 3, Theorem 1). We shall define a family $(B_i)_{i \in I}$ of open sets in $X$, by transfinite induction, such that (i) $\overline{B}_i \subset A_i$ for each $i \in I$; (ii) for each $i \in I$, the family formed by the $B_\lambda$ such that $\lambda \leq i$ and by the $A_\lambda$ such that $\lambda > i$ is an open covering of $Y$. Suppose that we have defined the $B_i$ for $i < \gamma$ so that (i) and (ii) are satisfied for all $i < \gamma$, and let us show that we can define $B_\gamma$ in such a way that (i) and (ii) are also satisfied for $i = \gamma$. Let us first show that the $B_i$ for which $i < \gamma$ and the $A_i$ for which $i \geq \gamma$ form a covering of $Y$. By hypothesis, for each $x \in Y$ there is only a finite number of indices $\lambda \in I$ such that $x \in A_\lambda$, say $\lambda_1 < \lambda_2 < \cdots < \lambda_n$; let $\lambda_h$ be the greatest of the $\lambda_i$ such that $\lambda_i < \gamma$; if $h < n$ we have $x \in A_{\lambda_n}$ and $\lambda_n \geq \gamma$, and if $h = n$ the inductive hypothesis shows that $x$ belongs to some $B_\lambda$ such that $\lambda \leq \lambda_n < \gamma$, and our assertion follows.

Now put $C = (\complement Y) \cup \left( \bigcup_{i < \gamma} B_i \right) \cup \left( \bigcup_{i > \gamma} A_i \right)$; $C$ is open, and from what has been said we have $\complement A_\gamma \subset C$; by virtue of Axiom $(O''_V)$ for normal spaces, there is therefore an open set $V$ such that $\complement A_\gamma \subset \overline{V} \subset V \subset C$. If we put $B_\gamma = \complement \overline{V}$, we have $\overline{B}_\gamma \subset \complement V \subset A_\gamma$ and $B_\gamma \cup C = X$, so that the $B_i$ such that $i \leq \gamma$ and the $A_i$ such that $i > \gamma$ cover $Y$.