An Equivalent Condition Of Normality

In this post I would like to present a useful condition of normality in terms of point-finite open cover. I include it here for use in subsequent discussion.

An open cover $\mathcal{V}=\lbrace{V_\alpha:\alpha \in S}\rbrace$ of a space $X$ is shrinkable if there exists an open cover $\mathcal{W}=\lbrace{W_\alpha:\alpha \in S}\rbrace$ such that $Cl(W_\alpha) \subset V_\alpha$ for each $\alpha \in S$ . The open cover $\mathcal{W}$ is called a shrinking of $\mathcal{V}$ . We prove the following lemma.

Lemma. A space $X$ is normal if and only if every point-finite open cover of $X$ is shrinkable.

Proof of Lemma. Suppose $X$ is normal. Let $\mathcal{V}=\lbrace{V_\alpha:\alpha < \mathcal{K}}\rbrace$ be a point-finite open cover of $X$ where $\mathcal{K}$ is some cardinal. Inductively define $\lbrace{W_\alpha:\alpha < \mathcal{K}}\rbrace$ as follows.

Let $C_0=X-\bigcup_{\alpha>0} V_\alpha$ . The set $C_0$ is a closed set and $C_0 \subset V_0$ . By normality, we can find an open set $W_0$ such that $C_0 \subset W_0 \subset Cl(W_0) \subset V_0$ .

Suppose that $W_\beta$ has been defined for all $\beta < \alpha$ . Just to capture what we have at this stage, for each $\beta < \alpha$ , we have:
$C_\beta=X-[S_\beta \cup T_\beta]$ where $S_\beta=\bigcup_{\tau < \beta}W_\tau$ and $T_\beta=\bigcup_{\tau > \beta} V_\tau$ ,
$C_\beta \subset W_\beta \subset Cl(W_\beta) \subset V_\beta$ .

Let $S=\bigcup_{\beta < \alpha} W_\beta$ . Furthermore, let $T=\bigcup_{\tau > \alpha} V_\tau$ . Now let $C_\alpha=X-[S \cup T)]$ . The set $C_\alpha$ is closed and $C_\alpha \subset V_\alpha$ . To see $C_\alpha \subset V_\alpha$ , let $x \in C_\alpha$ . Since $x \notin T$ , $x \in \bigcup_{\beta \leq \alpha}V_\beta$ . Since $\mathcal{V}$ is point-finite, the point $x$ is in at most finitely many $V_\beta$ , say $V_{\beta(0)},...,V_{\beta(n)}$ . Assume that $\beta(n)$ is the maximum of all the $\beta(i)$ . We claim that $\beta(n)=\alpha$ . Suppose that $\beta(n) < \alpha$ .

Note that $x \notin S$ . Thus $x \notin W_\beta$ for all $\beta < \alpha$ . This means $x \notin C_\beta$ for all $\beta < \alpha$ . In particular, $x \notin C_{\beta(n)}$ . It follows that $x \in \bigcup_{\tau > \beta(n)}V_\tau$ . But this is impossible. So $\beta(n)=\alpha$ and $x \in V_\alpha$ .

By normality, there is an open set $W_\alpha$ such that $C_\alpha \subset W_\alpha \subset Cl(W_\alpha) \subset V_\alpha$ .

All that remains to show is that the open sets $W_\alpha$ form a cover of $X$ . Once we have this, $\lbrace{W_\alpha:\alpha < \mathcal{K}}\rbrace$ is a shrinking of $\mathcal{V}$ . Let $x \in X$ . The point $x$ can only belong to finitely many $V_\alpha$ , say $V_{\alpha(0)},V_{\alpha(1)},...,V_{\alpha(n)}$ . Let $\alpha$ be the maximum of $\alpha(0),...,\alpha(n)$ . This means that $x \notin V_\tau$ for all $\tau > \alpha$ . If $x \notin W_\beta$ for all $\beta < \alpha$ , then $x \in C_\alpha \subset W_\alpha$ . Otherwise $x \in W_\beta$ for some $\beta < \alpha$ . In either cases, $x \in W_\beta$ for some $\beta \leq \alpha$ . Thus the open sets $W_\alpha$ form a cover and a shrinking of $\mathcal{V}$ .

For the converse, let $H$ and $K$ be disjoint closed subsets of $X$ . The open cover $\lbrace{X-H,X-K}\rbrace$ is point-finite. So it has a shrinking $\lbrace{U,V}\rbrace$ . Then $X-Cl(U)$ and $X-Cl(V)$ are two disjoint open sets containing $H$ and $K$ , respectively.