# 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.