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.

1 thought on “An Equivalent Condition Of Normality

  1. Pingback: The product of a normal countably compact space and a metric space is normal | Dan Ma's Topology Blog

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