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 of a space is shrinkable if there exists an open cover such that for each . The open cover is called a shrinking of . We prove the following lemma.

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

**Proof of Lemma**. Suppose is normal. Let be a point-finite open cover of where is some cardinal. Inductively define as follows.

Let . The set is a closed set and . By normality, we can find an open set such that .

Suppose that has been defined for all . Just to capture what we have at this stage, for each , we have:

where and ,

.

Let . Furthermore, let . Now let . The set is closed and . To see , let . Since , . Since is point-finite, the point is in at most finitely many , say . Assume that is the maximum of all the . We claim that . Suppose that .

Note that . Thus for all . This means for all . In particular, . It follows that . But this is impossible. So and .

By normality, there is an open set such that .

All that remains to show is that the open sets form a cover of . Once we have this, is a shrinking of . Let . The point can only belong to finitely many , say . Let be the maximum of . This means that for all . If for all , then . Otherwise for some . In either cases, for some . Thus the open sets form a cover and a shrinking of .

For the converse, let and be disjoint closed subsets of . The open cover is point-finite. So it has a shrinking . Then and are two disjoint open sets containing and , respectively.

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