In this post we discuss a lemma (Lemma 1 below) that is useful for proving normality. In some cases, it is more natural using this lemma to prove that a space is normal than using the definition of normality. The proof of Lemma 1 is not difficult. Yet it simplifies some proofs of normality. One reason is that the derivation of two disjoint open sets that are to separate two disjoint closed sets is done in the lemma, thus simplifying the main proof at hand. The lemma is well known and is widely used in the literature. See Lemma 1.5.15 in . Two advanced examples of applications are  and . After proving the lemma, we give three elementary applications of the lemma. One of the applications is a characterization of perfectly normal spaces. This characterization is, in some cases, easier to use, e.g. making it easy to show that perfectly normal implies hereditarily normal.
In this post, we only consider spaces that are regular and . A space is regular if for each open set and for each , there exists an open with . A space is if every set with only one point is a closed set.
A space is a normal space if the following condition (Condition 1) is satisfied:
- For each closed subset of , and for each open subset of with , there exists a sequence of open subsets of such that and for each .
Proof of Lemma 1
Suppose the space satisfies condition 1. Let and be disjoint closed subsets of the space . Consider . Using condition 1, there exists a sequence of open subsets of the space such that and for each . Consider . Similarly, there exists a sequence of open subsets of the space such that and for each .
For each positive integer , define the open sets and as follows:
Let and . It is clear and are open and that and . We claim that and are disjoint. Suppose . Then for some and for some . Assume that . The fact that implies . The fact that implies that for all . In particular, , a contradiction. Thus . This completes the proof that the space is normal.
Spaces with nice bases
One application is that spaces with a certain type of bases satisfy condition 1 and thus are normal. For example, spaces with bases that are countable and spaces with bases that are -locally finite. Spaces with these bases are metrizable. The proof that these spaces are metrizable will be made easier if they can be shown to be normal first. The Urysohn functions (the functions described in Urysohn’s lemma) can then be used to embed the space in question into some universal space that is known to be metrizable. Using regularity and Lemma 1, it is straightforward to verify the following three propositions.
Let be a regular space with a countable base. Then is normal.
Let be a regular space with a -locally finite base. Then is normal.
Let be a regular space with a -discrete base. Then is normal.
A characterization of perfectly normal spaces
Another application of Lemma 1 is that it leads naturally to a characterization of perfect normality. Recall that a space is perfectly normal if is normal and perfect. A space is perfect if every closed subset of is a set (i.e. the intersection of countably many open subsets of ). Equivalently a space is perfect if and only if every open subset of is an set, i.e., the union of countably many closed subsets of . We have the following theorem.
A space is perfectly normal if and only if the following condition holds.
- For each open subset of , there exists a sequence of open subsets of such that and for each .
Clearly, condition 2 is strongly than condition 1.
Proof of Theorem 5
Suppose that the space is perfectly normal. Let be a non-empty open subset of . Then where each is a closed subset of . Using normality of , for each , there exists open subset of such that . Then consition 2 is satisfied.
Suppose condition 2 holds, which implies condition 1 of Lemma 1. Then is normal. It is clear that condition 2 implies that every open subset of is an set.
The characterization of perfectly normal spaces in Theorem 5 is hereditary. This means that any subspace of a perfectly normal space is also perfectly normal. In particular, perfectly normal implies hereditarily normal. Thus we have the following theorem.
Condition 2 in Theorem 5 is hereditary, i.e., if a space satisfies Condition 2, every subspace satisfies Condition 2. Therefore if the space is a perfectly normal space, then every subspace of is also perfectly normal. In particular, if is perfectly normal, then is hereditarily normal (i.e. every subspace of is normal).
Normality is hereditary with respect to subsets
Normality is not a hereditary notion. Lemma 1 can used to show that normality is hereditary with respect to subspaces.
Let be a normal space. Then every subspace of is normal.
Proof of Theorem 7
Let be a subspace of such that where each is closed subset of . Let be a closed subset of and let be an open subset of such that . We need to find , open in , such that and for all (closure of is within ).
Let be an open subset of such that . For each positive integer , let . Obviously is closed in . It is also the case that is closed in . To see this, let be a limit point of . Then is a limit point of . Hence since is closed in . We now have . The point is also a limit point of . Thus since is closed in . Now we have , proving that is closed in .
Now we have for all . By Lemma 1, for each , there exists a sequence of open subsets of such that and for all . Note that . Rename over all by the sequence . Then . It also follows that for all (closure of is within ). This completes the proof that the set is normal.
- Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
- Gruenhage, G., Normality in for complete , Trans. Amer. Math. Soc., 340 (2), 563-586, 1993.
- Nyikos, P., A compact nonmetrizable space such that is completely normal, Topology Proc., 2, 359-363, 1977.
Revised April 14, 2015