The purpose of this post is to present a proof of the Vedenissoff Theorem, a useful theorem that characterizes perfect normality. A space is perfectly normal if is normal and every closed subset of is a set. The Vedenissoff Theorem characterizes perfect normality in terms of zero sets and cozero sets. A subset of a space is a zero-set if there is a continuous such that . A set is a cozero-set if is a zero-set. The Vedenissoff Theorem were published in 1936 and 1940 (references found in [1]). All spaces are at least Hausdorff. The Urysohn’s Lemma is used as a basic tool to obtain continuous functions on a normal space and is stated without proof.

* Urysohn’s Lemma*. Let be a normal space. If and are disjoint closed sets, then there is a continuous such that for each and for each .

* The Vedenissoff Theorem*. Given a space , the following conditions are equivalent.

- The space is perfectly normal.
- Every closed subset of is a zero-set.
- Every open subset of is a cozero-set.
- If and are disjoint closed sets, then there is a continuous such that and .

* Proof*. . Let be a closed set and let where each is open. By the Urysohn’s Lemma, for each , there is a continuous such that for each and for each . Then define a function by . Clearly, for each . For , for some and . Thus . The remaining thing to show is that is continuous. Note that the sequence of continuous functions converges uniformly to the function where . This implies that the limit must be continuous.

is clear.

. Let and be two disjoint closed subsets of . There is a continuous such that . Likewise, there is a continuous such that . Define . This is a continuous function. It is easy to verify that and .

. Clearly, condition 4 implies that is normal. Now show that every closed set is . Let be a proper closed subset (i.e., there is ). Then we have continuous such that and . We have where .

*Reference*

- Engelking, R.,
*General Topology, Revised and Completed Edition*, 1989, Heldermann Verlag, Berlin.

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