The Vedenissoff Theorem

The purpose of this post is to present a proof of the Vedenissoff Theorem, a useful theorem that characterizes perfect normality. A space X is perfectly normal if X is normal and every closed subset of X is a G_\delta-set. The Vedenissoff Theorem characterizes perfect normality in terms of zero sets and cozero sets. A subset A of a space X is a zero-set if there is a continuous f:X \rightarrow I=[0,1] such that A=f^{-1}(0). A set A \subset X is a cozero-set if X-A 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 X be a normal space. If H \subset X and K \subset X are disjoint closed sets, then there is a continuous f:X \rightarrow I such that f(x)=0 for each x \in H and f(x)=1 for each x \in K.

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

  1. The space X is perfectly normal.
  2. Every closed subset of X is a zero-set.
  3. Every open subset of X is a cozero-set.
  4. If H \subset X and K \subset X are disjoint closed sets, then there is a continuous f:X \rightarrow I such that H=f^{-1}(0) and K=f^{-1}(1).

Proof. 1 \rightarrow 2. Let A \subset X be a closed set and let A=\bigcap_{n<\omega}U_n where each U_n is open. By the Urysohn’s Lemma, for each n, there is a continuous f_n:X \rightarrow I such that f_n(x)=0 for each x \in A and f_n(y)=1 for each y \in X-U_n. Then define a function f:X \rightarrow I by \displaystyle f(x)=\sum_{n<\omega} \frac{f_n(x)}{2^{n+1}}. Clearly, f(x)=0 for each x \in A. For x \notin A, x \in X-U_n for some n and f(x)> \frac{1}{2^n}. Thus A=f^{-1}(0). The remaining thing to show is that f is continuous. Note that the sequence of continuous functions \lbrace{g_n}\rbrace converges uniformly to the function f where \displaystyle g_n=\sum_{i=0}^n \frac{f_i}{2^{i+1}}. This implies that the limit f must be continuous.

2 \rightarrow 3 is clear.

3 \rightarrow 4. Let H and K be two disjoint closed subsets of X. There is a continuous f:X \rightarrow I such that f^{-1}(0)=H. Likewise, there is a continuous g:X \rightarrow I such that g^{-1}(0)=K. Define \displaystyle h=\frac{f}{f+g}. This is a continuous function. It is easy to verify that h^{-1}(0)=H and h^{-1}(1)=K.

4 \rightarrow 1. Clearly, condition 4 implies that X is normal. Now show that every closed set is G_\delta. Let A be a proper closed subset (i.e., there is z \in X-A). Then we have continuous f:X \rightarrow I such that f^{-1}(0)=A and f^{-1}(1)=\lbrace{z}\rbrace. We have \displaystyle A=\bigcap_{n<\omega}U_n where U_n=f^{-1}([0,\frac{1}{2^{n+1}})).

Reference

  1. Engelking, R., General Topology, Revised and Completed Edition, 1989, Heldermann Verlag, Berlin.
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One thought on “The Vedenissoff Theorem

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

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