# 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 ). 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.