Every perfectly normal space is completely normal.

Question

Let $X$ be a perfectly normal space. I want to show that $X$ is also completely normal.

Let $A$ and $B$ be disjoint subsets of $X$. Let $f,g : X \rightarrow [0,1]$ be continuous functions that vanish precisely on $\overline A$ and $\overline B$ respectively. Now we consider the continuous function $h = f-g$.

We see that $h(x) \leq 0$ when $x \in \overline A$ and $h(x) \geq 0$ when $x \in \overline B$.

How do we use this fact to separate $\overline A$ and $\overline B$?

Answer 1

1. It is well-known (and fairly obvious) that each perfectly normal space is normal.

2. Each subspace $Y$ of a perfectly normal space $X$ is perfectly normal:
   Let $A \subset Y$ be closed. There exists a map $f : X \to I = [0,1]$ such that $f^{-1}(0) = \overline A$. Then $g = f \mid_Y$ has the property $g^{-1}(0) = A$ because $\overline A \cap Y = A$.

3. Your claim follows.