A question here on perfectly normal spaces got me into investigating the definition of such a space. The definition on wikipedia says
A perfectly normal space is a topological space X in which every two disjoint non-empty closed sets $E$ and $F$ can be precisely separated by a continuous function $f$ from $X$ to the real line $\mathbb{R}$: the preimages of $\{0\}$ and $\{1\}$ under $f$ are, respectively, $E$ and $F$. (In this definition, the real line can be replaced with the unit interval $[0,1]$.)
It turns out that $X$ is perfectly normal if and only if $X$ is normal and every closed set is a $G_\delta$ set.
I took it for granted that a perfectly normal space was also assumed to be a normal space, which seems to be the case if you take the definition to be a perfect space where every closed set is $G_\delta$. If you take the other definition, I don't think you can assume automatically that the space is perfect. So this is kind of a follow up, from the first definition above, how can you conclude that a perfectly normal space is also normal?
Best Answer
Let $F$ and $G$ be disjoint closed sets in a perfectly normal space $X$ as defined in Wikipedia. There is a continuous function $f:X\to [0,1]$ such that $F = f^{-1}[\{0\}]$ and $G = f^{-1}[\{1\}]$. Let $$U = f^{-1}\left[\left[0,\frac13\right)\right]$$ and $$V = f^{-1}\left[\left(\frac23,1\right]\right].$$ Since $f$ is continuous, $U$ and $V$ are open in $X$. Clearly $U \cap V = \varnothing$, $F\subseteq U$, and $G\subseteq V$, so $U$ and $V$ separate the closed sets $F$ and $G$. It follows that $X$ is normal.