Perfectly normal space

Remarks

Equivalently, for all disjoint closed sets $A, B \subseteq X$ there is a continuous function $f \colon X \to [0, 1]$ with $f^{-1}(0) = A$ and $f^{-1}(1) = B$. Another equivalent form: every closed set is $f^{-1}(0)$ for some continuous $f \colon X \to \R$. A perfectly normal space is completely normal. A perfectly normal $T_1$ space is called a $T_6$ space.

Example: every metric space $(X, d)$ is perfectly normal. It is normal, and a closed set $A$ equals $\bigcap_{n \ge 1} \set{x : d(x, A) < 1/n}$, an intersection of countably many open sets.

Non-example: the space $[0, \omega_1]$ of countable ordinals together with $\omega_1$, with the order topology, is compact Hausdorff and so normal. The closed set $\set{\omega_1}$ is not a G-delta set: any countably many neighbourhoods $(\alpha_n, \omega_1]$ contain $(\sup_n \alpha_n, \omega_1]$, which has more than one point. So clause 2 fails.