Completely normal space

Remarks

Equivalently, $X$ is completely normal if and only if the following holds. For all $A, B \subseteq X$ with $A \cap \overline{B} = \varnothing$ and $\overline{A} \cap B = \varnothing$, there are disjoint open sets $U \supseteq A$ and $V \supseteq B$. It is enough to check that every open subspace is normal.

Example: every metric space $(X, d)$ is completely normal. Each subspace is again a metric space, and every metric space is normal.

Non-example: the Tychonoff plank is compact Hausdorff, hence normal, but it is not completely normal. Its subspace obtained by removing one corner point is not normal.

Every perfectly normal space is completely normal.