An equivalent condition to normality of topological spaces

Question

I wonder if this claim is true: Let $X$ be a topological space. Then $X$ is normal iff for every two open subsets $U,V$ such that $U\cup V=X$, there exist two closed subsets $A\subset U$, $B\subset V$ such that $A\cup B=X$.

Thanks in advance.

Answer 1

The idea is obvious: complements.

Suppose $X$ is normal If we have such $U$ and $V$, $U^c = X\setminus U$ and $V^c = X\setminus V$ are closed and disjoint (why?). Now apply normality.

The reverse is similar: if $X$ obeys the closed shrinking condition (as it's called) and $C$ and $D$ are closed and disjoint, their complements form an open cover, that we can shrink to a closed cover etc.