# Proving the following shrinking lemma for binary covers

Prove a topological space is normal if and only if for each pair of open sets $$U,V$$ such that $$U cup V=X$$, there exists open sets $$U’$$ and $$V’$$ such that $$overline U’ subset U$$ and $$overline V’ subset V$$ and $$U’ cup V’=X$$

Attempt:

$$(Rightarrow)$$ Since $$U cup V=X implies (X-U) cap (X-V)=varnothing$$. $$X-U$$ and $$X-V$$ are disjoint closed sets. Then by normality of $$X$$, there are disjoint open sets $$W,Z$$ with $$X-U subset W$$ and $$X-V subset Z$$, and an equivalent definition of normality states there are open sets $$W’,Z’$$ containing $$X-U$$ and $$X-V$$ with $$overline W’ subset W$$ and $$overline Z’ subset Z$$. Now I am unsure how to show that $$W’ cup Z’=X$$, which is what I need to show in the problem.

$$(Leftarrow)$$ Suppose the second part of the statement holds. Let $$A$$ be closed in $$X$$ and $$U$$ be any open set containing $$X$$. Then the existence of an open set $$U’$$ containing $$A$$ such that $$overline U’ subset U$$ implies $$X$$ is normal.

Comment:I am having difficulty understanding (why/if) $$W’ cup Z’=X$$ holds in the first direction. Also I am skeptical whether I started the second direction correctly by starting with a closed set, and an arbitrary open set containing this closed set. I also realize it may not be true that $$U’$$ contains the closed set $$A$$, which adds to my skepticism. Any ideas?