# Property of normal spaces

Let $X$ be a normal space and let $U_1$ and $U_2$ be open subsets of $X$ such that $X= U_1 \cup U_2$. Show that there are open sets $V_1$ and $V_2$ such that $\operatorname{cl}(V_1) \subseteq U_1$, $\operatorname{cl}(V_2) \subseteq U_2$, and $X=V_1 \cup V_2$.

I don't see how $X=V_1 \cup V_2$.

Hint: Essentially by applying de Morgan's Laws to the usual definition of normality we get the following equivalent formulation:

Whenever $U_1 , U_2$ are open subsets of $X$ such that $U_1 \cup U_2 = X$, there are closed sets $F_1 , F_2 \subseteq X$ such that $F_1 \subseteq U_1$ and $F_2 \subseteq U_2$ and $F_1 \cup F_2 = X$.

After using this, apply yet another equivalent formulation of normality (that you have most likely seen before) to a couple of closed sets and a couple of open sets including them.

(If you don't like the first step, consider first the closed sets $E_1 = X \setminus U_1, E_2 = X \setminus U_2$. Note that since $U_1 \cup U_2 = X$, then $E_1 \cap E_2 = ( X \setminus U_1 ) \cap ( X \setminus U_2 ) = \varnothing$. You will find yourself going back-and-forth a little bit with this method, but it could be considered a bit more basic.)

The basic picture is that though $U_1 , U_2$ cover $X$, there is some overlap. The property then says that we can shrink these open sets (to open sets whose closure is included in the original) such we don't miss anything; the shrinking only happens in the overlap. This is pretty much dual to the usual definition of normality, which says that if you have closed sets which are disjoint, then we can enlarge each of them to open sets which are still disjoint.