# Prove that any finite open cover of $T_{4}$ space has a refinement with a specific property

Prove that if $X$ is $T_4$ and $\Omega=\{\Omega_1\,\Omega_2,...,\Omega_n\}$ is an open cover of $X$, then there exists a refinement $\Omega'$ such that $\overline{\Omega'}=\{\overline{\Omega_1'},...,\overline{\Omega_n'}\}$ has the property that $\overline{\Omega_{i}'}\subseteq \Omega_i \forall i\in\{1,2,..,n\}$

My initial approach was to set $V_i=\Omega_i$\ $\cup_{j=1}^{i-1}\Omega_j$ and try to use any of the characteristics of normality such as Tychonoff or Urysohn's Theorems. However, they all involve some properties of closed sets. I tried to use the complements of $\Omega_i$ but I don't know what to do with them.

How do I approach this?

## 1 Answer

Fact 1: the shrinking lemma I showed here:

if $U_1,\ldots, U_n$ is an open cover of a $T_4$ space $X$ there is a closed cover $F_1, \ldots, F_n$ of $X$ such that $F_i \subseteq U_i$ for $i=1,\ldots,n$.

Now by normality we can again find $V_i$ open such that $F_i \subseteq V_i \subseteq \overline{V_i} \subseteq U_i$ and then you can use the intermediate $V_i$ as your open cover.