# Proof Verification: Compact Hausdorff Implies Normal

Here is the definition of "normal space" that I am working with:

A topological space $$(X,\tau)$$ is normal if for any two disjoint closed sets $$A$$ and $$B$$, there exists disjoint open sets $$U$$ and $$V$$ such that $$A \subseteq U$$ and $$B \subseteq V$$.

Prove that every compact Hausdorff space is a normal space.

Here is my attempt at a proof:

1. Suppose $$(X,\tau)$$ is a compact Hausdorff space. Let $$A$$ and $$B$$ be disjoint closed subsets in $$X$$. Since closed subsets of Hausdorff and compact spaces are Hausdorff and compact, $$A$$ and $$B$$ are also compact Hausdorff spaces.
2. Since $$A$$ and $$B$$ are compact, every open covering has a finite subcovering. Thus $$A \subseteq \displaystyle{\Pi_{i\in I}U_i}$$ has finite subcovering $$A \subseteq U_1 \cup ... \cup U_n$$ and $$B \subseteq \displaystyle{\Pi_{j\in J}V_j}$$ has finite subcovering $$B \subseteq V_1 \cup ... \cup V_m$$, where $$U_i$$ and $$V_j$$ are open sets for any $$i \in I$$, $$j \in J$$.
3. Let $$U=U_1 \cup ... \cup U_n$$ and $$V=V_1 \cup ... \cup V_m$$. Then, $$U \cap V = (U_1 \cup ... \cup U_n) \cap (V_1 \cup ... \cup V_m)= (U_1 \cap V_1) \cup ... \cup (U_n \cap V_m) = \phi$$ since $$U_i \cap V_j = \phi$$.

Thus $$U$$ and $$V$$ are disjoint open sets, and $$(X,\tau)$$ is normal.

There are a few doubts that I have concerning this proof:
Firstly, is it true that $$U_i \cap V_j = \phi$$ ? I think it is because $$(X,\tau)$$ is Hausdorff, so open neighbourhoods about distinct points are separated.
Secondly, did I distribute the intersection correctly in step 3?

Be carefull: In $$3$$ are more terms and isn't true $$U \cap V$$ will be no necesary empty.

One way is a follow:

As you say, every cover by open sets $$A \subseteq \cup_{i\in I}U_i$$ has a finite covering, i.e. $$A \subseteq U_1 \cup \ldots \cup U_n$$.

Then, for $$U_1$$, will be find a finite open cover of $$B$$ that is disjoint:

As $$X$$ its Hausdorff, for every $$y \in B$$ we can find a open set $$V_y^1$$ such that $$y \in V_y^1$$ and $$U_1 \cap V_y^1 = \emptyset$$. We can do this for every $$y \in B$$. It's clear that $$B \subseteq \cup_{j\in J}V_j^1$$. As $$B$$ it's Hausdorff (as you say) then we can extract a finite subset, i.e. $$B \subset V_{j_1}^1 \cup \ldots \cup V_{j_n}^1:= \tilde{V}^1$$. It's clear that $$V_{j_1}^1 \cup \ldots \cup V_{j_n}^1$$ and $$U_1$$ are disjoint by construction.

We can repeat this for every open set $$U_2, \ldots, U_n$$ of the finite covering of $$A$$. At the end we have $$n$$ finite coverings of $$B$$, called $$\tilde{V}^1, \ldots, \tilde{V}^n$$, and having the property that $$\tilde{V}^j \cap U_j = \emptyset$$ for all $$j$$.

Finally take $$W = \tilde{V}^1 \cap \ldots \cap \tilde{V}^n$$. As every $$\tilde{V}^j$$ is open, $$W$$ is open. As $$B \subset \tilde{V}^j$$ for every $$j$$ (this is by construction), $$B \subset W$$. Finally as $$\tilde{V}^j \cap U_j = \emptyset$$ for every $$j$$, then $$W \cap (U_1 \cup \ldots \cup U_n) = \emptyset$$.

If we define: $$U= U_1 \cup \ldots \cup U_n$$, then $$U$$ and $$W$$ are the disjoint open sets we were looking for.