# Normal Topological Space => can seperate closed sets such that the closures are distinct

Suppose that X is a normal topological space, and that $$F_1$$, $$F_2$$ are disjoint closed subsets in X. Prove that there exist open subsets $$W_1$$$$F_1$$, $$W_2$$$$F_2$$, whose closures are disjoint. Here, normal is defined as: A space is normal if for any pair of disjoint closed subsets $$F_1$$, $$F_2$$, there exist disjoint open subsets $$U_1$$$$F_1$$, $$U_2$$$$F_2$$.

The usual approach I take with these sort of problems is to break it down; by picking say a point x in $$F_1$$ and applying the definition to that, with hope of taking a union over $$F_1$$. However here I don't know if the singleton {x} is closed so have gotten myself pretty stuck. Looking for a hint.

## 1 Answer

Combine two ways of formulating normality:

$$X$$ is normal (in the sense of separating two disjoint closed sets by disjoint open sets) iff for all closed sets $$F$$ and open sets $$U$$ such that $$F \subseteq U$$ we have an open subset $$V$$ such that $$F \subseteq V \subseteq \overline{V} \subseteq U$$.

Proof: This is quite easy to see: suppose $$F$$ and $$U$$ with $$F \subseteq U$$ are given, then $$F$$ and $$U^\complement$$ are disjoint and closed so we have disjoint open sets $$O_1$$ and $$O_2$$ such that $$F \subseteq O_1$$ and $$U^\complement \subseteq O_2$$. But then $$O_1 \subseteq O_2^\complement$$ by disjointness of $$O_1$$ and $$O_2$$ and also $$O_2^\complement \subseteq U$$; as $$O_2^ \complement$$ is closed and contains $$O_1$$, $$F \subseteq O_1 \subseteq \overline{O_1} \subseteq O_2^ \complement \subseteq U$$ and we can take $$V=O_1$$.

For the reverse, of we have this property and two disjoint closed sets $$F_1$$ and $$F_2$$, we have $$F_2^ \complement$$ open and containing $$F_1$$ so we have $$V$$ open with $$F_1 \subseteq V \subseteq \overline{V} \subseteq F_2^\complement$$ and then $$O_1= V$$ and $$O_2=\overline{V}^ \complement$$ are the required disjoint open sets separating $$F_1$$ and $$F_2$$. QED.

Knowing this, it's easy: separate $$F_1$$ and $$F_2$$ by disjoint open sets $$O_1$$ and $$O_2$$ resp. and then apply the alternative formulations again for both $$F_i$$ and $$O_i$$ ($$i=1,2$$) to get open neighbourhoods $$U_i$$ of $$F_i$$ with closure inside the $$O_i$$, so a fortiori disjoint too.

• Thanks for your reply! I'm struggling to see how $O_2^C ⊆ U$ follows though – beelal May 18 at 1:20
• @beelal It's equivalent to $U^\complement \subseteq O_2$. Taking complements on both sides reverses inclusions: $A \subseteq B \leftrightarrow B^\complement \subseteq A^\complement$ – Henno Brandsma May 18 at 6:45