# Separating points in a compact Hausdorff space.

Suppose that $$X$$ is a compact Hausdorff space and let $$\{x_1,\ldots,x_n\}$$ be a finite collection of points in $$X$$. It it possible to find open neighbourhoods $$U_i$$ of $$x_i$$ such that such that $$x_j \notin \overline{U_i}$$ for all $$j \ne i$$ and $$\overline{U_i} \cap \overline{U_j} = \varnothing$$ for all $$i \ne j$$? If not what extra separation axioms are needed?

I know for example that $$X$$ being compact Hausdorff implies that $$X$$ is normal, but that doesn't seem strong enough.

Use regularity of $$X$$ to obtain disjoint open sets $$U$$ and $$V$$ such that $$x_1 \in U$$ and $$\{x_2, \ldots, x_n\} \subseteq V$$. Then $$U \subseteq V^c$$ so that $$\overline{U} \subseteq V^c$$. In particular, $$\overline{U} \cap V = \varnothing$$. Now use normality of $$X$$ to obtain disjoint open neighbourhoods $$W$$ of $$\overline{U}$$ and $$Y$$ of $$\{x_2, \ldots, x_n\}$$. Then $$Y \subseteq W^c$$ so that $$\overline Y \subseteq W^c \subseteq \overline{U}^c$$. Hence, $$\overline{Y} \cap \overline{U} = \varnothing$$. A similar process can be undertaken to separate off the other points.
You can use Hausdorffness to separate the $$x_i$$ into pairwise disjoint $$U_i$$ such that $$x_i \in U_i$$, $$i=1,\ldots,n$$ (this follows by a simple induction on $$n$$ that this can be done in any Hausdorff space) and then regularity to find $$x_i \in V_i \subseteq U_i$$, $$V_i$$ open, with $$\overline{V_i} \subseteq U_i$$. No normality is needed, just Hausdorffness and regularity. And compact Hausdorff spaces are regular.