# separation properties in Hausdorff, compact spaces

Suppose $X$ is a compact Hausdorff topological space, $C\subseteq X$ a closed subset and $x\notin C$ a point. I have to prove that there exists a compact neighborhood of $x$ which is disjoint from $C$.

Here's what I did: since $X$ is Hausdorff, for every $y\in C$ we have two disjoint neighbourhoods of $y$ and $x$, respectively, say $U_y$ and $V_y$. In this way we get an open covering of $C$: $$C\subseteq\bigcup_{y \in C}U_y$$ and by compactness, we can find $y_1,\ldots,y_n$ such that $$C\subseteq\bigcup_{i=1}^n U_{y_i}.$$

At the same time, we note that $B=\bigcap_{i=1}^n V_{y_i}$ is a open neighbourhood of $x$ and it's disjoint from $\bigcup_{i=1}^n U_{y_i}$. I would like to finish the proof saying that the closure of $B$ is compact (since closed in compact is compact) and we are through, but I should justify why the closure should still be disjoint from $C$. Could you help me with that? Thanks.

-
$\bigcap_i V_{y_i}$ is disjoint from $\bigcup_i U_{y_i}$. Does this help you? –  martini Feb 13 '12 at 11:47
Can I say: suppose $p\in\bar{B}\cap C$. Then $p\in U_{y_i}$ for some $i$, so any small open neighbourhood of $p$ contained in $U_{y_1}$ intersects both $U_{y_i}$ and $C$ (as $p$ must be on the boundary of $C$) and this is a contradiction because $U_{y_i}$ was supposed to be disjoint from $C$? Sorry but I got terribly confused with this proof! –  fatoddsun Feb 13 '12 at 11:59

In general, if you have two disjoint open sets $U, V$, then $\overline{V} \cap U = \emptyset$ too.
Assume this was not the case: let $x \in \overline{V} \cap U$. Then, by definition of the closure $\overline{V}$, for every open set $W$ containing $x$, we would have $W \cap V \neq \emptyset$. But $U$ is such an open set: $x\in U$. Then we should have $U \cap V \neq \emptyset$.