# Proof with compact sets in Hausdorff space

1. Prove that every compact set in Hausdorff space is closed.

2. Let $(X,\tau)$ be Hausdorff space and $A,B$ compact, disjont subsets of $(X,\tau)$. Prove that exist two disjoint sets $V,W$ open in $(X,\tau)$, so $A \subset V$, $B \subset W$.

I'have got an idea with 1.

Let K be a compact set in Hausdorff space $(X,\tau )$ and let $a \not\in K$.

$\forall x \in K$ we choose a pair of disjoint open sets $V (x), W(x) \in \tau$ : $a \in V (x)$ and $x \in W(x)$. Because of K compactness, we can choose $x_1, . . . , x_n \in K$ : $K \subset W(x_1) \ \cup . . . \cup \ W(x_n) = W$. Then $V = V (x_1) \cap . . . \cap V (x_n)$ is a neighbourhood of $a$ disjoint with $W$, so with $K$ as well. Therefore $a \not\in K$, and finally $\overline{K} = K$

Is this correct? Actually I don't have idea about 2nd part. I think this proof might be useful.

HINT: Your argument for (1) is correct. You can indeed use the same idea for (2). Note that (1) essentially proves the special case of (2) in which $A=\{a\}$ for some $a\in X$. In general, then, you can separate each $a\in A$ from $B$ with disjoint open sets. And then?
• Well, generally speaking, I think I can replace point $a$ with $A$ set.
• @MatJ: Yes, but how? Specifically, how can you adapt your argument for (1)? HINT: Think of $B$ as taking the place of the point $a$ in (1), and $A$ as taking the place of $K$. Commented Nov 11, 2014 at 22:15
Why at the end of your proof do you write $a\notin K$? Don't you assume that at the beginning? And how do you conclude that $\bar K = K$? It is certainly true, but maybe it's easier to reason that $X\setminus K$ is open, since you have just demonstrated that around any point $a\in X\setminus K$ has an open nbhd $V$ which doesn't intersect $K$ (as it doesn't intersect $W$).