# Thread: Compactness in a Hausdorff space

1. ## Compactness in a Hausdorff space

Let $x$ be a Hausdorff space and let $A, B$ be disjoint compact subsets of $X$. Show that there exist disjoint open subsets $U, V$of $X$such that $A \subseteq U$ and $B \subseteq V$.

2. Originally Posted by Amanda1990
Let $x$ be a Hausdorff space and let $A, B$ be disjoint compact subsets of $X$. Show that there exist disjoint open subsets $U, V$of $X$such that $A \subseteq U$ and $B \subseteq V$.
Try to first do this lemma.
If $A$ is a compact subset of a Hausdroff space $X$ and $
p \in X\backslash A$
then there exists open sets $G\;\& \,H$ such that $p \in G\;,\,A \subseteq H\;\& \;G \cap H = \emptyset$.

Once you have done that lemma, take note $q \in B\; \Rightarrow \;q \in X\backslash A
$
and build an open set for each $q \in B$ that is an open cover for $B$.
Remember these are compact sets. Also the intersection of finitely many open sets is open.

3. OK I can see how your lemma would help us here but there are two issues:

1) I'm sure you're thinking of an equally valid alternative method, but once we've proved your lemma isn't it more natural to buid an open set for every element in B and then take the union of all of these? (Then even infinite unions of open sets are still open).

2) I can visualise why the lemma "should" work but this is quite different to proving it! It seems like the same kind of idea as te question itself...I'm trying to use the compactness by using open sets and then showing that only finitely of them are needed to cover the open set, then use the Hausdorff property, but I can't get near a formal proof.

4. If $A$ is a compact subset of a Hausdroff space $X$ and $
p \in X\backslash A$
then there exists open sets $G\;\& \,H$ such that $p \in G\;,\,A \subseteq H\;\& \;G \cap H = \emptyset$.
Here is a quick outline of a proof.
$\left( {\forall x \in A} \right)\left( {\exists O_x \;\& \;Q_x } \right)\left[ {x \in O_x \;,\;p \in Q_x \;\& \;O_x \cap Q_x = \emptyset } \right]$, these are open sets.
The O’s cover A. There is a finite collection $\left\{ {O_{x_1 } ,O_{x_2 } , \cdots ,O_{x_n } } \right\}$ which also covers A.
Look the set of corresponding Q’s: $\left\{ {Q_{x_1 } ,Q_{x_2 } , \cdots ,Q_{x_n } } \right\}$.
Let $H = \bigcup\limits_n {O_{x_n } } \;\& \;G = \bigcap\limits_n {Q_{x_n } }$ now it should be clear to you that $A \subseteq H\;,\;p \in G\;\& \;H \cap G = \emptyset$.