Let A and B be disjoint compact subsets of a Hausdorff topological space X. Show that there exist disjoint open sets U and V containing A and B, respectively.

Printable View

- Apr 8th 2009, 09:50 AMr2dee6Compactness and Hausdorff Topology Question
Let A and B be disjoint compact subsets of a Hausdorff topological space X. Show that there exist disjoint open sets U and V containing A and B, respectively.

- Apr 8th 2009, 10:27 AMclic-clac
Hi

Let $\displaystyle K$ and $\displaystyle L$ be such compacts. Take one point $\displaystyle k$ in $\displaystyle K.$

For every point $\displaystyle l$ of $\displaystyle L,$ there are two open sets $\displaystyle U_l$ and $\displaystyle V_l$ such that $\displaystyle k\in U_l$ and $\displaystyle l\in V_l $ and $\displaystyle U_l\cap V_l=\emptyset$ .

$\displaystyle K\subseteq \bigcup\limits_{l\in L}V_l$ so $\displaystyle K\subseteq \bigcup\limits_{l\in I_k}V_l$ for a finite subset $\displaystyle I_k$ of $\displaystyle L$.

Now $\displaystyle O_k:=\bigcap\limits_{l\in I_k}U_l$ is an open set disjoint from $\displaystyle L$ (it is disjoint from $\displaystyle \bigcup\limits_{l\in I_k}V_l$ ) which contains $\displaystyle k .$

Repeat that operation for every $\displaystyle k\in K .$ You obtain a set of open sets $\displaystyle \{O_k;\ k\in K\}$ such that $\displaystyle K\subseteq\bigcup\limits_{k\in K}O_k$ and $\displaystyle \bigcup\limits_{k\in K}O_k$ is disjoint from $\displaystyle L$ (more precisely from $\displaystyle \bigcap\limits_{k\in K}\bigcup\limits_{l\in I_k}V_l$ ).

So same properties (and even better) for $\displaystyle \bigcup\limits_{k\in J}O_k$ for a finite subset $\displaystyle J$ of $\displaystyle K :$

what can you say of $\displaystyle \bigcap\limits_{k\in J}\bigcup\limits_{l\in I_k}V_l$ and $\displaystyle \bigcup\limits_{k\in J}O_k $ ? - Apr 21st 2009, 11:22 AMmonkey.brains
hey could you please explain the answer to me a little bit. As i am having a little bit of difficult following it

- Apr 21st 2009, 12:26 PMPlato
This is a standard two-part proof of this theorem.

First suppose that $\displaystyle L$ is a compact set and $\displaystyle p \notin L$. Now prove that there two open sets, $\displaystyle O~\&~Q$, such that $\displaystyle p \in O,~L \subseteq Q\,\& \,O \cap Q = \emptyset $.

That is the first part of the proof above.

Now in the second part of the proof we have two disjoint compact sets, $\displaystyle K~\&~L$.

Because they are disjoint, $\displaystyle \left( {\forall p \in K} \right) \Rightarrow \quad p \notin L$.

So apply the first part of the theorem. Because $\displaystyle K$ is compact we get a finite collections of open sets the union of which is an open set containing $\displaystyle K$.

Intersecting the corresponding $\displaystyle Q’s$ , we get an open set containing $\displaystyle L$.