Compactness and disjoint open sets Let $C1$ and $C2$ be two disjoint compact subsets of $R^n$. Prove that there exist disjoint open sets $U_1$ and $U_2$ such that $C_1 \subseteq U_1$ and $C_2 \subseteq U_2$.
Intuitively, this seems to be correct, however, I could not figure out the importance of compactness for this problem - why arbitrary subsets of $R^n$ could hold such property for some open sets $U_1$ and $U_2$? Is it due to the fact of boundedness and closeness of the sets $C1$ and $C2$?
 A: Here is a second tack.  The function $(x,y)\mapsto d(x,y)$ is a continuous function on the compact set $C_1\times C_2$.  It therefore has a positive minimum value $v$.  Let
$$U_k = \bigcup_{x\in C_k} B_{v/2}(x), \qquad k = 0, 1.$$
These two sets do the job for you.
A: Actually, the assertion holds even if $C_1$ and $C_2$ are just closed and not necessarily bounded, but the proof is a bit more tricky in that case (you'd probably have to use properties of the distance function to a set).
For compact sets, you can use the definition of compactness alone (together with the so-called Hausdorff property of $\mathbb R^n$). Divide the proof in 2 steps.

*

*Show that for every $c_2\in C_2$ there are disjoint open sets $V_1(c_2),V_2(c_2)$ such that $C_1\subseteq V_1(c_2)$ and $c_2\in V_2(c_2)$.

*Apply 1 and the definition of the compactness of $C_2$ to prove the assertion.

In order to show 1, apply that for every $c_1\in C_1$ there are disjoint open sets $W_1(c_1),W_2(c_1)$ such that $c_1\in W_1(c_1)$ and $c_2\in W_2(c_1)$ and the definition of compactness of $C_1$.
A: It is because every open cover of a compact set possesses finite subcover. To proceed your question, first prove that in $\mathbb{R}^n$, you can separate a point and a compact set:
Let $C$ be our compact set and $x\not\in C$. Due to $\mathbb{R}^n$ is Hausdorff, for each point $c\in C$, you can find open sets $U_c$ and $V_{x_c}$ in $\mathbb{R}^n$ such that $c\in U_c$, $x\in V_{x_c}$ and $U_c\cap V_{x_c}=\emptyset$.
Next, consider the open cover $\{U_c: c\in U_c\}$ of $C$. Due to the compactness of $C$, this open cover possesses finite subcover. Thus, we can find natural number $k$ such that
$$C\subseteq \bigcup\limits_{i=1}^k U_{c_i}.$$
Note that $U_{c_i} \cap V_{x_{c_i}}=\emptyset$ for each $i$. Now consider $V=\bigcap\limits_{i=1}^k V_{x_{c_i}}$. Note that $\bigcup\limits_{i=1}^k U_{c_i}\cap V =\emptyset$. Thus, we can separate a compact set and a point (not in the compact set) in $\mathbb{R}^n$ by open sets.
Now, coming to your question:
For each $c\in C_1$ consider the pair of open sets $U_c$ and $V_c$ which separates $c$ and $C_2$. Indeed, this is ensured by the forgoing arguments. Now, again, consider the open cover
$$\{U_c:c\in C_1\}$$
for $C_1$. Due to the compactness, it possesses finite subcover. Thus, we can find natural number $k$ such that
$$C_1\subseteq \bigcup\limits_{i=1}^k U_{c_i}.$$
Note that $U_{c_i} \cap V_{c_i}=\emptyset$ for each $i$. Now, consider $V=\bigcap\limits_{i=1}^k V_{c_i}$. Note that $\bigcup\limits_{i=1}^k U_{c_i}\cap V =\emptyset$. Thus, we can separate two disjoint compact sets in $\mathbb{R}^n$ by open sets.
EDIT Observe that apart from Hausdorff property of $\mathbb{R}^n$, we have not used anything (even nicer properties) of $\mathbb{R}^n$. Thus, can you guess, is this valid in any Hausdorff topological vector space?
