If $K_1$ & $K_2$ are disjoint nonempty compact sets ,show that there exist $k_i$ $\in$ $K_i$ such that
$|k_1 - k_2|$=inf{$|x_1 - x_2|$: $x_i$ $\in$ $K_i$}.
They are all subsets of $\mathbb R$
I am able to proof that the set is bounded but I am trying to proof that the set is closed using sequence but I can't. Give me some hint.
Hint: $d:\mathbb R\times \mathbb R\to \mathbb R:(x,y)\mapsto |x-y|$ is continuous, and $K_1\times K_2$ is compact.
Here is a proof using sequences: in fact, we can prove a stronger result: assume $K$ is compact and $C$ is closed.
The $\inf$ exists because $S=\{|x-y|:x\in K;\ y\in C\}$ is bounded below. Therefore, there is a sequence $(x_k,y_k)\in K\times C$ such that $|x_k-y_k|\to \inf S=s$. Without loss of generality, $|x_k-y_k|<s+1$. Since $K$ is compact, we get a subsequence $x_{k_i}\to p\in K.$ On the other hand, $K$ is bounded so it lies in some ball of radius $R$, so $|y_k|\le |y_k-x_k|+|x_k|\le s+1+R$. Then, $y_{k_i}$ is bounded, so it also has a convergent subsequence, $y_{k_{i_j}}\to q\in C\ $ (because $C$ is closed.) But then $s=\lim |x_{k_{i_j}}-y_{k_{i_j}}|=|p-q|.$