# If $K_1$ & $K_2$ are disjoint nonempty compact sets ,show that there exist $k_i$ $\in$ $K_i$

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: Use the Extreme Value Theorem – YuiTo Cheng Jan 31 at 13:52

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, $$|x_k-y_k|. Since $$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 $$s=\lim |x_{k_{i_j}}-y_{k_{i_j}}|=|p-q|.$$