Compact sets of real number

Question

I don't know how to prove the following statement:

If $K_1, K_2$ are non empty disjoint compact subset of the real numbers, prove that there exist $k_1\in K_1$, $k_2\in K_2$ such that $|k_1-k_2|= \inf \{\,|x_1-x_2| : x_1\in K_1, x_2 \in K_2 \,\}$.

I'd like to prove that $K= \{\,|x_1-x_2| : x_1\in K_1, x_2 \in K_2\,\}$ is a compact subset of the real numbers, but I have not find the way to do it.

Note: I can't use that a continuos image of a compact set is compact.

Thanks!

Answer 1

Reposting my comment as an answer:

To prove $K_1 - K_2$ is compact, we will prove that every open cover has a finite open subcover. Let $\Omega$ be an open cover of $K_1 - K_2$. Since $K_1$ is compact, for each $y \in K_2$, we may extract a finite subcover $\mathcal{O}_y$ of $K_1 - y = \{x - y\, | \, x\in K_1\}$. Since $K_2$ is compact, we may extract a finite subset $\{y_1,\ldots,y_m\} \subset K_2$ such that $\mathcal{O} = \bigcup_{i=1}^m \mathcal{O}_{y_i}$ is an open cover of $x-K_2 = \{x-y \, | \, y\in K_2\}$ for each $x$, and hence $\mathcal{O}$ is an open cover of $K_1 - K_2$.

Since each $\mathcal{O}_{y_i}$ is a finite collection of open sets, and $\mathcal{O}$ is a finite union of finite collections of open sets, $\mathcal{O}$ is a finite subcover.

Hence every open cover of $K_1-K_2$ admits a finite open subcover. Hence $K_1 - K_2$ is compact.

Answer 2

Compact sets are closed and bounded, so both $K_1$ and $K_2$ are closed and bounded. Hence boundedness implies there exist $M_1,M_2,m_1,m_2$ such that $x_1 ∈ [m_1,M_1]$ and $x_2 ∈ [m_2,M_2]$. So $|x_1-x_2|$ is always finite. This means your set $K$ is bounded. The end values $M_1,M_2,m_1,m_2$ are attained since $K_1$ and $K_2$ are closed. By the continuity of the distance function ($|$ . $|$) and $K_1$ and $K_2$ being closed, K must also be closed.