Prove if $C$ and $D$ are non empty disjoint closed subsets of $M$ such that $C$ or $D$ is compact, then $d(C,D)$ > 0

Question

Prove if $C$ and $D$ are non empty disjoint closed subsets of $M$ such that $C$ or $D$ is compact, then $d(C,D)$ > 0

In $R$, two closed subsets are compact and their distance is definitely greater than 0 as long as they're disjoint, how can I generalize this any Metric Space?

Answer 1

Suppose that $d(C,D)=0$ and that $C$ is compact. So there exists a sequence of elements $c_n\in C$ such that $d(c_n,D)\to 0$.

Because $C$ is compact, $\{c_n\}$ admits a convergent subsequence, say $\{c_{n_k}\}$, that converges to $c\in C$. But then $$ d(c,D)=\lim d(c_n,D)=0. $$ That is, there exists a sequence $\{d_n\}$ in $D$ with $d_n\to c$. As $D$ is closed, $c\in D$. So $c\in C\cap D$, contradicting the fact that $C$ and $D$ are disjoint.

In conclusion, $d(C,D)>0$.

Answer 2

Assume $C$ is compact. Take a covering of $C$ by balls $B_\rho(x)$ such that $x$ is in the complement of $D$ and also the ball $B_\rho(x)$ is in the complement of $D$. This is possible as the complement of $D$ is open.

Now you can select a finite number of such balls covering $C$. These balls do not touch $D$. Thus the distance between $C$ and $D$ is larger than the smallest radius of this finite number of balls, which is positive.