Distance between Disjoint Compact Set and Closed Set in Metric Space is Positive

Theorem
Let $\struct {X, d}$ be a metric space.

Let $A$ and $B$ be disjoint subsets of $X$ such that $A$ is compact and $B$ is closed.

Then $\map d {A, B} > 0$, where $\map d {A, B}$ is the distance between $A$ and $B$.

Proof
Define $f : X \to \R$ by:


 * $\map f x = \map d {x, B}$

for each $x \in X$.

From the definition of the distance between $A$ and $B$, we have:


 * $\ds \map d {A, B} = \inf_{a \mathop \in A} \map f a$

From Compact Subspace of Hausdorff Space is Closed and Metric Space is Hausdorff, $A$ is closed and hence contains all its limit points.

From Point at Zero Distance from Subset of Metric Space is Limit Point or Element we then have:


 * $\map f a > 0$

for each $a \in A$.

From Distance from Point to Subset is Continuous Function, $f$ is continuous.

From Restriction of Continuous Mapping is Continuous: Metric Spaces, the restriction $f \restriction_A$ is continuous.

From Continuous Image of Compact Space is Compact: Corollary 3, we have that $f$ attains its bounds.

That is, there exists $\alpha \in A$ such that:


 * $\ds \map d {\alpha, B} = \inf_{a \mathop \in A} \map f a$

giving:


 * $\ds \inf_{a \in A} \map f a > 0$

and so:


 * $\map d {A, B} > 0$