In a metric space compactness forces two things at once: the set is closed, and it is bounded. Neither implication needs anything new — a finite subcover is produced and read. The converse is where the trouble starts: closed and bounded is not enough in a general metric space, and the line is special.

A field that can always be inspected by finitely many patrols must have a fence and a finite size. Having a fence and a finite size does not, on its own, mean finitely many patrols suffice — that depends on the terrain.

An infinite set with the discrete metric: every pair is at distance $1$, so the whole space is bounded, and it is closed in itself. The singletons form an open cover with no finite subcover, so it is not compact. Closed and bounded is not enough outside $\mathbb{R}^n$.

Let $(X,d)$ be a metric space and $K\subseteq X$ compact. Then $K$ is closed in $X$ and bounded — that is, contained in some ball. The proofs are two covers, each thinned once. The converse fails in general and holds in $$, which is the Heine–Borel theorem two lessons from here.

• Bounded: the balls $B(x_0,n)$ for $n\ge1$ cover $K$, and finitely many of them have a largest, which contains .

Take a point outside the set. Around each point of the set put a ball of radius half the distance to the chosen point; these balls cover the set, and each of them stays away from the chosen point by at least that half-distance. Thin the cover to finitely many, take the smallest of their half-distances, and the ball of that radius around the chosen point meets none of them — so it misses the set entirely. Every point outside the set therefore has room, which is openness of the complement.

Take a point outside the set; its distance to each point of the set is positive.

Each point of the set lies in its own ball, so these balls cover it.

Compactness thins the cover to finitely many balls.

A minimum of finitely many positive numbers is positive — this is where finiteness is used.

The triangle inequality keeps that ball out of every chosen ball, so it misses the set and the complement is open.

Compactness of a subset of a metric space gives closed and bounded, each by thinning one cover.

A continuous real function on a compact space has compact image, so the image is closed and bounded and contains its own supremum.

• In a metric space, compact implies closed and bounded, each by thinning one cover.
• The converse fails in general: an infinite discrete metric space is the counterexample.
• A compact metric space is complete.
• A continuous real function on a compact space attains its largest and smallest values.

Great work! You now know how to:

• prove that a compact subset of a metric space is closed and bounded;
• give a closed bounded set that is not compact;
• use that a compact metric space is complete;
• apply the extreme value theorem and say where it fails.