In the line and in every finite-dimensional real space, compactness has a description anyone can check: closed and bounded. That is the Heine–Borel theorem. It is a theorem about those spaces and not a definition of compactness — two lessons ago gave a closed bounded metric space that is not compact — and keeping the two apart is the point of this lesson.

In one country a passport is proof of citizenship. That is a fact about the country, not the meaning of citizenship, and travellers who forget the difference get into trouble at other borders.

The route to Heine–Borel in the plane: $[-a,a]\times[-a,a]$ is a product of two compact intervals and so is compact, a bounded set sits inside such a box, and a closed subset of a compact space is compact. The corners are drawn solid because they belong — without them the set would not be closed, and the theorem would not apply.

A subset of $\mathbb{R}^n$ is compact exactly when it is closed and bounded. One direction is the metric lesson of this chapter. The other assembles three facts: a closed interval is compact, a finite product of compact spaces is compact, and a closed subset of a compact space is compact.

• $[a,b]$ is compact. Taken on trust, as it has been since the chapter opened: the proof needs the least upper bound property of $\mathbb{R}$, which belongs to the analysis course.
• A product of finitely many compact spaces is compact. Taken on trust as well: the proof for two factors is the tube lemma, and the general case is Tychonoff's theorem, which is beyond this course.
• With those two, a closed box $[-a,a]^n$ is compact, and a closed bounded is a closed subset of such a box — so it is compact by the theorem of the lesson on compact subsets.

Boundedness puts the set inside a closed box. Each side of the box is a closed interval, which is compact — the fact this course takes on trust — and a finite product of compact spaces is compact, which is the second fact on trust. So the box is compact. The set is closed in the whole space, so it is closed in the box, and a closed subset of a compact space is compact. That last step is the one this chapter proved, and the two borrowed facts are named where they are used.

Boundedness means the set fits inside a ball, and a ball fits inside a box.

The one-dimensional fact, taken on trust because its proof needs the least upper bound property.

A finite product of compact spaces is compact, the second fact taken on trust.

A closed set stays closed in a subspace containing it, since closed sets of a subspace are traces.

A closed subset of a compact space is compact — proved earlier in this chapter.

In $\mathbb{R}^n$ — and only in spaces like it — compactness is exactly closed and bounded.

• In $\mathbb{R}^n$: compact exactly when closed and bounded.
• The proof uses a compact interval and a finite product, both on trust, and a closed subset of a compact space.
• The theorem fails in a general metric space, and the counterexamples are ordinary.
• In $\mathbb{R}^n$ compactness, sequential compactness and closed-and-bounded all agree.

Great work! You now know how to:

• state Heine–Borel and the three facts its proof uses;
• decide compactness of a subset of the plane;
• name the two facts this chapter takes on trust;
• say why closed and bounded is not the definition of compactness.