Heine-Borel Theorem

A Practical Test for Compactness

In “Closed Intervals,” we proved compactness directly from the open-cover definition. The Heine–Borel Theorem gives a broader test: for subsets of the real line, compactness is equivalent to being both closed and bounded. It can often replace an open-cover argument with two simpler checks. The criterion is special to the real line; one should not assume that closedness and boundedness characterize compactness in every setting.

This is the theorem established in “Compactness in the Real Line.” We use it here as a criterion rather than re-proving it. Recall that compactness means every open cover has a finite subcover. In one direction, compact subsets of \(\mathbb{R}\) are closed and bounded; in the other, closed bounded subsets are compact. The empty set is included: it is closed and bounded, and it is compact because the empty subfamily covers it.

The theorem gives a quick method. To prove a set compact, establish both that it contains all its limit points (or equivalently is closed) and that it lies within some bounded interval. To prove it is not compact, it is enough to show that it fails either condition. The theorem does not say that every closed set or every bounded set is compact by itself.

Using the Criterion Carefully

Closedness and boundedness are independent requirements. Closedness prevents a set from omitting limit points that its own points approach. Boundedness prevents it from extending without limit in either direction. Neither property alone controls both possible failures of compactness.

A Uniform Scale for Open Covers

Compactness gives more than the existence of a finite subcover. For any open cover of a compact set, there is one positive radius that works uniformly at every point: around each point, a ball of that radius, restricted to the compact set, lies inside one member of the cover. This is the Lebesgue Number Lemma. The radius need not work for the entire ball in \(\mathbb{R}\), only for its intersection with the set being covered.

Proof. If \(K=\varnothing\), take \(\delta=1\); the assertion about every \(y\in K\) is then vacuously true. Suppose \(K\ne\varnothing\). For each \(x\in K\), choose \(U_x\in\mathcal{U}\) with \(x\in U_x\). Because \(U_x\) is open, there is \(r_x>0\) such that \(B_{2r_x}(x)\subseteq U_x\).

The balls \(B_{r_x}(x)\), for \(x\in K\), form an open cover of \(K\). Compactness gives a finite subcover \(B_{r_{x_1}}(x_1),\ldots,B_{r_{x_m}}(x_m)\). Define \(\delta=\min\{r_{x_1},\ldots,r_{x_m}\}\), which is positive because it is the minimum of finitely many positive numbers.

Take any \(y\in K\). Choose an index \(i\) such that \(y\in B_{r_{x_i}}(x_i)\). If \(z\in B_\delta(y)\cap K\), then \(|z-y|<\delta\leq r_{x_i}\) and \(|y-x_i|<r_{x_i}\). The triangle inequality gives \(|z-x_i|\leq|z-y|+|y-x_i|<\delta+r_{x_i}\leq2r_{x_i}\). Thus \(z\in B_{2r_{x_i}}(x_i)\subseteq U_{x_i}\). This holds for every such \(z\), so \(B_\delta(y)\cap K\subseteq U_{x_i}\). The same \(\delta\) works for every \(y\in K\), proving the lemma. \(\square\)

The finite subcover is what makes a uniform radius possible. Each individual point has some suitable radius, but those radii might vary and might become arbitrarily small across an unbounded or noncompact set. Compactness reduces the covering to finitely many balls, after which the minimum of finitely many positive radii is still positive.

Bounded Infinite Sets Have Accumulation Points

Another consequence combines Heine–Borel with the open-cover definition. A bounded infinite set cannot have all its points isolated from the rest of the set: at least one point of the real line must be an accumulation point. This is the Bolzano–Weierstrass Theorem for subsets of \(\mathbb{R}\).

Proof. Let \(E\) be bounded and infinite, and let \(K=\overline{E}\), its closure. Since \(E\) is bounded, it is contained in some closed bounded interval \([-M,M]\). That interval is closed and contains \(E\), so the closure-as-smallest-closed-superset theorem gives \(K\subseteq[-M,M]\). The closure \(K\) is closed, and it is bounded by this inclusion. The Heine–Borel Theorem therefore shows that \(K\) is compact.

Suppose, for a contradiction, that no point of \(K\) is an accumulation point of \(E\). For each \(x\in K\), the definition of accumulation point then gives some \(r_x>0\) such that \(B_{r_x}(x)\cap(E\setminus\{x\})=\varnothing\). Thus \(B_{r_x}(x)\cap E\subseteq\{x\}\), so each such ball contains at most one point of \(E\). The balls \(B_{r_x}(x)\), \(x\in K\), form an open cover of \(K\). Since \(K\) is compact, finitely many of them cover \(K\). Their union can contain only finitely many points of \(E\), because each ball contains at most one. But \(E\subseteq K\), so these balls must cover the infinite set \(E\), a contradiction. Therefore some point of \(K\) is an accumulation point of \(E\), as required. \(\square\)

When the Theorem Is Useful

Heine–Borel is a decision tool and a source of compactness in proofs. If a set is visibly a closed bounded interval, the result from “Closed Intervals” gives compactness directly; for a more general subset of \(\mathbb{R}\), Heine–Borel often gives a shorter route. Once compactness is known, results such as the Lebesgue Number Lemma and Bolzano–Weierstrass can supply uniform control or accumulation points.

A common pitfall is to check only one of the two hypotheses. The interval \((0,4]\) is bounded but not closed, and \(\mathbb{Z}\) is closed but not bounded. Neither is compact. Another pitfall is to treat the criterion as universal across mathematics: the equivalence stated here concerns subsets of \(\mathbb{R}\). Always check the setting and the hypotheses before applying a compactness criterion.