Heine-Borel Theorem

The Heine-Borel Theorem is a fundamental characterization of compactness in Euclidean spaces. It states that a subset of $\mathbb{R}$ (or $\mathbb{R}^n$ ) is compact if and only if it is closed and bounded. This provides a simple criterion for compactness and is essential for proving the Extreme Value Theorem, Uniform Continuity, and many other results in analysis.

Statement

Theorem3.1Heine-Borel Theorem

Let $K \subseteq \mathbb{R}$ . Then $K$ is compact if and only if $K$ is closed and bounded.

RemarkOnly in finite dimensions

The Heine-Borel theorem holds in $\mathbb{R}^n$ but not in general infinite-dimensional spaces. For example, in the Banach space $\ell^\infty$ , the closed unit ball is closed and bounded but not compact.

Proof

Proof(⇒) Compact implies closed and bounded

Bounded: Suppose $K$ is compact. Consider the open cover $\{(-n, n) \mid n \in \mathbb{N}\}$ . By compactness, finitely many of these intervals cover $K$ , say $K \subseteq (-N, N)$ for some $N$ . Thus $K$ is bounded.

Closed: We show the complement $\mathbb{R} \setminus K$ is open. Let $x \in \mathbb{R} \setminus K$ . For each $y \in K$ , the sets

$U_y = \left(-\infty, \frac{x+y}{2}\right) \quad \text{and} \quad V_y = \left(\frac{x+y}{2}, \infty\right)$

are open, $y \in U_y$ , and $x \in V_y$ . The collection $\{U_y \mid y \in K\}$ is an open cover of $K$ . By compactness, finitely many cover $K$ , say $K \subseteq U_{y_1} \cup \cdots \cup U_{y_n}$ . Let $V = V_{y_1} \cap \cdots \cap V_{y_n}$ (a finite intersection of open sets, hence open). Then $x \in V$ and $V \cap K = \varnothing$ (since each $V_{y_i}$ is disjoint from $U_{y_i}$ ). Thus $V$ is a neighborhood of $x$ contained in $\mathbb{R} \setminus K$ , so $\mathbb{R} \setminus K$ is open.

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Proof(⇐) Closed and bounded implies compact

Suppose $K$ is closed and bounded. Since $K$ is bounded, $K \subseteq [a, b]$ for some interval $[a, b]$ .

Claim: $[a, b]$ is compact.

Let $\{U_\alpha\}_{\alpha \in I}$ be an open cover of $[a, b]$ . Define

$S = \{x \in [a, b] \mid [a, x] \text{ can be covered by finitely many } U_\alpha\}.$

Then $a \in S$ (since $a \in U_\alpha$ for some $\alpha$ ). Also, $S$ is bounded above by $b$ . By completeness, $c = \sup S$ exists. We show $c = b$ and $c \in S$ , completing the proof.

Since $c \in [a, b]$ , there exists $\alpha_0$ such that $c \in U_{\alpha_0}$ . Since $U_{\alpha_0}$ is open, there exists $\epsilon > 0$ such that $(c - \epsilon, c + \epsilon) \subseteq U_{\alpha_0}$ .

Since $c = \sup S$ , there exists $x \in S$ with $c - \epsilon < x \leq c$ . By definition of $S$ , $[a, x]$ can be covered by finitely many $U_\alpha$ . Adding $U_{\alpha_0}$ to this collection, we see that $[a, \min(c + \epsilon/2, b)]$ can be covered by finitely many $U_\alpha$ . Thus $\min(c + \epsilon/2, b) \in S$ . If $c < b$ , this contradicts $c = \sup S$ . Thus $c = b$ and $b \in S$ , so $[a, b]$ is covered by finitely many $U_\alpha$ .

Conclusion: Since $[a, b]$ is compact and $K \subseteq [a, b]$ is closed, $K$ is compact (closed subsets of compact sets are compact).

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Applications

ExampleExtreme Value Theorem

If $f : [a, b] \to \mathbb{R}$ is continuous, then $f$ attains its maximum and minimum on $[a, b]$ .

Proof: $[a, b]$ is compact (by Heine-Borel), so $f([a, b])$ is compact (continuous images of compact sets are compact). Thus $f([a, b])$ is closed and bounded, hence $\sup f([a, b])$ and $\inf f([a, b])$ exist and are attained.

ExampleUniform continuity on compact sets

If $K$ is compact and $f : K \to \mathbb{R}$ is continuous, then $f$ is uniformly continuous on $K$ . This uses Heine-Borel to show that any continuous function on a closed bounded interval is uniformly continuous.

ExampleLebesgue number lemma

If $K$ is compact and $\{U_\alpha\}$ is an open cover of $K$ , there exists $\delta > 0$ (the Lebesgue number) such that every subset of $K$ with diameter $< \delta$ is contained in some $U_\alpha$ . This is proved using Heine-Borel.

Summary

The Heine-Borel theorem characterizes compactness in $\mathbb{R}$ :

Compact $\Leftrightarrow$ closed and bounded.
Proof uses completeness (least upper bound property) for closed and bounded $\Rightarrow$ compact.
Applications: Extreme Value Theorem, uniform continuity, Lebesgue number lemma.

See Compactness for more on compact sets, and Continuity for applications to continuous functions.