The Heine-Borel Theorem

The Extreme Value Theorem — every continuous real-valued function on a nonempty compact set attains a maximum and minimum — is proved directly from Heine–Borel: continuous images of compact sets are compact, and a compact subset of $\mathbb{R}$ is closed and bounded, hence contains its supremum and infimum. This single fact underlies optimization theory, the existence of best approximations, and the foundations of the calculus of variations. In real analysis, uniform continuity of continuous functions on closed bounded domains (a strengthening of pointwise continuity essential for interchange-of-limits arguments and for defining the Riemann integral rigorously) follows from Heine–Borel via a covering argument. In differential geometry, compactness of closed bounded submanifolds of $\mathbb{R}^n$ (spheres, tori embedded in Euclidean space, level sets of proper functions) guarantees existence of extremal geodesics, critical points of curvature functionals, and well-posedness of many variational problems. In numerical analysis and optimization, compactness of feasible regions (closed, bounded constraint sets in $\mathbb{R}^n$) guarantees that optimization algorithms searching for minima are searching over a space where a minimum actually exists, a prerequisite for convergence proofs of algorithms like gradient descent on constrained domains. In complex analysis, Montel's theorem and normal families arguments rely on compactness properties of closed bounded families of functions, echoing the same closed-and-bounded intuition in function space (though there compactness needs an extra equicontinuity hypothesis, precisely because function spaces are infinite-dimensional, as flagged by the $\ell^2$ counterexample above). In statistics and machine learning, compactness of parameter spaces (often closed bounded boxes in $\mathbb{R}^n$) is a standard regularity assumption ensuring maximum likelihood estimators or empirical risk minimizers exist. In cryptography and coding theory, finite-dimensional lattice-based arguments frequently invoke closed-bounded-compact reasoning when analyzing shortest-vector problems over bounded regions.

Ask the tutor: How does the equivalence between compactness and sequential compactness in metric spaces extend or fail to extend the Heine–Borel picture to infinite-dimensional settings?

The word "compact" should evoke the idea that a set cannot "leak" mass off to infinity and cannot have "holes" or "missing endpoints" that a sequence could march toward without ever arriving. Boundedness rules out escape to infinity: an unbounded set like $\mathbb{R}$ itself can be covered by intervals $(-n,n)$ for $n=1,2,3,\dots$, and no finite subcollection of these ever covers all of $\mathbb{R}$, because any finite subcollection has a largest $n$ and misses everything beyond it. Closedness rules out missing boundary points: the open interval $(0,1)$ can be covered by the sets $(1/n, 1)$ for $n \ge 2$, and again no finite subcollection works, because points arbitrarily close to $0$ are always left out by any finite choice of $n$'s. The genuinely deep part of the theorem is the converse direction: that closed and bounded is enough. The intuition is a divide-and-conquer, or "no escape by bisection," argument. Suppose a closed bounded set could be covered by open sets with no finite subcover. Enclose it in a large closed cube. Cut the cube into $2^n$ smaller sub-cubes (in each dimension, cut in half). If every sub-cube's intersection with the set had a finite subcover, then gluing these finitely many finite subcovers together would give a finite subcover of the whole set — a contradiction. So at least one sub-cube must also fail to have a finite subcover. Repeating this indefinitely produces an infinite nested sequence of shrinking cubes, each stubbornly resisting a finite subcover, whose diameters shrink to zero. Nested closed bounded sets in $\mathbb{R}^n$ with diameters going to zero must intersect in exactly one point (this is where completeness of $\mathbb{R}$ enters), and that point lies in some single open set of the original cover — which then covers an entire small cube from some stage onward, giving the finite subcover after all. This contradiction is the heart of the proof.

Example 1: The set $K = \{0\} \cup \{1/n : n \in \mathbb{N}\} \subseteq \mathbb{R}$. We verify compactness both via Heine–Borel and directly via covers, to see the theorem in action. Closed: The set of limit points of $K$ is exactly $\{0\}$ (since $1/n \to 0$ and all other points are isolated: for $n\ge 2$, the interval $(1/n - \delta, 1/n+\delta)$ with $\delta$ small enough contains no other point of $K$). Since $0 \in K$, $K$ contains all its limit points, so $K$ is closed. Bounded: Every element of $K$ lies in $[0,1]$, so $K \subseteq \overline{B}(0,1)$; bounded. By Heine–Borel, $K$ is compact. Let's confirm directly with an open cover, which illustrates why the point $0$ being included is essential. Let $\mathcal{U} = \{U_\alpha\}$ be any open cover of $K$. Since $0 \in K$, some $U_{\alpha_0}$ contains $0$, and being open it contains an interval $(-\varepsilon,\varepsilon)$ for some $\varepsilon>0$. By the Archimedean property, choose $N$ with $1/N < \varepsilon$; then $1/n \in U_{\alpha_0}$ for all $n \ge N$ (infinitely many points swallowed by a single set). Only the finitely many remaining points $1, 1/2, \dots, 1/(N-1)$ need individual open sets from the cover, say $U_{\alpha_1},\dots,U_{\alpha_{N-1}}$ with $1/i \in U_{\alpha_i}$. Then

a finite subcover. This matches the theorem's prediction exactly, and shows concretely how boundedness (all points in $[0,1]$) and closedness (the accumulation point $0$ is included) conspire to make the "escaping mass" get captured by one open set.

Example 2: The closed unit disk boundary union a spiraling arc: $K = \{(r\cos\theta, r\sin\theta) : r = 1, \ \theta \in [0, 4\pi]\} \subseteq \mathbb{R}^2$, i.e. the unit circle traced twice — but let's instead take a genuinely asymmetric non-trivial set. Let

This is a solid ellipse (semi-axes $1$ and $2$) with a small open disk removed near the point $(2,0)$ — but $(2,0)$ is outside the ellipse entirely (since the ellipse only reaches $x \in [-1,1]$), so in fact the removed disk doesn't intersect the ellipse at all, and $K$ is just the closed elliptical region:

(We include this seemingly-elaborate description precisely to illustrate a subtlety: one must actually check whether a "closed set minus an open set" is closed, since removing an open set from a closed set removes a closed set of points and can be checked as intersecting with a closed complement.) In general, for a closed set $A$ and open set $V$, $A \setminus V = A \cap (\mathbb{R}^2\setminus V)$ is an intersection of two closed sets, hence closed. Here, since the removed disk is disjoint from $A$ anyway, $K = A$ exactly. Closed: $K = \{(x,y) : x^2 + y^2/4 - 1 \le 0\}$ is the preimage of the closed set $(-\infty,0]$ under the continuous function $f(x,y) = x^2+y^2/4 - 1$, hence closed (preimages of closed sets under continuous maps are closed). Bounded: If $(x,y)\in K$ then $x^2 \le 1$ and $y^2 \le 4$, so $|x|\le 1, |y|\le 2$, giving $\|(x,y)\| \le \sqrt{1+4}=\sqrt5$. So $K \subseteq \overline B(0,\sqrt5)$. By Heine–Borel, $K$ is compact. As a check of the theorem's use: consider $g(x,y) = x^3 - 3xy^2$ (real part of $z^3$) restricted to $K$. Since $g$ is continuous and $K$ is compact, the Extreme Value Theorem (itself proved using Heine–Borel: continuous image of compact is compact, and compact subsets of $\mathbb{R}$ are closed and bounded hence contain their sup and inf) guarantees $g$ attains a maximum and minimum on $K$ — a conclusion that requires exactly the closed-and-bounded package Heine–Borel supplies, and would fail if we instead used the open elliptical region $\{x^2+y^2/4<1\}$, where $g$ has no maximum on the set (values increase toward the missing boundary without attaining a top value there).