Heine–Borel Theorem · aiwiki.page

The Heine–Borel theorem characterizes compactness in Euclidean space: a subset of $\mathbb{R}^n$ , with its usual topology, is compact if and only if it is closed and bounded. It connects a property defined through arbitrary open coverings with two comparatively elementary geometric conditions. The same characterization holds in $\mathbb{C}^n$ , identified with $\mathbb{R}^{2n}$ . (www2.math.upenn.edu)

Statement and terminology

For any positive integer $n$ and any subset $K\subseteq\mathbb{R}^n$ ,

K\text{ is compact} \quad\Longleftrightarrow\quad K\text{ is closed and bounded}.

Here closed means that $K$ is a closed set in the ambient space $\mathbb{R}^n$ , and bounded means that some finite $R$ satisfies

\|x\|\le R\qquad\text{for every }x\in K,

where $\|\cdot\|$ is the Euclidean norm. The ambient space and its usual topology are essential parts of the statement. (math.uchicago.edu)

An open cover of $K$ is a family $\{U_\alpha\}_{\alpha\in A}$ of open sets whose union contains $K$ . Compactness means that every such cover has a finite subcover:

K\subseteq U_{\alpha_1}\cup\cdots\cup U_{\alpha_m}.

The selected sets must come from the original cover. The cover itself need not be finite or countable. (dpmms.cam.ac.uk)

A common interval formulation states that every open cover of a closed bounded interval $[a,b]$ has a finite subcover. This one-dimensional result is also called the Heine–Borel theorem and serves as a starting point for the general Euclidean version. (maths.tcd.ie)

Proof structure

Compact sets are closed and bounded

This implication holds in every metric space, not just Euclidean space. For boundedness, cover a compact set by open balls of positive integer radii centered at a fixed point. A finite subcover places the set inside the ball with the largest selected radius. (math.uchicago.edu)

For closedness, take a point $y\notin K$ . For each $x\in K$ , let $r_x=d(x,y)/3$ . Finitely many balls $B(x_i,r_{x_i})$ cover $K$ . If

\delta=\min_i r_{x_i}>0,

the triangle inequality shows that $B(y,\delta)$ meets none of these balls. Thus the complement of $K$ is open, so $K$ is closed. This is a direct proof of the general metric-space implication. (math.uchicago.edu)

Closed bounded intervals are compact

Let $\mathcal U$ cover $[a,b]$ , and define

S=\{t\in[a,b]:[a,t]\text{ has a finite subcover from }\mathcal U\}.

The set $S$ contains $a$ and is bounded above. By the least-upper-bound property of the real numbers, $s=\sup S$ exists.

Choose $U\in\mathcal U$ containing $s$ , and choose $\varepsilon>0$ such that

(s-\varepsilon,s+\varepsilon)\subseteq U.

There is $t\in S$ with $t>s-\varepsilon$ . A finite cover of $[a,t]$ , together with $U$ , therefore covers $[a,s]$ and, if $s<b$ , extends beyond $s$ . The latter would contradict the definition of $s$ . Hence $s=b$ , and the same construction gives a finite cover of $[a,b]$ . (maths.tcd.ie)

Passage to higher dimensions

A finite Cartesian product of compact intervals is compact. Consequently, every closed box

[a_1,b_1]\times\cdots\times[a_n,b_n]

is compact. Any bounded subset of $\mathbb{R}^n$ lies in such a box, and a closed subset of a compact space is compact. These facts complete the converse implication. (www2.math.upenn.edu)

Sequential formulation

In metric spaces, compactness is equivalent to sequential compactness: every sequence has a subsequence converging to a point of the space. Thus the Euclidean theorem can also be stated as

K\text{ is closed and bounded} \quad\Longleftrightarrow\quad \text{every sequence in }K\text{ has a subsequence converging in }K.

The requirement that the limit belong to $K$ is indispensable. (dpmms.cam.ac.uk)

This formulation is closely related to the Bolzano–Weierstrass theorem, which asserts that every bounded sequence in $\mathbb{R}^n$ has a convergent subsequence. Closedness ensures that the resulting limit remains in the set. Conversely, sequential compactness forces both boundedness and closedness. (www2.math.upenn.edu)

Examples and applications

Applying the characterization gives compact sets such as closed intervals, closed balls, spheres, finite sets, and the Cantor set. Compactness requires neither a nonempty interior nor connectedness. By contrast, $(0,1)$ is bounded but not closed, while $\mathbb{R}$ is closed but not bounded; neither is compact. (www2.math.upenn.edu)

The open interval illustrates why excluding boundary points matters. The sets

U_m=(1/m,1),\qquad m=2,3,\ldots,

cover $(0,1)$ , but no finite subfamily does: a finite selection misses points sufficiently close to zero. This is an explicit open-cover witness to its noncompactness. (www2.math.upenn.edu)

The theorem makes compactness available in many arguments about continuous functions:

Attainment of extrema. By the extreme value theorem, a continuous real-valued function on a nonempty closed bounded Euclidean set attains its minimum and maximum.
Uniform continuity. A continuous map from a compact metric space to a metric space is uniformly continuous. In particular, this holds on closed bounded Euclidean domains.
Continuous images. The continuous image of a compact set is compact; if the image lies in Euclidean space, it is therefore closed and bounded. (maths.ox.ac.uk)

For optimization, the first consequence supplies an existence result: a continuous objective on a nonempty closed bounded feasible set has a global minimizer. It does not establish uniqueness or provide an algorithm for finding that minimizer. (maths.ox.ac.uk)

Scope and limitations

Closedness and boundedness do not characterize compactness in arbitrary metric spaces. For an explicit example, consider the standard unit vectors $e_1,e_2,\ldots$ in the Hilbert space $\ell^2$ . They lie in its closed unit ball, but

\|e_i-e_j\|=\sqrt2\qquad(i\ne j).

The sequence has no convergent subsequence, so that closed bounded ball is not compact. This illustrates the failure of the Euclidean criterion in infinite-dimensional settings. (math.uchicago.edu)

The general metric-space replacement is

K\text{ is compact} \quad\Longleftrightarrow\quad K\text{ is complete and totally bounded},

where $K$ carries its induced metric. Complete means that every Cauchy sequence converges in $K$ . Totally bounded means that, for every $\varepsilon>0$ , finitely many radius- $\varepsilon$ balls cover $K$ . Ordinary boundedness controls overall size; total boundedness supplies finite coverings at every scale. (dpmms.cam.ac.uk)

In a complete metric space, the corresponding subset criterion is “closed and totally bounded.” Euclidean space satisfies the stronger-looking Heine–Borel formulation because bounded Euclidean sets are totally bounded. (ncatlab.org)

Historical development

The theorem is named after Eduard Heine and Émile Borel, but its modern statement emerged through several contributions. Heine’s 1872 work concerned uniform continuity on closed intervals. In 1895, Borel published a finite-covering theorem for countable interval covers. Subsequent work by mathematicians including Arthur Schönflies and Henri Lebesgue extended the covering formulation beyond the countable case. The historical naming therefore reflects related proof methods and successive generalizations, rather than a joint publication by Heine and Borel. (mathshistory.st-andrews.ac.uk)

References

Compactness and Heine-Borel — Advanced Analysiswww2.math.upenn.edu
MATH 395 Notesmath.uchicago.edu
Compactness and the Heine-Borel Theorem (continued)maths.tcd.ie
Advanced Analysis — Compactness and Connectednesswww2.math.upenn.edu
Part A Mathematics Synopses 2023–24maths.ox.ac.uk
Heine-Borel theorem in nLabncatlab.org
Eduard Heine — MacTutor History of Mathematicsmathshistory.st-andrews.ac.uk