Compactness in R

Compactness is one of the most important concepts in analysis. A set is compact if every open cover has a finite subcover. In $\mathbb{R}$ , the Heine-Borel theorem characterizes compact sets as precisely those that are closed and bounded. Compact sets behave like "finite sets" in many ways — continuous functions attain maxima and minima, sequences have convergent subsequences, and uniform continuity holds.

Definitions

Definition3.1Open cover

An open cover of a set $K \subseteq \mathbb{R}$ is a collection $\{U_\alpha\}_{\alpha \in I}$ of open sets such that

$K \subseteq \bigcup_{\alpha \in I} U_\alpha.$

Definition3.2Compact set

A set $K \subseteq \mathbb{R}$ is compact if every open cover of $K$ has a finite subcover. That is, for every collection $\{U_\alpha\}_{\alpha \in I}$ of open sets with $K \subseteq \bigcup_{\alpha} U_\alpha$ , there exist finitely many indices $\alpha_1, \ldots, \alpha_n$ such that

$K \subseteq U_{\alpha_1} \cup \cdots \cup U_{\alpha_n}.$

RemarkIntuition

Compactness says that no matter how you cover $K$ with open sets, you can always extract a finite subcover. This is a finiteness property: compact sets "look finite" from the perspective of covers.

ExampleFinite sets are compact

Every finite set $K = \{x_1, \ldots, x_n\}$ is compact. Given an open cover $\{U_\alpha\}$ , for each $x_i$ , choose $U_{\alpha_i}$ containing $x_i$ . Then $K \subseteq U_{\alpha_1} \cup \cdots \cup U_{\alpha_n}$ .

Example[a, b] is compact

The closed bounded interval $[a, b]$ is compact. This is the content of the Heine-Borel theorem (see Heine-Borel Theorem).

Example(a, b) is not compact

The open interval $(0, 1)$ is not compact. Consider the cover $\{(1/n, 1) \mid n \geq 2\}$ . Every point $x \in (0, 1)$ lies in some $(1/n, 1)$ (choose $n > 1/x$ ), so this is an open cover. But no finite subcover exists: if we take finitely many $(1/n_1, 1), \ldots, (1/n_k, 1)$ , their union is $(1/N, 1)$ where $N = \max(n_1, \ldots, n_k)$ , which does not contain $(0, 1/N)$ .

ExampleUnbounded sets are not compact

$\mathbb{R}$ is not compact. The cover $\{(-n, n) \mid n \in \mathbb{N}\}$ has no finite subcover (any finite union is bounded).

Heine-Borel Theorem

Theorem3.1Heine-Borel Theorem

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

RemarkOnly in finite dimensions

The Heine-Borel theorem is special to $\mathbb{R}^n$ (and finite-dimensional normed spaces). In infinite-dimensional spaces, closed and bounded sets need not be compact. For example, the closed unit ball in $\ell^\infty$ is not compact.

Sequential compactness

Definition3.3Sequential compactness

A set $K \subseteq \mathbb{R}$ is sequentially compact if every sequence in $K$ has a subsequence converging to a point in $K$ .

Theorem3.2Equivalence of compactness and sequential compactness

For subsets of $\mathbb{R}$ , the following are equivalent:

$K$ is compact (every open cover has a finite subcover).
$K$ is sequentially compact (every sequence has a convergent subsequence in $K$ ).
$K$ is closed and bounded.

RemarkBolzano-Weierstrass

Sequential compactness is closely related to the Bolzano-Weierstrass theorem: every bounded sequence in $\mathbb{R}$ has a convergent subsequence. For closed bounded sets, the limit stays in the set (by closedness).

ExampleSequences in [a, b]

Let $(x_n)$ be any sequence in $[a, b]$ . By Bolzano-Weierstrass, $(x_n)$ has a convergent subsequence $x_{n_k} \to L$ . Since $a \leq x_{n_k} \leq b$ for all $k$ , taking limits gives $a \leq L \leq b$ , so $L \in [a, b]$ . Thus $[a, b]$ is sequentially compact.

Properties of compact sets

Theorem3.3Closed subsets of compact sets

If $K$ is compact and $F \subseteq K$ is closed (in the subspace topology), then $F$ is compact.

Theorem3.4Continuous images of compact sets

If $K$ is compact and $f : K \to \mathbb{R}$ is continuous, then $f(K)$ is compact.

RemarkExtreme Value Theorem

Theorem 3.4 immediately implies the Extreme Value Theorem: if $f : [a, b] \to \mathbb{R}$ is continuous, then $f(K)$ is a compact subset of $\mathbb{R}$ , hence closed and bounded. Thus $f$ attains its maximum and minimum on $[a, b]$ .

ExampleContinuous function attains maximum

Let $f : [0, 1] \to \mathbb{R}$ be $f(x) = x^2$ . Then $f([0, 1]) = [0, 1]$ is compact. The maximum is $\max f = 1$ (attained at $x = 1$ ), and the minimum is $\min f = 0$ (attained at $x = 0$ ).

ExampleFailure on open intervals

Let $f : (0, 1) \to \mathbb{R}$ be $f(x) = 1/x$ . Then $(0, 1)$ is not compact, and $f((0, 1)) = (1, \infty)$ is unbounded (not compact). The function $f$ does not attain a maximum on $(0, 1)$ .

Summary

Compactness is a central concept in analysis:

Compact sets: every open cover has a finite subcover.
In $\mathbb{R}$ : compact $\Leftrightarrow$ closed and bounded (Heine-Borel).
Sequential compactness: every sequence has a convergent subsequence.
Continuous images of compact sets are compact (Extreme Value Theorem).

See Heine-Borel Theorem for the proof, and Continuity for applications.