Compactness
1 Compactness
Compactness is the right generalisation of “closed and bounded” to settings beyond \(\mathbb{R}^n\). Informally, a compact set cannot escape to infinity or have missing boundary points: any infinite collection of points in it must pile up somewhere inside it. It is the property that lets us extract limits, and it underlies the extreme value theorem, the fact that \(\overline{\mathbb{R}}\) is a compact metric space (see Metric Spaces), and tightness and Prokhorov’s theorem in probability (see Tightness).
2 Definitions
A subset \(K\) of a metric space \(X\) is (sequentially) compact if every sequence in \(K\) has a subsequence that converges to a point of \(K\).
An open cover of \(K\subseteq X\) is a family \(\{U_\alpha\}\) of open sets with \(K\subseteq\bigcup_\alpha U_\alpha\). The set \(K\) is compact if every open cover has a finite subcover, that is, finitely many \(U_{\alpha_1},\dots,U_{\alpha_n}\) with \(K\subseteq\bigcup_iU_{\alpha_i}\).
In a metric space the two definitions agree (proved below). The open cover definition is the one used in general Topology, where there are no sequences to speak of.
- \([0,1]\) and every closed bounded subset of \(\mathbb{R}^n\) is compact, as is any finite set.
- \((0,1)\) is not compact: the sequence \(1/n\) has no limit in \((0,1)\), and the open cover \(\{(1/n,1)\}_{n\geq2}\) has no finite subcover.
- \(\mathbb{R}\) is not compact: the sequence \(n\) has no convergent subsequence, and \(\{(-n,n)\}\) has no finite subcover.
- In infinite dimensions closed and bounded is not enough. The closed unit ball in \((C([0,1]),d_\infty)\) is not compact: \(f_n(x)=x^n\) has no uniformly convergent subsequence, as its pointwise limit is discontinuous at \(1\).
3 Basic Properties
A compact subset \(K\) of a metric space is closed and bounded.
Proof. Bounded: if not, fix \(x_0\in K\) and choose \(x_n\in K\) with \(d(x_n,x_0)>n\). Any subsequence is unbounded, so none converges, contradicting compactness. Closed: let \(x_n\in K\) with \(x_n\to x\). By compactness a subsequence converges to some \(y\in K\), but also converges to \(x\), so \(x=y\in K\) by uniqueness of limits. By the sequential criterion for closed sets in Metric Spaces, \(K\) is closed.
A closed subset \(F\) of a compact set \(K\) is compact.
Proof. A sequence in \(F\) lies in \(K\), so has a subsequence converging in \(K\); its limit lies in \(F\) since \(F\) is closed.
4 Heine-Borel
Every bounded sequence in \(\mathbb{R}^n\) has a convergent subsequence.
Proof. For \(n=1\): every real sequence has a monotone subsequence (call \(m\) a peak if \(x_m\geq x_k\) for all \(k>m\); infinitely many peaks give a decreasing subsequence, and otherwise after the last peak we can repeatedly find larger terms, giving an increasing one). A bounded monotone sequence converges, see Sequence Convergence. For \(\mathbb{R}^n\) pass to a subsequence converging in the first coordinate, then a further one converging in the second, and so on; after \(n\) steps all coordinates converge.
A subset of \(\mathbb{R}^n\) is compact if and only if it is closed and bounded.
Proof. (\(\Rightarrow\)) is the theorem above. (\(\Leftarrow\)) A sequence in a bounded set is bounded, so has a convergent subsequence by Bolzano-Weierstrass, and the limit lies in the set since it is closed.
5 Equivalence of the Definitions
For a subset \(K\) of a metric space, \(K\) is open cover compact if and only if it is sequentially compact.
Proof. (\(\Rightarrow\)) Let \(\{x_n\}\subseteq K\) have no subsequence converging in \(K\). Then every \(y\in K\) has a ball \(B_y\) containing \(x_n\) for only finitely many \(n\) (otherwise we could choose \(n_1<n_2<\cdots\) with \(x_{n_k}\to y\)). The balls \(B_y\) cover \(K\), so finitely many of them do, and then only finitely many indices \(n\) exist, a contradiction.
(\(\Leftarrow\)) Sketch. Sequential compactness gives (i) total boundedness: for every \(\epsilon>0\) finitely many \(\epsilon\)-balls cover \(K\) (otherwise choose points at distance at least \(\epsilon\) from each other, a sequence with no convergent subsequence), and (ii) a Lebesgue number \(\delta>0\) for any open cover, such that every ball of radius \(\delta\) lies in a single set of the cover. Covering \(K\) by finitely many \(\delta\)-balls then yields a finite subcover.
A metric space is thus compact if and only if it is complete and totally bounded.
6 Continuous Functions on Compact Sets
If \(f:K\to Y\) is continuous and \(K\) is compact, then \(f(K)\) is compact.
Proof. Let \(y_n=f(x_n)\) be a sequence in \(f(K)\). By compactness a subsequence \(x_{n_k}\to x\in K\), and by continuity \(y_{n_k}=f(x_{n_k})\to f(x)\in f(K)\).
A continuous function \(f:K\to\mathbb{R}\) on a nonempty compact set attains its maximum and minimum.
Proof. \(f(K)\) is compact in \(\mathbb{R}\), hence closed and bounded by Heine-Borel, so \(\sup f(K)\) and \(\inf f(K)\) are finite and belong to \(f(K)\).
A continuous function on a compact metric space is also uniformly continuous (Heine-Cantor). Continuous images of compact sets explain why a homeomorphism \(\varphi:\overline{\mathbb{R}}\to[-1,1]\) shows \(\overline{\mathbb{R}}\) is compact.