Compactness in Finite Dimensions

Compactness Beyond Completeness

The previous tutorial showed that every finite-dimensional normed space is complete: each Cauchy sequence converges to a vector in the space. Compactness gives a stronger kind of control on subsets. Rather than asking whether a sequence whose terms become close has a limit, compactness ensures that every sequence in the set has a convergent subsequence whose limit remains in the set.

Throughout, let \(V\) be a finite-dimensional real normed vector space with norm \(N\). An open set in \(V\) means an open set for the metric \(d(x,y)=N(x-y)\). Compactness is defined by open covers, but in finite dimensions it can be recognized by sequences. This makes compactness practical in problems about limits and continuous functions.

The empty set is compact, since the empty subcollection covers it. The results about maxima and minima below will assume \(K\) is nonempty. For nonempty sets, the finite-dimensional Heine–Borel Property established in the tutorial Finite-Dimensional Normed Spaces says that a set is compact if and only if it is closed and bounded. We will use that theorem, not re-prove it. The new criterion below replaces the open-cover definition with a condition on sequences.

A Sequential Test for Compactness

Proof. First suppose \(K\) is compact, and let \((x_n)\) be a sequence in \(K\). We will find a point \(x\in K\) such that every open neighborhood of \(x\), relative to \(K\), contains \(x_n\) for infinitely many indices \(n\). If such a point exists, we can choose indices \(n_1<n_2<\cdots\) so that \(N(x_{n_j}-x)<1/j\). Then \(x_{n_j}\to x\), as required.

Suppose, to the contrary, that no such point exists. For every \(x\in K\), there is then a relative open neighborhood \(U_x\) of \(x\) that contains \(x_n\) for only finitely many indices. The sets \(U_x\), as \(x\) ranges over \(K\), cover \(K\). Compactness gives a finite subcover \(U_{x_1},\ldots,U_{x_m}\). Each of these sets contains terms with only finitely many indices, so their finite union contains terms with only finitely many indices. But the union covers \(K\), and every \(x_n\) lies in \(K\). It must therefore contain \(x_n\) for every index \(n\), a contradiction. This proves that the desired subsequence exists.

Now suppose every sequence in \(K\) has a subsequence converging to a point of \(K\). We show that \(K\) is closed and bounded in \(V\), then apply the finite-dimensional Heine–Borel Property.

To see that \(K\) is bounded, suppose it were unbounded. For each positive integer \(n\), choose \(x_n\in K\) with \(N(x_n)>n\). By hypothesis, some subsequence \(x_{n_j}\) converges to a point of \(K\). A convergent sequence is bounded, but \(N(x_{n_j})>n_j\) and \(n_j\to\infty\), which is impossible. Hence \(K\) is bounded.

To see that \(K\) is closed, let \(x\) belong to the closure of \(K\). For each \(n\geq1\), choose \(x_n\in K\) such that \(N(x_n-x)<1/n\). Then \(x_n\to x\). The hypothesis gives a subsequence \(x_{n_j}\) converging to some \(y\in K\). Since that subsequence also converges to \(x\), uniqueness of norm limits gives \(x=y\). Thus \(x\in K\), so \(K\) is closed. The Heine–Borel Property now shows that \(K\) is compact. \(\square\)

This argument also explains why the limit must be required to lie in \(K\). A subsequence may converge in the ambient space without its limit belonging to the set. Closedness prevents that failure. In the other direction, compactness rules out sequences that escape without any subsequence settling near a point of the set.

Continuous Functions Attain Their Extreme Values

The sequential characterization turns compactness into a useful tool for optimizing continuous functions. In one real variable, the Extreme-Value Theorem applies to compact subsets of \(\mathbb{R}\). The same conclusion holds when the domain is a compact subset of any finite-dimensional normed space.

Proof. First we prove that \(f\) is bounded. If it were unbounded, we could choose a sequence \((x_n)\) in \(K\) such that \(|f(x_n)|>n\). By the sequential characterization, some subsequence \(x_{n_j}\) converges to a point \(x\in K\). Continuity implies \(f(x_{n_j})\to f(x)\), so the real sequence \((f(x_{n_j}))\) is bounded. This contradicts \(|f(x_{n_j})|>n_j\to\infty\). Thus \(f(K)\) is bounded.

Let \(M=\sup f(K)\), which exists as a finite real number because \(f(K)\) is nonempty and bounded. For each \(n\geq1\), the definition of supremum gives \(x_n\in K\) such that \(M-1/n<f(x_n)\leq M\). Choose a subsequence \(x_{n_j}\) converging to \(x_{\max}\in K\). By continuity, \(f(x_{n_j})\to f(x_{\max})\). The inequalities

then imply \(f(x_{\max})=M\). Applying the same argument to the continuous function \(-f\) shows that \(-f\) attains its maximum on \(K\), so \(f\) attains its minimum there. This proves the result. \(\square\)

Uniform Continuity and a Common Pitfall

Compactness provides more than boundedness and attainment of extrema. A continuous function on a compact subset of a finite-dimensional normed space is uniformly continuous, even when that subset is not an interval. The proof again uses subsequences.

Proof. Suppose \(f\) were not uniformly continuous. Then there would be an \(\varepsilon_0>0\) such that, for every \(n\geq1\), there are \(x_n,y_n\in K\) satisfying

The sequential characterization gives a subsequence \(x_{n_j}\) converging to some \(x\in K\). The triangle inequality yields

Thus \(y_{n_j}\to x\) as well. Continuity at \(x\) implies \(f(x_{n_j})\to f(x)\) and \(f(y_{n_j})\to f(x)\). Hence \(|f(x_{n_j})-f(y_{n_j})|\to0\), contradicting the lower bound \(\varepsilon_0\). Therefore \(f\) is uniformly continuous. \(\square\)

A common pitfall is to confuse completeness with compactness. Completeness says that Cauchy sequences converge; it does not say that every bounded sequence has a convergent subsequence. The previous tutorial established completeness of the whole finite-dimensional space. Here, compactness of a subset gives the stronger subsequence property for every sequence in that subset. In finite dimensions, closedness and boundedness identify exactly which subsets have this stronger property, by the Heine–Borel Property.