Compactness as a Finite-Subcover Property
Completeness, considered in the previous tutorial, asks whether every Cauchy sequence has a limit in the space. Compactness is a different kind of control: it says that every open cover can be reduced to finitely many sets. In \(\mathbb{R}^n\), compactness also has a useful sequential description. Establishing that connection lets us use the sequence results developed earlier while keeping the open-cover definition as the foundation.
A set open in the subspace \(K\) has the form \(K\cap V\), where \(V\) is open in \(\mathbb{R}^n\). Equivalently, compactness can be tested using covers of \(K\) by open subsets of \(\mathbb{R}^n\): a finite subcover in either formulation gives one in the other. The empty set is compact, since the empty subfamily covers it.
The definition requires a finite subcover for every open cover. Showing that one particular cover has a finite subcover is not enough. Conversely, to prove that a set is not compact, it suffices to exhibit a single open cover that has no finite subcover.
Compactness and Subsequences
Call \(K\) sequentially compact if every sequence in \(K\) has a subsequence converging to a point of \(K\). In general topological spaces, this condition need not be equivalent to compactness. The equivalence does hold in \(\mathbb{R}^n\), using both the metric and the countability properties established earlier in the course.
Proof. First suppose \(K\) is compact, and let \((x^{(k)})_{k\geq1}\) be a sequence in \(K\). We show that some point \(x\in K\) has the property that every open ball about \(x\) contains \(x^{(k)}\) for infinitely many indices \(k\). If there were no such point, then for every \(x\in K\) there would be an open ball \(B_{r_x}^{(2)}(x)\) containing terms with only finitely many indices. These balls cover \(K\). Compactness gives a finite subcover. Each ball in that finite subcover contains terms at only finitely many indices, so their union contains terms at only finitely many indices. But every \(x^{(k)}\) belongs to the union, which would mean that all positive integer indices form a finite set. This is impossible.
Choose a point \(x\in K\) with the stated property. We construct a subsequence converging to \(x\). For each positive integer \(j\), the ball \(B_{1/j}^{(2)}(x)\) contains terms for infinitely many indices. We can therefore choose indices \(k_1<k_2<\cdots\) such that \(x^{(k_j)}\in B_{1/j}^{(2)}(x)\). Then \[ \|x^{(k_j)}-x\|_2<\frac{1}{j}, \] so \(x^{(k_j)}\to x\in K\). Thus \(K\) is sequentially compact.
For the converse, suppose every sequence in \(K\) has a subsequence converging to a point of \(K\). If \(K\) is empty, it is compact, so suppose \(K\) is nonempty. Earlier results show that \(\mathbb{R}^n\) is second-countable and that every subspace of a second-countable space is second-countable. By the theorem that every second-countable space is Lindelöf, every open cover of \(K\) has a countable subcover. Write such a subcover as \(U_1,U_2,\ldots\); if it is finite, the required finite subcover already exists.
Assume, seeking a contradiction, that this countable cover has no finite subcover. For every positive integer \(m\), choose \[ x^{(m)}\in K\setminus(U_1\cup\cdots\cup U_m). \] By sequential compactness, some subsequence \(x^{(m_j)}\) converges to a point \(x\in K\), where the indices \(m_j\) increase without bound. Since the \(U_i\) cover \(K\), there is an index \(r\) with \(x\in U_r\). The set \(U_r\) is open in \(K\), so convergence implies that \(x^{(m_j)}\in U_r\) for all sufficiently large \(j\). But \(m_j\geq r\) for all sufficiently large \(j\), and the choice of \(x^{(m_j)}\) puts it outside \(U_r\) whenever \(m_j\geq r\). This contradiction proves that the cover has a finite subcover. Hence \(K\) is compact. \(\square\)
The two directions use different features of Euclidean space. Compactness produces a subsequence by covering the set with neighborhoods; the reverse direction uses second-countability to reduce an arbitrary cover to a countable one. In later arguments, the sequential form is often easier to apply when a sequence is already available.
Worked Example: Finite Sets Are Compact
Let \(K=\{p_1,\ldots,p_m\}\subseteq\mathbb{R}^n\), where \(m\) is a positive integer, and let \(\mathcal{U}\) be any open cover of \(K\). For each \(i\), choose a member \(U_i\in\mathcal{U}\) containing \(p_i\). Then \(U_1,\ldots,U_m\) cover every point of \(K\), so they form a finite subcover. Thus \(K\) is compact. If \(K\) is empty, compactness holds by the definition. The argument depends on having only finitely many points: it does not show that an infinite set is compact.
Worked Example: A Closed Cube Is Compact
Fix real numbers \(a_i\leq b_i\) for \(i=1,\ldots,n\), and consider the cube \[ Q=[a_1,b_1]\times\cdots\times[a_n,b_n]. \] Take any sequence \(x^{(k)}\) in \(Q\). Its first coordinate is a bounded real sequence, so the Bolzano–Weierstrass Theorem for real sequences gives a subsequence along which the first coordinate converges. From that subsequence, select a further subsequence along which the second coordinate converges. Continue through the finitely many coordinates. The resulting subsequence has a limit in every coordinate. Each coordinate limit lies in its interval \([a_i,b_i]\), since that interval is closed. By the Coordinate Criterion for Convergence, the subsequence converges in \(\mathbb{R}^n\) to a point of \(Q\). Thus \(Q\) is sequentially compact, and the theorem shows that it is compact.
If \(a_i=b_i\) for one or more coordinates, those coordinates are constant throughout the sequence and their limits remain in the corresponding one-point intervals. If all coordinates are degenerate, \(Q\) is a singleton and the same conclusion holds.
Compact Subsets Are Closed and Bounded
Compactness places strong restrictions on a subset of Euclidean space. Since \(\mathbb{R}^n\) is Hausdorff, the earlier theorem “Compact Subsets of Hausdorff Spaces Are Closed” already implies that every compact \(K\subseteq\mathbb{R}^n\) is closed. Boundedness follows directly from the open-cover definition.
Proof. For each positive integer \(m\), let \(B_m^{(2)}(0)\) be the open Euclidean ball of radius \(m\) centered at the origin. The family \[ \{B_m^{(2)}(0):m\geq1\} \] covers \(\mathbb{R}^n\), and therefore covers \(K\). If \(K\) is compact, finitely many of these balls cover it. Let \(M\) be the largest radius index among those finitely many balls. Since the balls are nested, their union is \(B_M^{(2)}(0)\). Consequently \(K\subseteq B_M^{(2)}(0)\), so \(K\) is bounded. The empty set is bounded as well. \(\square\)
Worked Example: Euclidean Space Is Not Compact
For each positive integer \(m\), let \(U_m=B_m^{(2)}(0)\). These open balls cover \(\mathbb{R}^n\), since for every \(x\in\mathbb{R}^n\) an integer \(m>\|x\|_2\) satisfies \(x\in U_m\). A finite selection of these balls has a largest radius, say \(M\), and their union is just \(B_M^{(2)}(0)\). It does not cover \(\mathbb{R}^n\): for example, the vector \((M+1,0,\ldots,0)\) has Euclidean norm \(M+1\) and lies outside that ball. Hence this open cover has no finite subcover, and \(\mathbb{R}^n\) is not compact.
This agrees with the boundedness theorem: \(\mathbb{R}^n\) is unbounded when \(n\geq1\). The open-cover argument, however, proves noncompactness directly from the definition.
Worked Example: The Open Unit Ball Is Not Compact
For \(n\geq1\), let \(B_1^{(2)}(0)=\{x\in\mathbb{R}^n:\|x\|_2<1\}\), and let \(e_1=(1,0,\ldots,0)\). For each integer \(k\geq2\), define \(x^{(k)}=(1-1/k)e_1\). Since \[ \|x^{(k)}\|_2=1-\frac{1}{k}<1, \] every term belongs to the open unit ball. In the ambient Euclidean space, \(x^{(k)}\to e_1\), because \[ \|x^{(k)}-e_1\|_2=\frac{1}{k}\longrightarrow0. \] Every subsequence has the same ambient limit \(e_1\), which is not in the open unit ball because \(\|e_1\|_2=1\). By uniqueness of limits, no subsequence can converge to a point of the ball. The compactness and sequential compactness theorem therefore shows that the open unit ball is not compact.
How to Use Compactness Carefully
The open-cover definition and the sequential criterion serve different purposes. Use the definition when a cover is given or when a direct cover can be constructed. Use sequential compactness when you can produce a sequence and need a convergent subsequence. The equivalence proved here licenses switching between these approaches for subsets of \(\mathbb{R}^n\).
A frequent pitfall is to confuse completeness with compactness. Completeness controls Cauchy sequences, while sequential compactness requires a convergent subsequence from every sequence. For example, \(\mathbb{R}^n\) is complete, but it is not compact. Another pitfall is to assume that boundedness alone guarantees compactness: the open unit ball is bounded, yet it is not compact. The next tutorial will identify the precise relationship between compactness, closedness, and boundedness in \(\mathbb{R}^n\).
Check Your Understanding
Use the definition and results in this tutorial to answer each question.
- What must be true of every open cover of a compact set?
- In the proof that compactness implies sequential compactness, why would a finite subcover by neighborhoods containing only finitely many sequence indices lead to a contradiction?
- Where is second-countability used to prove that sequential compactness implies compactness?
- Why does the Bolzano–Weierstrass Theorem in each coordinate give a convergent subsequence in the closed cube?
- How does the family of balls \(B_m^{(2)}(0)\) show that \(\mathbb{R}^n\) is not compact?
- Why does the sequence approaching \(e_1\) show that the open unit ball is not compact?