Why Compactness Supplies Missing Limits
A Cauchy sequence is one whose terms eventually become arbitrarily close to one another. That condition does not by itself guarantee that the sequence converges within its space: its terms may be approaching a point that is absent. Compactness prevents this particular failure in metric spaces. The proof has two parts: compactness gives a cluster point for the sequence, and the Cauchy condition then forces every sufficiently late term to be close to that point.
Throughout, \((X,d)\) is a metric space with its metric topology. A space is compact if every open cover has a finite subcover. A sequence \((x_n)\) is Cauchy if for every \(\varepsilon>0\), there is \(N\in\mathbb{N}\) such that \(d(x_m,x_n)<\varepsilon\) whenever \(m,n\ge N\). The space is complete if every Cauchy sequence in it converges to a point of the space.
Compactness Produces a Cluster Point
Proof. Suppose, to the contrary, that the sequence has no cluster point. Then for each \(x\in X\), the negation of the definition gives an open neighborhood \(U_x\) of \(x\) and an index \(N_x\in\mathbb{N}\) such that \(x_n\notin U_x\) for every \(n\ge N_x\). The sets \(U_x\), as \(x\) ranges over \(X\), form an open cover of \(X\). Compactness gives a finite subcover \(U_{x_1},\ldots,U_{x_r}\). Let \(N=\max\{N_{x_1},\ldots,N_{x_r}\}\). Since the sets cover \(X\), the point \(x_N\) belongs to at least one of them, say \(U_{x_i}\). But \(N\ge N_{x_i}\), so the defining property of \(U_{x_i}\) says \(x_N\notin U_{x_i}\), a contradiction. Therefore the sequence has a cluster point. \(\square\)
The cluster point is enough to construct a convergent subsequence. For each \(k\in\mathbb{N}\), the ball \(B(p,1/k)\) around a cluster point \(p\) contains terms with arbitrarily large indices. Choose indices \(n_1<n_2<\cdots\) so that \(x_{n_k}\in B(p,1/k)\). The choice can be made successively: after choosing \(n_{k-1}\), use the cluster-point property with \(N=n_{k-1}+1\). Then \(d(x_{n_k},p)<1/k\), which tends to zero, so \(x_{n_k}\to p\).
The Cauchy Condition Controls the Whole Sequence
Proof. If \(X\) is empty, there are no sequences in \(X\), so the definition of completeness holds vacuously. Otherwise, let \((x_n)\) be any Cauchy sequence in \(X\). By the lemma, it has a cluster point \(p\in X\). We show that \(x_n\to p\). Let \(\varepsilon>0\). By the Cauchy condition, there is \(N\in\mathbb{N}\) such that \(d(x_m,x_n)<\varepsilon/2\) for all \(m,n\ge N\). Since \(p\) is a cluster point, the neighborhood \(B(p,\varepsilon/2)\) contains some \(x_{n_k}\) with \(n_k\ge N\). Thus \(d(x_{n_k},p)<\varepsilon/2\). For any \(n\ge N\), both \(n\) and \(n_k\) are at least \(N\), so the triangle inequality gives $$ d(x_n,p)\le d(x_n,x_{n_k})+d(x_{n_k},p)<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon. $$ This holds for every \(n\ge N\), which is precisely \(x_n\to p\). Since \(p\in X\), every Cauchy sequence in \(X\) converges in \(X\), and \(X\) is complete. \(\square\)
The two ingredients do different jobs. Compactness provides a point \(p\) that the sequence returns arbitrarily close to at late indices. The Cauchy condition says that once one sufficiently late term is close to \(p\), all sufficiently late terms are close to that term and hence to \(p\).
Worked Example: A Compact Set with an Accumulation Point
Let \(K=\{0\}\cup\{1/n:n\in\mathbb{N}\}\), with the metric inherited from \(\mathbb{R}\). First verify compactness directly. Take any open cover of \(K\), and choose a member \(U_0\) containing \(0\). By the subspace topology, there is \(\delta>0\) such that \(K\cap(-\delta,\delta)\subseteq U_0\). For every \(n\) with \(n>1/\delta\), we have \(0<1/n<\delta\), so \(1/n\in U_0\). Only finitely many positive integers satisfy \(n\le 1/\delta\). For each corresponding point \(1/n\), choose one member of the cover containing it. Together with \(U_0\), these finitely many sets cover \(K\). Thus \(K\) is compact.
Now take \(x_n=1/n\). Given \(\varepsilon>0\), choose \(N\in\mathbb{N}\) with \(1/N<\varepsilon\). For \(m,n\ge N\), both \(1/m\) and \(1/n\) lie in \((0,1/N]\), so $$ |x_m-x_n|=\left|\frac1m-\frac1n\right|\le\frac1N<\varepsilon. $$ Thus \((x_n)\) is Cauchy. Also, for \(n\ge N\), \(|x_n-0|=1/n\le 1/N<\varepsilon\), so \(x_n\to0\), and \(0\in K\). For the specific choice \(\varepsilon=1/10\), \(N=11\) works because \(1/11<1/10\). This verifies the estimates at a concrete threshold as well as the general argument.
Check: \(0\) is a cluster point of this sequence: every ball \(B(0,r)\), with \(r>0\), contains \(1/n\) for every integer \(n>1/r\). The limit obtained by the Cauchy estimate is therefore the point supplied by compactness.
Worked Example: A Finite Compact Metric Space
Let \(X=\{0,2\}\) with the metric inherited from \(\mathbb{R}\). It is compact: from any open cover, choose one covering set containing \(0\) and one containing \(2\); those two sets cover \(X\). Consider \(x_n=2\) for \(n=1,2,3\) and \(x_n=0\) for every \(n\ge4\). For any \(\varepsilon>0\), take \(N=4\). If \(m,n\ge4\), then \(x_m=x_n=0\), and \(d(x_m,x_n)=|0-0|=0<\varepsilon\). So the sequence is Cauchy. For the same \(N\), \(d(x_n,0)=|0-0|=0<\varepsilon\) whenever \(n\ge4\), so it converges to \(0\).
In fact, every Cauchy sequence in this \(X\) is eventually constant. Apply the Cauchy condition with \(\varepsilon=1\). After some index \(N\), any two terms are at distance less than \(1\). Distinct points of \(X\) have distance \(|2-0|=2\), which is not less than \(1\). Hence all terms from index \(N\) onward are equal, and the sequence converges to that common value. This gives a direct verification of completeness in this particular finite space.
Check: the computed late-term distance is \(0\), and \(0<1\); the distance between the two distinct points is \(2\), and \(2\not<1\). The strict inequalities in the Cauchy condition are satisfied exactly as required.
Why the Compactness Hypothesis Matters
Worked Example: A Cauchy Sequence with No Limit in the Space
Give \(X=(0,1)\) the metric inherited from \(\mathbb{R}\), and set \(x_n=1/(n+1)\). For \(m,n\ge N\), both terms lie in \((0,1/(N+1)]\), so $$ |x_m-x_n|\le\frac{1}{N+1}. $$ Given \(\varepsilon>0\), choose \(N\) with \(1/(N+1)<\varepsilon\). The displayed inequality proves that \((x_n)\) is Cauchy.
It does not converge to a point of \(X\). Indeed, suppose it converged to some \(y\in(0,1)\). Choose \(n\) large enough that \(1/(n+1)<y/2\), which is possible because \(1/(n+1)\to0\) in \(\mathbb{R}\). Then \(|x_n-y|=y-x_n>y/2\). But convergence to \(y\), applied with the positive tolerance \(y/2\), requires \(|x_n-y|<y/2\) for all sufficiently large \(n\). These inequalities contradict each other. The sequence approaches \(0\), which is not in \(X\).
This space is not compact. The sets \(U_n=(1/n,1)\), for \(n\ge2\), are open in the subspace \(X\) and cover \(X\): for each \(x\in(0,1)\), choose \(n>1/x\), giving \(1/n<x<1\). Any finite collection of these sets has a largest index \(M\), and its union is \(U_M=(1/M,1)\), because the sets increase with \(n\). It misses \(1/(2M)\in(0,1)\). Thus this open cover has no finite subcover.
Common pitfall. A Cauchy sequence need not converge merely because its terms approach a real number. In the preceding example, saying “\(1/(n+1)\) converges to \(0\), so it converges in \((0,1)\)” uses a limit that is not a point of the space. Convergence in a metric space requires the limit to belong to that space. Compactness matters because it ensures the sequence has a cluster point inside the space; the Cauchy condition then turns that cluster point into the limit of the whole sequence.
Check Your Understanding
For each question, state which space and metric you are using, and keep track of whether the proposed limit belongs to that space.
- In the proof that compactness gives a sequence cluster point, why does a finite subcover contradict the assumed absence of a cluster point?
- Explain how to choose a subsequence converging to a cluster point \(p\) using the balls \(B(p,1/k)\).
- In the theorem proof, why is it important that the index \(n_k\) can be chosen at least as large as the Cauchy index \(N\)?
- Show that the sequence \(x_n=1-1/(n+1)\) is Cauchy in \((0,1)\), and determine whether its limit belongs to that space.
- For \(X=\{0,2\}\) with the inherited metric, explain why a Cauchy sequence cannot have both values occurring arbitrarily far along the sequence.