Compactness Implies Sequential Compactness

From Open Covers to Convergent Subsequences

Compactness is defined using open covers, while sequential compactness is phrased in terms of sequences. In a metric space, the bridge between them is a pair of useful metric properties: compactness gives total boundedness and completeness. Total boundedness lets us repeatedly find infinitely many terms inside smaller and smaller balls; completeness then supplies the limit of the resulting Cauchy subsequence.

Throughout, \((X,d)\) is a metric space, with its metric topology. It is compact if every open cover has a finite subcover. It is sequentially compact if every sequence in \(X\) has a subsequence converging to a point of \(X\). 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 \(X\) converges to a point of \(X\).

Definition: A metric space is totally bounded if for every \(\varepsilon>0\), it can be covered by finitely many open balls of radius \(\varepsilon\). Equivalently, for every \(\varepsilon>0\) there are \(p_1,\dots,p_r\in X\) such that \(X\subseteq\bigcup_{i=1}^r B(p_i,\varepsilon)\). The centers \(p_i\) are an \(\varepsilon\)-net for \(X\).

Compactness Gives Total Boundedness

Proposition: Every compact metric space is totally bounded.

Proof. If \(X=\varnothing\), the empty collection of balls is a finite cover for every \(\varepsilon>0\), so the conclusion holds. Otherwise, fix \(\varepsilon>0\). The collection \(\{B(x,\varepsilon):x\in X\}\) is an open cover of \(X\): each ball is open in the metric topology, and \(x\in B(x,\varepsilon)\) because \(d(x,x)=0<\varepsilon\). Compactness gives a finite subcover \(B(p_1,\varepsilon),\dots,B(p_r,\varepsilon)\). These finitely many balls cover \(X\), so \(X\) is totally bounded. \(\square\)

The centers of the finite cover can be chosen from \(X\), as they are the centers of the balls in the original cover. This matters when the cover is used to locate terms of a sequence in \(X\).

Compactness Gives Completeness

We use a consequence of compactness about closed sets. A family of sets has the finite intersection property if every finite subfamily has nonempty intersection.

Lemma: In a compact space, any family of closed sets with the finite intersection property has nonempty intersection.

Proof. Suppose a family of closed sets has the finite intersection property but its whole intersection is empty. The complements of the sets are open and cover the space, since every point fails to belong to at least one of the closed sets. A finite subcover would imply that the corresponding finite collection of closed sets has empty intersection. This contradicts the finite intersection property. Therefore the whole intersection is nonempty. \(\square\)

Proposition: Every compact metric space is complete.

Proof. If \(X=\varnothing\), there are no sequences in \(X\), so the completeness condition holds. Otherwise, let \((x_n)\) be a Cauchy sequence in \(X\). For each \(k\in\mathbb{N}\), define \(F_k=\overline{\{x_n:n\ge k\}}\), where closure is taken in \(X\). Each \(F_k\) is nonempty and closed, and \(F_{k+1}\subseteq F_k\). Thus every finite intersection of the \(F_k\) is nonempty: for any finite list, its intersection is the set with the largest index in that list. The lemma gives a point \(x\in\bigcap_{k\ge1}F_k\).

We show that \(x_n\to x\). Given \(\varepsilon>0\), choose \(N\) so that \(d(x_m,x_n)<\varepsilon/2\) whenever \(m,n\ge N\). Since \(x\in F_N\), the open ball \(B(x,\varepsilon/2)\) meets \(\{x_n:n\ge N\}\). Hence there is \(j\ge N\) with \(d(x_j,x)<\varepsilon/2\). For every \(n\ge N\), the triangle inequality gives \(d(x_n,x)\le d(x_n,x_j)+d(x_j,x)<\varepsilon/2+\varepsilon/2=\varepsilon\). This is convergence to \(x\in X\), so \(X\) is complete. \(\square\)

Totally Bounded Spaces Produce Cauchy Subsequences

Lemma: Every sequence in a totally bounded metric space has a Cauchy subsequence.

Proof. Let \((x_n)\) be a sequence in a totally bounded space \(X\). We construct nested infinite sets of indices \(A_1\supseteq A_2\supseteq\cdots\), and choose \(n_k\in A_k\) with \(n_1<n_2<\cdots\). At stage \(k\), total boundedness gives a finite cover of \(X\) by open balls of radius \(1/(2k)\). The infinitely many terms whose indices belong to \(A_{k-1}\) are distributed among finitely many balls. At least one ball contains terms with infinitely many such indices; let \(A_k\subseteq A_{k-1}\) be those indices. For the first stage, take \(A_0=\mathbb{N}\). Since \(A_k\) is infinite, it contains an index larger than \(n_{k-1}\), so the increasing choice of indices is possible.

Fix \(k\), and take arbitrary \(m,n\ge k\). Nesting gives \(n_m\in A_m\subseteq A_k\) and \(n_n\in A_n\subseteq A_k\). By the definition of \(A_k\), both \(x_{n_m}\) and \(x_{n_n}\) lie in the same selected ball of radius \(1/(2k)\). If \(a\) is its center, then \(d(x_{n_m},a)<1/(2k)\) and \(d(a,x_{n_n})<1/(2k)\). The triangle inequality yields \(d(x_{n_m},x_{n_n})<1/(2k)+1/(2k)=1/k\).

Given \(\varepsilon>0\), choose \(k\in\mathbb{N}\) with \(1/k<\varepsilon\). The preceding bound holds for every \(m,n\ge k\), so \((x_{n_k})\) is Cauchy. \(\square\)

Theorem: Every compact metric space is sequentially compact.

Proof. If \(X=\varnothing\), there are no sequences in \(X\), and the assertion holds vacuously. Otherwise, let \((x_n)\) be any sequence in \(X\). Compactness makes \(X\) totally bounded, so the lemma provides a Cauchy subsequence \((x_{n_k})\). Compactness also makes \(X\) complete, so this Cauchy subsequence converges to some point of \(X\). We have found a convergent subsequence of an arbitrary sequence in \(X\), which is exactly sequential compactness. \(\square\)

Worked Examples

Worked Example: A Finite Metric Space

Let \(X=\{p,q,r\}\) with \(d(p,q)=1\), \(d(q,r)=1\), and \(d(p,r)=2\), together with \(d(x,x)=0\) and symmetry. These distances satisfy the triangle inequality: the only largest distance is \(2\), and the other two distances on that triple add to \(1+1=2\). Thus they define a metric.

This space is compact: from any open cover, choose one cover member containing each of the three points. Those at most three sets are a finite subcover. For the sequence \(x_{3j-2}=p,\ x_{3j-1}=q,\ x_{3j}=r\), the subsequence with indices \(n_k=3k-2\) is constant at \(p\). Its terms converge to \(p\), since for every \(\varepsilon>0\), \(d(p,p)=0<\varepsilon\).

Check: For \(\varepsilon=1/2\), the three balls \(B(p,1/2),B(q,1/2),B(r,1/2)\) are respectively \(\{p\},\{q\},\{r\}\): each other point is at distance at least \(1>1/2\). They form a finite cover, as total boundedness requires at this radius.

Worked Example: The Unit Interval

Give \(I=[0,1]\) its usual metric \(d(x,y)=|x-y|\). To verify compactness directly, let \(\mathcal{U}\) be an open cover of \(I\), and let \(S\) be the set of \(t\in I\) for which \([0,t]\) is covered by finitely many members of \(\mathcal{U}\). A cover member containing \(0\) contains \([0,\delta)\cap I\) for some \(\delta>0\), by openness in the subspace topology. Thus \(t=\min(\delta/2,1)\) belongs to \(S\), so \(S\) is nonempty. Let \(c=\sup S\).

If \(c<1\), choose a cover member \(U\) containing \(c\). Openness gives \(\delta>0\) such that \((c-\delta,c+\delta)\cap I\subseteq U\). By the supremum property, there is \(s\in S\) with \(s>c-\delta/2\). The finite cover of \([0,s]\), together with \(U\), covers \([0,\min(c+\delta/2,1)]\). Since \(c<1\), this right endpoint is greater than \(c\), contradicting that \(c\) is an upper bound of \(S\). Hence \(c=1\). A cover member containing \(1\) contains \((1-\eta,1]\) for some \(\eta>0\). Choose \(s\in S\) with \(s>1-\eta\); a finite cover of \([0,s]\), together with that member, covers \(I\). Therefore \(I\) is compact.

Now take \(x_n=1/(n+1)\). The subsequence with \(n_k=k\) is the whole sequence. For any \(\varepsilon>0\), choose \(N\in\mathbb{N}\) with \(N+1>1/\varepsilon\). For \(k\ge N\), \(d(x_k,0)=1/(k+1)\le1/(N+1)<\varepsilon\). Thus this sequence converges to \(0\in I\), as the theorem guarantees a convergent subsequence must exist.

Check: The computed first terms are \(x_1=1/2,\ x_2=1/3,\ x_3=1/4\), each in \(I\). The distance formula is \(d(x_k,0)=|1/(k+1)-0|=1/(k+1)\), giving the stated convergence estimate.

Worked Example: A Compact Set with Infinitely Many Points

Let \(K=\{0\}\cup\{1/n:n\in\mathbb{N}\}\) with the metric inherited from \(\mathbb{R}\). To check compactness, take an open cover of \(K\) and choose a member \(U\) containing \(0\). Since \(U\) is open in \(K\), there is \(\varepsilon>0\) such that \(B_{\mathbb{R}}(0,\varepsilon)\cap K\subseteq U\). Choose \(N\in\mathbb{N}\) with \(N>1/\varepsilon\). For every \(n\ge N\), \(0<1/n\le1/N<\varepsilon\), so \(1/n\in U\). Only the finitely many points \(1,1/2,\dots,1/(N-1)\) remain; choose one cover member for each of them. Together with \(U\), these give a finite subcover. If \(N=1\), there are no remaining points to add. Thus \(K\) is compact.

For the sequence \(x_n=1/n\) in \(K\), choose the subsequence \(x_{n_k}=1/k\), with \(n_k=k\). Given \(\varepsilon>0\), choose \(N\in\mathbb{N}\) with \(N>1/\varepsilon\). For \(k\ge N\), \(d(x_{n_k},0)=|1/k-0|=1/k\le1/N<\varepsilon\). Hence this subsequence converges to \(0\in K\).

Check: The first three terms are \(1,\ 1/2,\ 1/3\), obtained by substituting \(n=1,2,3\) into \(1/n\). They are all in \(K\), and the displayed estimate verifies convergence of the full sequence, not merely of a selected further subsequence.

A Common Pitfall in the Cauchy Estimate

To prove a subsequence is Cauchy, the estimate must compare two arbitrary sufficiently late terms. Merely comparing each term with one fixed term does not give the claimed pairwise bound. For example, on the real line, both \(-0.9\) and \(0.9\) are within distance \(1\) of \(0\), but their mutual distance is \(|-0.9-0.9|=1.8\), which is not less than \(1\).

In the extraction proof, the correction is to take arbitrary \(m,n\ge k\), not just one later index and the fixed index \(k\). Nesting ensures both \(n_m\) and \(n_n\) belong to \(A_k\), so both corresponding terms lie in the same ball chosen at stage \(k\). If its radius is \(1/(2k)\), the triangle inequality gives the pairwise estimate \(d(x_{n_m},x_{n_n})<1/k\). Since \(1/k\) can be made smaller than any prescribed \(\varepsilon>0\), this is exactly the Cauchy condition.

Check Your Understanding

Use the definitions and proof steps above to answer these questions.

Why does the family of all balls \(B(x,\varepsilon)\), with \(x\in X\), cover \(X\) for every \(\varepsilon>0\)?
In the completeness proof, why do the closed sets \(F_k=\overline{\{x_n:n\ge k\}}\) have the finite intersection property?
In the subsequence construction, explain why one ball in the finite cover must contain terms with infinitely many indices from \(A_{k-1}\).
Why must the Cauchy estimate compare \(x_{n_m}\) and \(x_{n_n}\) for arbitrary \(m,n\ge k\)?
Where does the proof use completeness, and why does total boundedness alone not finish the argument?