Compact Implies Sequentially Compact

From Open Covers to Subsequences

Open-cover compactness and sequential compactness describe different kinds of control. Compactness starts with an open cover and guarantees a finite subcover. Sequential compactness starts with a sequence and guarantees a subsequence that converges to a point of the space. In a metric space, the first kind of control implies the second: if a sequence had no point approached by infinitely many of its terms, small open balls would give a cover with no finite subcover.

The argument has two parts. First, compactness forces every sequence to have a cluster point, meaning that every neighborhood of that point contains terms from infinitely many indices. Second, choosing terms successively closer to the cluster point produces a convergent subsequence. The infinite-index condition matters: a ball containing a term that appears only once does not by itself provide terms for a subsequence.

Definition: Let \((x_n)\) be a sequence in a metric space \((X,d)\). A point \(x\in X\) is a cluster point of the sequence if, for every \(r>0\), the set \(\{n\in\mathbb{N}:x_n\in B_r(x)\}\) is infinite. Equivalently, every open ball centered at \(x\) contains terms \(x_n\) for infinitely many indices \(n\).

This definition concerns indices, not just distinct values. If a sequence takes the same value infinitely often, that value is a cluster point, even though the sequence may have only finitely many distinct values. Conversely, if a point occurs just once, that occurrence does not make it a cluster point.

Compactness Forces a Cluster Point

Theorem (Every Sequence in a Compact Metric Space Has a Cluster Point): Let \((X,d)\) be a compact metric space. Every sequence in \(X\) has a cluster point in \(X\).

Proof. Take any sequence \((x_n)\) in \(X\). Suppose, for contradiction, that it has no cluster point in \(X\). Then for each \(x\in X\), the definition of cluster point fails. Thus there is a radius \(r_x>0\) for which only finitely many indices \(n\) satisfy \(x_n\in B_{r_x}(x)\).

The collection of balls \(\{B_{r_x}(x):x\in X\}\) is an open cover of \(X\), since each point \(x\) belongs to its own ball. Compactness provides a finite subcover, say

X\subseteq B_{r_{x_1}}(x_1)\cup\cdots\cup B_{r_{x_m}}(x_m).

For each \(i\), only finitely many indices \(n\) have \(x_n\in B_{r_{x_i}}(x_i)\). The union of these \(m\) finite sets of indices is finite. But every term \(x_n\) belongs to at least one ball in the finite subcover, so every positive integer \(n\) must belong to that union. This would make the set of all positive integers finite, a contradiction. Therefore \((x_n)\) has a cluster point in \(X\). \(\square\)

The key move is to turn the failure of a sequence property into an open cover. The cover is not chosen in advance: its balls are selected specifically so that each one accounts for only finitely many terms by index. A finite subcover would then account for only finitely many indices altogether, which cannot cover an infinite sequence.

Extracting a Convergent Subsequence

Theorem (Compact Metric Spaces Are Sequentially Compact): Every compact metric space is sequentially compact.

Proof. Let \((X,d)\) be compact, and take any sequence \((x_n)\) in \(X\). By the preceding theorem, it has a cluster point \(x\in X\). We construct increasing indices \(n_1<n_2<\cdots\) such that \(d(x_{n_k},x)<1/k\) for every \(k\).

Because \(x\) is a cluster point, the ball \(B_1(x)\) contains terms for infinitely many indices. Choose one such index as \(n_1\). Suppose \(n_{k-1}\) has been chosen. The ball \(B_{1/k}(x)\) again contains terms for infinitely many indices. An infinite set of positive integers cannot be contained in the finite set \(\{1,\ldots,n_{k-1}\}\), so there is an index \(n_k>n_{k-1}\) with \(x_{n_k}\in B_{1/k}(x)\). This completes the recursive construction.

For every \(k\), the construction gives \(d(x_{n_k},x)<1/k\). Since \(1/k\to0\), it follows that \(x_{n_k}\to x\). The limit belongs to \(X\), because the cluster point was found in \(X\). Thus every sequence in \(X\) has a subsequence converging to a point of \(X\), which is precisely sequential compactness. \(\square\)

The metric is used in the extraction step: balls of radii \(1/k\) provide a sequence of neighborhoods whose sizes shrink to zero. At each stage, infinitely many available indices ensure that one can choose a new index larger than the previous one. This guarantees a genuine subsequence rather than merely a list of terms that might repeat the same index.

Worked Applications

Worked Example: Sequences in a Compact Interval

The Heine–Borel Theorem says that a closed and bounded interval in \(\mathbb{R}\), with its usual metric, is compact. In particular, \([0,1]\) is compact. Consider the sequence

x_n=n\sqrt{2}-\lfloor n\sqrt{2}\rfloor,\qquad n\geq1,

where \(\lfloor t\rfloor\) denotes the greatest integer less than or equal to \(t\). The defining property of the floor gives

\lfloor n\sqrt{2}\rfloor\leq n\sqrt{2}<\lfloor n\sqrt{2}\rfloor+1,

so \(0\leq x_n<1\), and hence \(x_n\in[0,1]\) for every \(n\). By the theorem, there are increasing indices \(n_k\) and a point \(x\in[0,1]\) such that \(x_{n_k}\to x\). No formula for the indices or the limit is needed for the existence conclusion. The point is that compactness guarantees a limit in the interval, even though the sequence was specified by a less transparent rule.

Worked Example: A Sequence on the Unit Circle

Let

S^1=\{(a,b)\in\mathbb{R}^2:a^2+b^2=1\}

with the Euclidean metric. This set is bounded because every \((a,b)\in S^1\) satisfies \(a^2+b^2=1\), so its Euclidean distance from the origin is \(1\). It is closed because it is the inverse image of the closed set \(\{1\}\) under the continuous function \((a,b)\mapsto a^2+b^2\). The map \(t\mapsto(\cos t,\sin t)\) is continuous on the compact interval \([0,2\pi]\) and has image \(S^1\); hence \(S^1\) is compact, since continuous images of compact spaces are compact.

For each positive integer \(n\), define \(p_n=(\cos(n^2),\sin(n^2))\). The identity \(\cos^2(t)+\sin^2(t)=1\) gives

\cos^2(n^2)+\sin^2(n^2)=1,

so \(p_n\in S^1\). Compactness yields a convergent subsequence \(p_{n_k}\to(a,b)\) with \((a,b)\in S^1\). In particular, both coordinates converge: the absolute difference in either coordinate is at most the Euclidean distance between the points. Taking limits in the circle equation gives \(a^2+b^2=1\). Thus the subsequence does not merely converge somewhere in the plane; its limit remains on the circle.

Worked Example: A Finite Metric Space

Let \(X=\{u,v,w\}\) have any metric, and consider an arbitrary sequence \((x_n)\) in \(X\). At least one of the three values must occur infinitely many times. Otherwise, each value would occur only finitely many times, and the union of those three finite sets of indices would be finite, even though it would have to contain every positive integer. Let \(q\in X\) be a value occurring infinitely often, and choose increasing indices \(n_1<n_2<\cdots\) with \(x_{n_k}=q\) for every \(k\). Then

d(x_{n_k},q)=d(q,q)=0

for every \(k\), so \(x_{n_k}\to q\). This direct argument illustrates the cluster-point mechanism: every ball centered at \(q\) contains the terms at all the selected indices. It also shows why repeated values pose no difficulty for sequential compactness.

For completeness, \(X\) is compact: from any open cover, choose one cover member containing each of the three points. Those at most three members form a finite subcover. The general theorem applies, while the argument above makes its conclusion explicit in this finite case.

Why the Limit Must Stay in the Space

The cluster-point theorem guarantees a point of \(X\), not merely a point in some larger space where \(X\) happens to sit. This is part of the conclusion of sequential compactness. For example, the sequence \(1/n\) in \((0,1)\) converges in \(\mathbb{R}\) to \(0\), but \(0\notin(0,1)\). The open interval is not compact, and this sequence has no subsequence converging to a point of the interval. Compactness rules out this kind of escape because the finite-subcover argument produces its cluster point inside the compact space itself.

A useful proof strategy is therefore to keep track of where each limit is obtained. In the theorem above, each selected term lies in \(X\), and the cluster point is defined to lie in \(X\); the convergence is consequently convergence in the given metric space. If one instead obtains a limit in a surrounding space, an additional argument may be needed to show that the limit belongs to the subset under consideration.

The implication proved here goes from compactness to sequential compactness for metric spaces. Earlier in this course, the theorems Complete and Totally Bounded Metric Spaces Are Compact and Sequential Compactness and Completeness with Total Boundedness gave related characterizations using completeness and total boundedness. The proof here has a different emphasis: it uses the definition of compactness directly and converts a sequence with no cluster point into an open cover that cannot have a finite subcover. The converse implication, from sequential compactness to open-cover compactness, is a separate direction.

Check Your Understanding

Use the cluster-point definition and the finite-subcover argument to answer the following questions.

Why does failure to be a cluster point give a ball containing terms for only finitely many indices?
In the cluster-point proof, why does a finite subcover account for only finitely many sequence indices?
When extracting a subsequence, why can the next index be chosen larger than the preceding index?
Why is the limit in the compactness theorem guaranteed to belong to the metric space?
What feature of a metric space lets the cluster-point argument produce a subsequence converging to its cluster point?
Why does a value occurring infinitely often give a convergent subsequence in a finite metric space?