Sequences as a Test for Compactness
Compactness is defined using open covers, but in metric spaces it is often useful to ask a different question: must every sequence have a convergent subsequence? This property is called sequential compactness. It is important to check not only whether terms approach a limit, but also whether that limit belongs to the space being studied.
Throughout, a metric space \((X,d)\) is given the topology induced by its metric. A sequence in \(X\) is a function from \(\mathbb{N}\) to \(X\), written \((x_n)\). A subsequence is a sequence \((x_{n_k})\) whose indices satisfy \(n_1<n_2<\cdots\). Convergence \(x_{n_k}\to x\) means that for every \(\varepsilon>0\), there is \(K\in\mathbb{N}\) such that \(d(x_{n_k},x)<\varepsilon\) whenever \(k\ge K\). In particular, the limit \(x\) must be a point of \(X\).
The definition is about every sequence, not just sequences whose terms are distinct or whose limits are easy to guess. Repeated terms are allowed. A constant subsequence is often enough to prove sequential compactness for a finite space; to disprove it, it is enough to find one sequence with no subsequence converging in the space.
Closed Real Intervals
The fundamental example is a closed interval in \(\mathbb{R}\) with its standard topology and usual metric. The proof below constructs a subsequence by repeatedly keeping a half-interval that contains infinitely many terms. It uses the least-upper-bound property of \(\mathbb{R}\), in particular the existence of suprema of nonempty bounded sets of real numbers.
Proof. If \(a=b\), every sequence in \([a,b]=\{a\}\) is constant and converges to \(a\). Now suppose \(a<b\), and let \((x_n)\) be any sequence in \([a,b]\). Set \(I_0=[a,b]\). We construct nested closed intervals \(I_k=[\alpha_k,\beta_k]\) such that \(I_k\) contains terms with infinitely many indices and has length \((b-a)/2^k\). Given \(I_{k-1}\), divide it into its two closed halves. If each half contained terms with only finitely many indices, their union \(I_{k-1}\) would also contain terms with only finitely many indices, contrary to the construction. Choose a half containing terms with infinitely many indices and call it \(I_k\). Since \(I_k\subseteq I_{k-1}\), the intervals are nested. Let \(S_k=\{n\in\mathbb{N}:x_n\in I_k\}\); each \(S_k\) is infinite, so choose recursively \(n_k\in S_k\) with \(n_k>n_{k-1}\) (and choose any \(n_1\in S_1\)). The left endpoints \(\alpha_k\) are nondecreasing and bounded above by \(b\). Let \(c=\sup\{\alpha_k:k\in\mathbb{N}\}\). For every \(k\), \(c\le\beta_k\): all later left endpoints lie in \(I_k\), and all earlier left endpoints are at most \(\alpha_k\le\beta_k\). Also \(\alpha_k\le c\), so \(c\in I_k\). Both \(x_{n_k}\) and \(c\) belong to \(I_k\), giving \(d(x_{n_k},c)=|x_{n_k}-c|\le\beta_k-\alpha_k=(b-a)/2^k\). The right-hand side tends to zero, so for every \(\varepsilon>0\) it is less than \(\varepsilon\) for all sufficiently large \(k\). Thus \(x_{n_k}\to c\), and \(c\in[a,b]\). The sequence has a subsequence converging in the interval, as required. \(\square\)
Worked Example: A Sequence in the Closed Unit Interval
Consider \(X=[0,1]\) with the usual metric and the sequence \(x_n=n/(n+1)\), for \(n\in\mathbb{N}\). For each \(n\ge1\), \(n>0\) and \(n<n+1\), so \(0<x_n<1\); every term lies in \(X\).
Take the subsequence with \(n_k=k\), which is the original sequence. Its distance from \(1\) is \(|x_{n_k}-1|=|k/(k+1)-1|=1/(k+1)\). Given \(\varepsilon>0\), choose \(K\in\mathbb{N}\) with \(K+1>1/\varepsilon\). For \(k\ge K\), \(k+1\ge K+1>1/\varepsilon\), so \(1/(k+1)<\varepsilon\). Hence \(x_{n_k}\to1\), and \(1\in X\).
Check: Substituting \(k=1\) gives \(x_1=1/2\), which lies in \([0,1]\), and the distance from \(x_1\) to \(1\) is \(1/2=1/(1+1)\), as in the formula. The limit is an allowed point of the space, so this subsequence verifies the definition.
Finite Spaces and Infinite Discrete Spaces
Finite metric spaces pass the test for a reason different from the interval argument: one point must occur infinitely often. By contrast, an infinite discrete space has a sequence of distinct points that cannot converge. Here “discrete” means that every singleton is open.
Proof. Let \(X\) be finite and let \((x_n)\) be a sequence in \(X\). If \(X\) is empty there is no such sequence. Otherwise, at least one point \(x\in X\) occurs infinitely often: if every point occurred only finitely often, the union of the finitely many finite sets of indices at which each point occurs would be finite, although it must contain every index in \(\mathbb{N}\). Choose increasing indices \(n_1<n_2<\cdots\) with \(x_{n_k}=x\) for every \(k\). Then \(d(x_{n_k},x)=0\) for all \(k\), so this subsequence converges to \(x\in X\). \(\square\)
Worked Example: A Finite Metric Space
Let \(X=\{p,q,r\}\) with the discrete metric \(d(u,v)=0\) if \(u=v\), and \(d(u,v)=1\) if \(u\ne v\). Consider the sequence \(p,q,r,p,q,r,\ldots\), so \(x_{3k}=r\) for every \(k\ge1\). The indices \(3,6,9,\ldots\) are strictly increasing, and the resulting subsequence is \(r,r,r,\ldots\).
For any \(\varepsilon>0\), every term of this subsequence has distance \(d(r,r)=0<\varepsilon\) from \(r\). Thus the subsequence converges to \(r\in X\). This computation illustrates the general finite-space argument: repetitions supply a constant subsequence.
Worked Example: An Infinite Discrete Metric Space
Let \(X=\mathbb{N}\), with the discrete metric \(d(m,n)=0\) when \(m=n\) and \(d(m,n)=1\) when \(m\ne n\). The open ball \(B(n,1/2)=\{n\}\), so every singleton is open and the induced topology is discrete. Consider \(x_n=n\).
Every subsequence has the form \(x_{n_k}=n_k\), where \(n_1<n_2<\cdots\), so its terms are distinct. Fix any proposed limit \(m\in X\). At most one term of this subsequence equals \(m\); every other term has distance \(1\) from \(m\). In particular, for \(\varepsilon=1/2\), the terms are not eventually within \(\varepsilon\) of \(m\). No subsequence converges to \(m\). Since this holds for every \(m\in X\), the sequence has no convergent subsequence in \(X\), and \(X\) is not sequentially compact.
Check: The conclusion does not say that no sequence in an infinite discrete space can have a convergent subsequence. A sequence that repeats one point infinitely often does have a constant subsequence. The distinct sequence above is enough to disprove sequential compactness, whose definition requires the property for every sequence.
The Real Line and the Importance of the Limit Point
The usual metric space \(\mathbb{R}\) is not sequentially compact. Its unboundedness provides a direct sequence witnessing the failure. The issue for an open interval is different: a sequence can converge in \(\mathbb{R}\), while its limit is not in the interval.
Worked Example: The Real Line
In \(\mathbb{R}\) with the usual metric, take \(x_n=n\). Any subsequence has terms \(x_{n_k}=n_k\) with \(n_1<n_2<\cdots\). Since these are strictly increasing positive integers, \(n_k\ge k\) for every \(k\): the first is at least \(1\), and each later integer is at least one greater than its predecessor.
If this subsequence converged to some \(L\in\mathbb{R}\), then the definition of convergence with \(\varepsilon=1\) would give an index \(K\) such that \(n_k<L+1\) for all \(k\ge K\). Choose an integer \(k\ge K\) with \(k>L+1\). Then \(n_k\ge k>L+1\), a contradiction. Thus no subsequence converges in \(\mathbb{R}\), and the real line is not sequentially compact.
Worked Example: An Open Interval
Let \(X=(0,1)\) with its subspace metric from \(\mathbb{R}\), and set \(x_n=1/(n+1)\). Since \(n\ge1\), \(n+1\ge2\), so \(0<1/(n+1)\le1/2<1\). Every term is therefore in \(X\). For any subsequence with increasing indices \(n_k\), we have \(n_k\ge k\), and hence \(0<x_{n_k}=1/(n_k+1)\le1/(k+1)\). Given \(\varepsilon>0\), choose \(K\) such that \(1/(K+1)<\varepsilon\). Then for \(k\ge K\), \(0<x_{n_k}\le1/(k+1)\le1/(K+1)<\varepsilon\). Every subsequence therefore converges to \(0\) in \(\mathbb{R}\), but \(0\notin X\). No subsequence converges to a point of \(X\), so \((0,1)\) is not sequentially compact.
A Common Pitfall: Checking the Space of Convergence
Convergence depends on the ambient metric space, and sequential compactness requires a limit inside the space. A related error is to choose a sequence that does not actually lie in the proposed space. For example, \(1/n\) is not a sequence entirely in \((0,1)\): its first term is \(1/1=1\), and \(1\notin(0,1)\). The sequence \(1/(n+1)\) fixes that membership error: its first term is \(1/2\), and every term lies strictly between \(0\) and \(1\).
Common pitfall. One might calculate \(1/(n+1)\to0\) and conclude that the sequence has a convergent subsequence in \((0,1)\) because \(0\) is its limit. The calculation gives convergence in \(\mathbb{R}\), not convergence to a point of \((0,1)\). As shown above, every subsequence also tends to \(0\), which is outside the space. Thus none of those subsequences converges in \((0,1)\). When testing sequential compactness, always name the space and check that the subsequential limit belongs to it.
Sequential compactness is a sequence-based condition, whereas compactness is defined by finite subcovers of open covers. These definitions ask different questions. The results here establish the sequence test for the specified metric spaces; no equivalence between the two notions is needed for these examples.
Check Your Understanding
For each question, specify the metric space and verify that any proposed limit belongs to it.
- In \([2,5]\) with the usual metric, find a convergent subsequence of \(x_n=2+3n/(n+1)\), and verify its limit.
- Why must some point occur infinitely often in every sequence in a finite set?
- In an infinite discrete metric space, explain why a sequence of distinct points has no convergent subsequence.
- Give a sequence in \((0,1)\) whose subsequences all converge in \(\mathbb{R}\) to a point outside \((0,1)\).
- Does the sequence \(x_n=(-1)^n\) in \(\mathbb{R}\) itself show that \(\mathbb{R}\) is not sequentially compact? Find a convergent subsequence or explain why none exists.