Characterisations of compactness

Theorem

Characterisations of compactness

T-080Home MU-301Threads space

Statement

Let \((X,d)\) be a metric space. Then \(X\) is compact — every open cover of \(X\) admits a finite subcover — if and only if \(X\) is sequentially compact — every sequence \((x_n)_{n\in\mathbb{N}}\) in \(X\) has a subsequence converging to a point of \(X\). The equivalence holds for \(X\) itself and, by restricting the metric, for any subset \(K\subseteq X\) (with the subspace metric) taken with its induced topology.

Why it matters

Compactness is defined topologically (via open covers) and makes sense in any topological space, but that definition is awkward to verify directly. Sequential compactness is defined analytically (via subsequences) and is exactly the tool used throughout analysis: Bolzano–Weierstrass, existence of extrema of continuous functions, Arzelà–Ascoli-type arguments, and convergence proofs all proceed by extracting convergent subsequences. This theorem certifies that in the metric setting — where almost all of undergraduate analysis lives — these two completely different-looking notions are interchangeable, so one may freely use whichever is more convenient in a given argument.

The result is also a cautionary tale: it is special to metric (indeed metrizable) spaces. In general topological spaces the two notions diverge, and conflating them is a classic source of error once one moves to function spaces with non-metrizable topologies (e.g. weak-\(*\) topologies on infinite-dimensional duals restricted to non-separable settings).

Hypotheses

\((X,d)\) is a metric space (not merely topological). Counterexample sketch if dropped: let \(X=[0,1]^{[0,1]}\) with the product topology (not metrizable). The subset of functions valued in \(\{0,1\}\) is compact by Tychonoff, but one can build a sequence of such functions (e.g. binary digit indicators) with no pointwise-convergent subsequence, so compact does not imply sequentially compact here.

Every open cover is genuinely allowed to be infinite, and "subcover" means a sub-collection of the same cover, not a re-derived one. Counterexample sketch if one silently restricts to countable covers: countable compactness (every countable open cover has a finite subcover) is strictly weaker than compactness in general topological spaces, though it coincides with compactness in metric spaces via second countability — a fact quietly used inside this proof.

The space carries a single fixed metric \(d\) throughout (completeness/total boundedness are metric, not topological, properties). Counterexample sketch: \((0,1)\) with the usual metric is not complete and not compact; but \((0,1)\) is homeomorphic to \(\mathbb{R}\) via \(\tan\), and compactness is a topological invariant while completeness is not — so completeness-flavoured steps of the proof are only meaningful relative to the chosen \(d\), not to the topology alone.

Proof

We prove both implications. Throughout, \(B(x,r)=\{y\in X: d(x,y)\lt r\}\) denotes the open ball.

\(\text{compact} \implies \text{sequentially compact}\)

Contrapositive: suppose \((x_n)\) is a sequence in \(X\) with no convergent subsequence. We show \(X\) is not compact by constructing an open cover with no finite subcover. B

\(\forall y \in X,\ \exists r_y \gt 0 \text{ such that } B(y,r_y) \text{ contains } x_n \text{ for only finitely many } n\)

If no such \(r_y\) existed for some \(y\), then every ball \(B(y,1/k)\) would contain infinitely many terms of the sequence, and picking one such term from each (with indices strictly increasing) would produce a subsequence converging to \(y\) — contradicting the hypothesis of step 1. This uses only the definition of convergence in a metric space. B

\(\mathcal{U} = \{B(y,r_y): y\in X\}\) is an open cover of \(X\) with no finite subcover

Each \(y\in X\) lies in its own ball \(B(y,r_y)\), so \(\mathcal{U}\) covers \(X\); each member is open by definition of the metric topology. Any finite subfamily \(B(y_1,r_{y_1}),\dots,B(y_k,r_{y_k})\) contains, by step 2, only finitely many terms of \((x_n)\) in total (a finite union of finite sets is finite) — yet \((x_n)_{n\in\mathbb{N}}\) has infinitely many terms (with repetition allowed we instead index by \(\mathbb{N}\), which is infinite), so the finite subfamily cannot cover all of \(\{x_n : n\in\mathbb{N}\}\subseteq X\), hence cannot cover \(X\). Thus \(\mathcal{U}\) is an open cover admitting no finite subcover, so \(X\) is not compact. This is the required contrapositive. A

\(\text{sequentially compact} \implies X \text{ is totally bounded: } \forall \varepsilon\gt0\ \exists\, x_1,\dots,x_m\in X \text{ with } X=\bigcup_{i=1}^m B(x_i,\varepsilon)\)

Lemma (total boundedness from sequential compactness), proved by contradiction. Suppose some \(\varepsilon\gt0\) admits no finite \(\varepsilon\)-net. Construct \((x_n)\) recursively: pick \(x_1\in X\) arbitrarily; having chosen \(x_1,\dots,x_n\), since they do not \(\varepsilon\)-cover \(X\) (by assumption no finite set does), choose \(x_{n+1}\in X\setminus\bigcup_{i\le n} B(x_i,\varepsilon)\). Then \(d(x_i,x_j)\ge\varepsilon\) for all \(i\ne j\). Such a sequence has no Cauchy subsequence, hence no convergent subsequence (convergent sequences in a metric space are Cauchy, by the triangle inequality), contradicting sequential compactness. C

\(\text{Lebesgue number lemma: if } X \text{ is sequentially compact and } \mathcal{U} \text{ is an open cover of } X, \text{ then } \exists\, \delta\gt0 \text{ such that } \forall x\in X,\ \exists\, U\in\mathcal{U} \text{ with } B(x,\delta)\subseteq U\)

Proved by contradiction. Suppose no such \(\delta\) exists; then for every \(n\in\mathbb{N}\), taking \(\delta=1/n\) fails, so there is some \(x_n\in X\) such that \(B(x_n,1/n)\) is contained in no single member of \(\mathcal{U}\). By sequential compactness, \((x_n)\) has a subsequence \(x_{n_k}\to x^*\) for some \(x^*\in X\). Since \(\mathcal{U}\) covers \(X\), \(x^*\in U_0\) for some \(U_0\in\mathcal{U}\); as \(U_0\) is open, \(B(x^*,\rho)\subseteq U_0\) for some \(\rho\gt0\). Choose \(k\) large enough that \(d(x_{n_k},x^*)\lt\rho/2\) and \(1/n_k\lt\rho/2\) (possible since \(n_k\to\infty\) and \(x_{n_k}\to x^*\)). Then for any \(y\in B(x_{n_k},1/n_k)\), the triangle inequality gives \(d(y,x^*)\le d(y,x_{n_k})+d(x_{n_k},x^*)\lt \rho/2+\rho/2=\rho\), so \(B(x_{n_k},1/n_k)\subseteq B(x^*,\rho)\subseteq U_0\) — contradicting the choice of \(x_{n_k}\). Hence a Lebesgue number \(\delta\) exists. C

\(\text{sequentially compact} \implies \text{compact}\)

Let \(\mathcal{U}\) be an open cover of \(X\). By step 5, fix a Lebesgue number \(\delta\gt0\) for \(\mathcal{U}\). By step 4 (total boundedness applied with \(\varepsilon=\delta/2\)), there exist \(x_1,\dots,x_m\in X\) with \(X=\bigcup_{i=1}^m B(x_i,\delta/2)\). B

\(\forall i \in \{1,\dots,m\},\ \exists\, U_i \in \mathcal{U} \text{ with } B(x_i,\delta/2)\subseteq U_i\)

Directly by the defining property of the Lebesgue number \(\delta\) from step 5, applied at the point \(x_i\) with radius \(\delta/2\lt\delta\). A

\(X = \bigcup_{i=1}^m B(x_i,\delta/2) \subseteq \bigcup_{i=1}^m U_i \subseteq X\)

The first equality is step 6, the containment is step 7 applied termwise, and the outer containment holds since each \(U_i\subseteq X\). Hence \(\{U_1,\dots,U_m\}\subseteq \mathcal{U}\) is a finite subcover of \(X\). As \(\mathcal{U}\) was an arbitrary open cover, \(X\) is compact. A

Result

\(X \text{ compact} \iff X \text{ sequentially compact} \quad (\text{in a metric space } (X,d))\)

Reading. In a metric space, "every open cover has a finite subcover" and "every sequence has a convergent subsequence" describe exactly the same phenomenon. The forward direction is a direct diagonal/contrapositive argument; the reverse direction routes through two auxiliary metric facts — total boundedness and the Lebesgue number lemma — both of which genuinely use the metric, not just the topology.

Scope. Valid for any metric space, hence for any metrizable topological space (transport the metric along a homeomorphism). It also applies verbatim to subspaces \(K\subseteq X\) with the induced metric. It does not extend to general topological spaces, where compactness, sequential compactness, and countable compactness are three genuinely distinct properties.

Corollaries & converses

Bolzano–Weierstrass theorem for \(\mathbb{R}^n\): closed bounded subsets of \(\mathbb{R}^n\) are compact (via Heine–Borel) and hence, by this theorem, sequentially compact — every bounded sequence has a convergent subsequence.
A metric space is compact iff it is complete and totally bounded (combine step 4 with the standard fact that sequentially compact spaces are complete: every Cauchy sequence has a convergent subsequence and hence converges).
Compact metric spaces are separable and second countable (finite \(1/n\)-nets for each \(n\) union to a countable dense set), which is what licenses restricting attention to countable subcovers in some treatments.
Continuous images of compact metric spaces are compact, and (using sequential compactness directly) continuous real-valued functions on them attain their supremum and infimum — the extreme value theorem.
Converse: both directions proved above are genuine "iff" statements — there is no separate converse to check; the theorem already asserts logical equivalence, established independently in each direction (steps 1–3 and steps 4–8).

Fails without

Dropping "metric" (general topology, forward direction fails): the space \(\{0,1\}^{[0,1]}\) with the product topology is compact (Tychonoff) but not sequentially compact — the sequence of functions \(f_n(t) = \) the \(n\)-th binary digit of \(t\) has no pointwise-convergent subsequence, because any candidate subsequence can be shown (via a diagonal/pigeonhole argument on a suitable uncountable set of coordinates) to fail to converge at some coordinate.
Dropping "metric" (general topology, reverse direction fails): the space of countable ordinals \([0,\omega_1)\) with the order topology is sequentially compact (every sequence is eventually bounded below some countable ordinal, which has compact closure) but not compact — the cover by initial segments \([0,\alpha)\) for \(\alpha\lt\omega_1\) has no finite subcover, since a finite union of initial segments is again an initial segment bounded below \(\omega_1\).
Dropping total boundedness alone (step 4) while keeping the space merely complete: the closed unit ball of \(\ell^2\) with its norm metric is complete but not totally bounded (the standard basis vectors \(e_n\) satisfy \(d(e_i,e_j)=\sqrt2\) for \(i\ne j\), admitting no finite \(1\)-net), so it is neither compact nor sequentially compact — confirming that total boundedness is not automatic even for complete metric spaces, and genuinely does the work in step 4.

Common errors

Assuming "closed and bounded" implies compact in an arbitrary metric space — this is Heine–Borel, special to \(\mathbb{R}^n\) (finite dimension); in \(\ell^2\) the closed unit ball is closed and bounded but not compact, as shown above.
Proving only the easy direction (compact \(\implies\) sequentially compact via step 1–3) and believing the converse is "obviously similar" — the reverse direction is strictly harder and requires total boundedness plus the Lebesgue number lemma; there is no short symmetric argument.
Misstating the Lebesgue number lemma as "every open set in the cover has radius at least \(\delta\)" rather than "every ball of radius \(\delta\) is contained in some member of the cover" — these are different statements, and only the latter is true and useful.
Forgetting that a subsequence of a Cauchy sequence converging forces the whole sequence to converge (not just the subsequence) when concluding completeness from sequential compactness — this step needs its own short triangle-inequality argument, silently skipped by many students.
Conflating "sequentially compact" with "every sequence converges" — it only guarantees a convergent subsequence exists, not that the original sequence converges (e.g. \((-1)^n\) in \([-1,1]\) does not converge but has convergent subsequences).

Discussion

The theorem is one of the clearest instances in analysis of two conceptually different definitions of the same idea turning out to coincide under an extra structural hypothesis (metrizability). Historically, sequential compactness (via Bolzano–Weierstrass-style arguments) predates the modern open-cover definition of compactness, which emerged from the general topology developed by Fréchet, Hausdorff, and Alexandroff in the early twentieth century as topologists sought a notion robust enough to survive without a metric. The open-cover definition turned out to be the "right" one for category-theoretic and topological purposes (it is preserved by continuous images, behaves well with products via Tychonoff, and needs no metric), while sequential compactness remained the practically useful one for analysts, since most classical arguments (extracting a convergent subsequence of approximating functions, minimising sequences, etc.) are literally sequence-based.

The proof structure reveals why metrizability is exactly the right hypothesis: total boundedness (step 4) and the Lebesgue number lemma (step 5) both manufacture, from a merely sequential hypothesis, control over arbitrarily small balls uniformly across the whole space — and "ball" only makes sense once a metric is present. Without a metric there is no notion of ball radius to make these arguments go through, and indeed both directions of the equivalence can fail, as the \(\{0,1\}^{[0,1]}\) and \([0,\omega_1)\) examples above show independently.

A more refined statement subsumes this theorem: in a metric space, compact, sequentially compact, countably compact, and complete-plus-totally-bounded are all equivalent — a chain of four mutually implying conditions, of which this theorem supplies one edge (and, via completeness, effectively the whole cycle once one notes sequentially compact \(\implies\) complete is immediate from Cauchy subsequences converging). This full equivalence is sometimes packaged as "in metric spaces, compactness has no exotic variants," in deliberate contrast to general topology where at least four inequivalent compactness-type properties coexist (compact, countably compact, sequentially compact, pseudocompact).

Common misconception: students sometimes believe compactness is "basically the same as closed and bounded" as a matter of definition rather than a theorem (Heine–Borel) special to \(\mathbb{R}^n\); the present theorem is frequently — and wrongly — assumed to make this automatic in any metric space, when in fact total boundedness (not mere boundedness) is the correct metric-space generalisation, and the \(\ell^2\) unit ball counterexample shows bounded plus closed plus complete is still not enough.

Worked examples

Show directly, using sequential compactness, that \(K=\{0\}\cup\{1/n : n\in\mathbb{N}\}\subseteq\mathbb{R}\) is compact.

Strategy: verify sequential compactness (easier here) and invoke the theorem to conclude open-cover compactness without checking covers by hand. A

Let \((x_n)\) be any sequence in \(K\). Either infinitely many terms equal some fixed \(1/m\), giving a constant (hence convergent) subsequence, or infinitely many distinct values \(1/n_k\) occur with \(n_k\to\infty\), giving a subsequence \(1/n_k \to 0 \in K\).

Pigeonhole on the countable set \(K\): if no value repeats infinitely often, the terms used must involve arbitrarily large denominators, forcing convergence to \(0\). Either way a convergent subsequence with limit in \(K\) is produced. B

Hence every sequence in \(K\) has a subsequence converging to a point of \(K\): \(K\) is sequentially compact.

Direct conclusion of step 2, covering both cases exhaustively. A

\(K \text{ is sequentially compact} \implies K \text{ is compact (by the theorem)}\)

Reading. Rather than exhibiting a finite subcover for an arbitrary open cover of \(K\) by hand (annoying, because a cover element containing \(0\) must absorb all but finitely many of the \(1/n\)), we used the theorem to transfer the easy sequential check into the compactness conclusion.

Show that \(X=(0,1]\) with the usual metric is not compact, using sequential compactness.

Strategy: exhibit a single sequence in \(X\) with no subsequence converging to a point of \(X\); by the theorem (contrapositive of sequentially compact \(\implies\) compact, equivalently compact \(\implies\) sequentially compact), this shows \(X\) fails to be compact. A

\(x_n = 1/n \in (0,1], \quad n\in\mathbb{N}\)

Define the candidate sequence explicitly. A

\(\text{Every subsequence } (x_{n_k}) \text{ of } (x_n) \text{ converges to } 0 \text{ in } \mathbb{R}, \text{ and } 0\notin X\)

Since \(n_k\to\infty\) for any subsequence, \(x_{n_k}=1/n_k\to0\) by the standard limit of \(1/n\) in \(\mathbb{R}\); this is the unique possible limit (limits in a metric space are unique), and \(0\notin(0,1]\), so no subsequence converges to a point of \(X\). B

\((0,1] \text{ is not sequentially compact} \implies (0,1] \text{ is not compact}\)

Reading. A single badly-behaved sequence, with no need to construct any open cover explicitly, certifies non-compactness via the contrapositive direction of the theorem — illustrating why the sequential criterion is often the practically easier one to check.

Problems

Prove directly (without citing this theorem) that \([0,1]\) is sequentially compact, using the Bolzano–Weierstrass theorem for bounded sequences in \(\mathbb{R}\).
Solution
Let \((x_n)\) be a sequence in \([0,1]\). It is bounded (by \(0\) and \(1\)), so by the Bolzano–Weierstrass theorem it has a convergent subsequence \(x_{n_k}\to L\) for some \(L\in\mathbb{R}\). Since \([0,1]\) is closed in \(\mathbb{R}\) and \(x_{n_k}\in[0,1]\) for all \(k\), the limit \(L\) also lies in \([0,1]\) (limits of sequences in a closed set stay in the set). Hence every sequence in \([0,1]\) has a subsequence converging to a point of \([0,1]\): \([0,1]\) is sequentially compact.
Using the theorem, deduce that \([0,1]\) is compact from the previous problem, then explain why this gives an independent proof of Heine–Borel for the interval \([0,1]\) (without invoking Heine–Borel itself).
Solution
By problem 1, \([0,1]\) is sequentially compact. The theorem (sequentially compact \(\implies\) compact, steps 4–8 of the proof) then gives directly that \([0,1]\) is compact in the open-cover sense. This is independent of Heine–Borel because the only inputs used were the Bolzano–Weierstrass theorem (a statement about \(\mathbb{R}\)) and the general metric-space machinery of total boundedness and the Lebesgue number lemma — Heine–Borel itself (closed+bounded \(\iff\) compact in \(\mathbb{R}^n\)) was never invoked; indeed this argument can be used to prove the "closed and bounded \(\implies\) compact" half of Heine–Borel in \(\mathbb{R}^n\) via generalising problem 1 to boxes.
Let \((X,d)\) be a metric space and \(K_1\supseteq K_2\supseteq K_3\supseteq\cdots\) a decreasing chain of nonempty compact subsets. Using sequential compactness, prove \(\bigcap_{n=1}^\infty K_n \ne \emptyset\) (Cantor's intersection property).
Solution
Pick \(x_n\in K_n\) for each \(n\) (possible since each \(K_n\) is nonempty). Since \(K_1\) is compact, it is sequentially compact by the theorem, and as \(x_n\in K_1\) for all \(n\), there is a subsequence \(x_{n_k}\to x^*\) with \(x^*\in K_1\) (limits in a compact, hence closed, set stay inside it — compact subsets of metric spaces are closed). Fix any \(m\); for all \(n_k\ge m\) we have \(x_{n_k}\in K_{n_k}\subseteq K_m\) since the chain is decreasing, so the tail of the subsequence lies in \(K_m\); as \(K_m\) is compact hence closed, the limit \(x^*\) also lies in \(K_m\). Since \(m\) was arbitrary, \(x^*\in\bigcap_{n=1}^\infty K_n\), so the intersection is nonempty.
Give an explicit open cover of \((0,1]\) that has no finite subcover, confirming the conclusion of Worked Example 2 directly from the open-cover definition.
Solution
Take \(\mathcal{U}=\{U_n : n\in\mathbb{N}\}\) with \(U_n=(1/n,\,2)\), each an open subset of \(\mathbb{R}\) intersected with \((0,1]\) to give open sets in the subspace \((0,1]\) (or simply note \(U_n\cap(0,1]=(1/n,1]\) is open in the subspace topology). For any \(x\in(0,1]\), choose \(n\gt 1/x\), so \(x\in U_n\); thus \(\mathcal{U}\) covers \((0,1]\). Given any finite subfamily \(U_{n_1},\dots,U_{n_k}\), let \(N=\max(n_1,\dots,n_k)\); the union of the subfamily equals \(U_N=(1/N,1]\) (the largest of the sets, since they are nested increasing), which excludes the point \(1/(N+1)\in(0,1]\). Hence no finite subfamily covers \((0,1]\), matching the non-compactness established via sequences.
Let \((X,d)\) be a metric space. Prove that if \(X\) is compact then \(X\) is complete (every Cauchy sequence converges), using the theorem together with the standard fact that a Cauchy sequence with a convergent subsequence converges to the same limit.
Solution
Let \((x_n)\) be a Cauchy sequence in \(X\). Since \(X\) is compact, by the theorem \(X\) is sequentially compact, so \((x_n)\) has a subsequence \(x_{n_k}\to L\) for some \(L\in X\). We claim \(x_n\to L\). Let \(\varepsilon\gt0\); since \((x_n)\) is Cauchy, choose \(N\) so that \(d(x_n,x_m)\lt\varepsilon/2\) for all \(n,m\ge N\); since \(x_{n_k}\to L\), choose \(k\) with \(n_k\ge N\) and \(d(x_{n_k},L)\lt\varepsilon/2\). Then for any \(n\ge N\), \(d(x_n,L)\le d(x_n,x_{n_k})+d(x_{n_k},L)\lt\varepsilon/2+\varepsilon/2=\varepsilon\) by the triangle inequality. Hence \(x_n\to L\), so every Cauchy sequence in \(X\) converges: \(X\) is complete.