Three Distinctions That Prevent False Proofs
Compactness arguments in metric spaces often combine several useful conditions: boundedness, total boundedness, completeness, sequential compactness, and compactness by open covers. These conditions are related, but they are not interchangeable. A bounded set need not be totally bounded; a complete metric space need not be compact; and the agreement between sequential and open-cover compactness depends on having a metric.
Throughout, a metric space is a pair \((X,d)\). A subset \(A\subseteq X\) is bounded if it is contained in some ball of finite radius. It is totally bounded if, for every \(\varepsilon>0\), there are finitely many points \(x_1,\ldots,x_m\in X\) such that \(A\subseteq\bigcup_{i=1}^m B(x_i,\varepsilon)\). A sequence \((x_n)\) is Cauchy if for every \(\varepsilon>0\) there is \(N\) such that \(d(x_n,x_m)<\varepsilon\) whenever \(n,m\ge N\). The space is complete if every Cauchy sequence converges to a point of \(X\). It is sequentially compact if every sequence in it has a convergent subsequence. It is compact if every open cover has a finite subcover.
Proof. First suppose \(X\) is compact, and fix \(\varepsilon>0\). The balls \(B(x,\varepsilon)\), as \(x\) ranges over \(X\), form an open cover of \(X\). Compactness gives a finite subcover, which is exactly a finite \(\varepsilon\)-net. Thus \(X\) is totally bounded.
To prove completeness, let \((x_n)\) be a Cauchy sequence in \(X\). For each \(k\in\mathbb N\), let \(F_k=\overline{\{x_n:n\ge k\}}\), where closure is taken in \(X\). Each \(F_k\) is nonempty and closed, and \(F_{k+1}\subseteq F_k\). Any finite intersection of these sets is nonempty: the intersection is the one with largest index among the finitely many indices, and that set is nonempty. In a compact space, a family of closed sets with this finite intersection property has nonempty intersection. Indeed, if the whole intersection were empty, their open complements would cover \(X\); a finite subcover would give an empty finite intersection, a contradiction. Choose \(x\in\bigcap_{k=1}^\infty F_k\).
Let \(\varepsilon>0\). Choose \(N\) so that \(d(x_n,x_m)<\varepsilon/2\) for \(n,m\ge N\). Since \(x\in F_N\), the open ball \(B(x,\varepsilon/2)\) meets \(\{x_m:m\ge N\}\). Choose \(m\ge N\) with \(d(x_m,x)<\varepsilon/2\). For every \(n\ge N\), the triangle inequality gives \(d(x_n,x)\le d(x_n,x_m)+d(x_m,x)<\varepsilon/2+\varepsilon/2=\varepsilon\). Therefore \(x_n\to x\), and \(X\) is complete. The empty space satisfies both conclusions vacuously. \(\square\)
Bounded Is Not the Same as Totally Bounded
Boundedness asks whether one ball of some finite radius contains the space. Total boundedness asks for finite covers at every positive scale, including arbitrarily small scales. The radius in the definition of boundedness cannot be reused as if it worked for all \(\varepsilon\).
Worked Example: An Infinite Bounded Metric Space That Is Not Totally Bounded
Let \(X=\mathbb N\) and define \(d(m,n)=0\) when \(m=n\), and \(d(m,n)=1\) when \(m\ne n\). This is a metric: positivity and symmetry follow from the definition, and if \(m\ne n\), then \(d(m,n)=1\le d(m,k)+d(k,n)\), since at least one of \(m\ne k\) or \(k\ne n\) holds.
Fix \(p\in X\). Every \(n\in X\) satisfies \(d(n,p)\le1<2\), so \(X\subseteq B(p,2)\) and \(X\) is bounded. Now take \(\varepsilon=1/2\). For each \(p\in X\), \(B(p,1/2)=\{p\}\), since \(d(p,n)=1\) for every \(n\ne p\). A finite collection of these balls covers only finitely many points, so no finite collection covers \(X\). Thus \(X\) is not totally bounded.
Check: the same scale calculation verifies the failure directly: each ball of radius \(1/2\) contains exactly one point, while \(X\) has infinitely many points.
This example is also complete but not compact, so it illustrates two separate errors at once. If a sequence in this space is Cauchy, use \(\varepsilon=1/2\) in the Cauchy condition. Beyond some index, all pairs of terms have distance less than \(1/2\), which forces all those terms to be equal. The sequence is eventually constant and therefore converges in \(X\). On the other hand, every singleton is open, and the cover \(\{\{n\}:n\in\mathbb N\}\) has no finite subcover. Hence \(X\) is not compact.
Common mistake. The false calculation is: “\(X\subseteq B(p,2)\), so one ball covers \(X\); therefore \(X\) is totally bounded.” The conclusion does not follow. For total boundedness the number of balls must be finite separately for every \(\varepsilon>0\). At \(\varepsilon=1/2\), the calculation above shows that finitely many balls cover only finitely many points.
Complete and Totally Bounded Are Different Requirements
Completeness concerns limits of Cauchy sequences: the limit must remain in the space. Total boundedness concerns the availability of finite covers at every scale. In metric spaces, both conditions together imply compactness, as established in the earlier material on completeness and total boundedness. The examples below show why neither condition can be dropped.
Worked Example: A Totally Bounded but Incomplete Space
Let \(X=\mathbb Q\cap[0,1]\) with the metric \(d(x,y)=|x-y|\), the restriction of the standard metric on \(\mathbb R\). For \(\varepsilon>0\), choose \(N\in\mathbb N\) with \(1/N<\varepsilon\). The \(N+1\) rational points \(k/N\), for \(k=0,\ldots,N\), lie in \(X\). For any \(q\in X\), choose \(k=\lfloor Nq\rfloor\). Then \(0\le k\le N\) and \(0\le q-k/N<1/N<\varepsilon\). Thus the balls of radius \(\varepsilon\) centred at these finitely many points cover \(X\). This proves total boundedness.
Set \(c=\sqrt{2}/2\), which lies strictly between \(0\) and \(1\), and define \(q_n=\lfloor 10^n c\rfloor/10^n\). Each \(q_n\) is rational and belongs to \([0,1]\), and \(0\le c-q_n<10^{-n}\). For \(n,m\in\mathbb N\), the triangle inequality gives \(|q_n-q_m|\le |q_n-c|+|q_m-c|<10^{-n}+10^{-m}\), so \((q_n)\) is Cauchy. It cannot converge in \(X\): in \(\mathbb R\), the displayed bound gives \(q_n\to c\), and limits in a metric space are unique, while \(c\notin\mathbb Q\). To verify irrationality, if \(\sqrt2=p/q\) in lowest terms, then \(p^2=2q^2\), so \(p\) is even; writing \(p=2r\) gives \(q^2=2r^2\), so \(q\) is even as well, contradicting lowest terms. Hence \(c\) is irrational.
Conclusion: \(X\) is totally bounded but incomplete. It is not compact, since the theorem above shows that every compact metric space is complete.
A tempting but invalid argument is “a finite \(\varepsilon\)-net exists at every scale, so every Cauchy sequence has its limit in \(X\).” The net condition controls where points of \(X\) lie, but does not add missing limit points. In this example the rational points approximate \(c\) ever more closely, while \(c\) is not in the space.
Sequential Compactness Needs the Metric Hypothesis
For metric spaces, compactness and sequential compactness are equivalent; the equivalence is one of the standard results of compactness in metric spaces. It is unsafe to use the sequential condition as a replacement for open-cover compactness in an arbitrary topological space. Here is a counterexample, with the order topology specified explicitly.
Worked Example: A Sequentially Compact Space That Is Not Compact
Let \(\omega_1\) denote the first uncountable ordinal, and let \(X=[0,\omega_1)\) be the set of all countable ordinals with the order topology. The order topology is generated by open intervals and open initial and final rays. We first show that every sequence in \(X\) has a convergent subsequence.
Take a sequence \((\alpha_n)\) in \(X\). If some value occurs infinitely often, the corresponding constant subsequence converges. Otherwise every value occurs only finitely often, and the range is infinite. Its supremum \(\beta\) is a countable ordinal: it is the supremum of countably many countable ordinals, hence is still below \(\omega_1\).
We may remove finitely many values from the range as follows. Whenever the supremum of the remaining range is a successor ordinal \(\delta+1\), the value \(\delta+1\) must occur in that range; otherwise every remaining value would be at most \(\delta\), contradicting the supremum. Remove that value. Since it occurred only finitely often, this discards only finitely many terms of the sequence. Repeat if the new supremum is again a successor. This process cannot continue infinitely: the successive suprema would form a strictly decreasing infinite sequence of ordinals, which is impossible because every nonempty set of ordinals has a least element. After finitely many removals, the remaining range is still infinite and has a limit ordinal \(\beta\) as its supremum.
Enumerate the elements of this countable limit ordinal as \(b_1,b_2,\ldots\), and put \(\gamma_k=\max\{b_1,\ldots,b_k\}+1\). Since \(\beta\) is a limit ordinal, \(\gamma_k<\beta\); also the \(\gamma_k\) are nondecreasing and cofinal in \(\beta\). For each \(k\), infinitely many terms in the remaining sequence have value greater than \(\gamma_k\). Otherwise, after finitely many such terms, all values would be at most \(\gamma_k\); the finitely many earlier values together with \(\gamma_k\) would then bound the range strictly below \(\beta\), a contradiction.
We can therefore choose indices \(n_1<n_2<\cdots\) so that \(\alpha_{n_k}>\gamma_k\) and \(\alpha_{n_k}>\alpha_{n_{k-1}}\) for \(k\ge2\). At each stage the threshold is below \(\beta\), and infinitely many terms exceed it, so an index larger than the previous one is available. The selected values are below \(\beta\) and cofinal in \(\beta\). In the order topology, every neighborhood of the limit ordinal \(\beta\) contains all ordinals in some interval \((\gamma,\beta]\) with \(\gamma<\beta\). Cofinality therefore gives \(\alpha_{n_k}\to\beta\). This proves sequential compactness.
Now consider \(U_\alpha=[0,\alpha)\) for each \(\alpha<\omega_1\). These are open initial rays in the order topology. They cover \(X\), since for each \(\xi<\omega_1\), the successor \(\xi+1\) is still countable and \(\xi\in U_{\xi+1}\). Any finite subcollection has a largest index \(\alpha\), and its union is \(U_\alpha\), which omits the point \(\alpha\in X\). Thus the cover has no finite subcover, so \(X\) is not compact.
Conclusion: this space is sequentially compact but not compact. The example rules out transferring the metric equivalence to general topological spaces without checking additional hypotheses.
A Checklist for Compactness Arguments
When an argument invokes compactness, identify which definition or theorem is doing the work. In a metric space, the earlier results on compactness give the equivalence of open-cover compactness and sequential compactness, and characterize compactness by completeness together with total boundedness. Neither boundedness alone nor completeness alone can replace those conditions. Outside the metric setting, do not infer open-cover compactness just from the existence of convergent subsequences.
| Condition | What it says | What it does not guarantee by itself |
|---|---|---|
| Bounded | One finite-radius ball contains the space | Total boundedness |
| Totally bounded | Finite ball covers exist at every positive scale | Completeness |
| Complete | Every Cauchy sequence converges in the space | Compactness |
| Sequentially compact | Every sequence has a convergent subsequence | Open-cover compactness without suitable hypotheses |
Check Your Understanding
For each question, state the space and topology or metric being used, and identify the hypothesis that supports your conclusion.
- In the infinite discrete metric space above, compute \(B(p,1/2)\) and explain why no finite collection of such balls covers the space.
- Why does total boundedness of \(\mathbb Q\cap[0,1]\) not imply that it contains the limit of the Cauchy sequence constructed in the example?
- Which part of the proof that compact metric spaces are complete uses compactness rather than just the Cauchy property?
- For the open cover \(U_\alpha=[0,\alpha)\) of \([0,\omega_1)\), why does every finite subcollection omit a point?
- Which implication in the metric compactness equivalences should not be transferred to arbitrary topological spaces without further hypotheses?