Completeness Is Not Enough
Completeness concerns the behavior of Cauchy sequences: it says that they have limits in the space. Total boundedness concerns whether the entire space can be covered by finitely many small metric balls. These are different requirements. A space may contain all the limits its Cauchy sequences need and still have infinitely many points that remain uniformly far apart.
Throughout, \((X,d)\) is a metric space. 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 \(X\). A sequence converges to \(x\in X\) if for every \(\varepsilon>0\), all sufficiently late terms lie in \(B(x,\varepsilon)=\{y\in X:d(x,y)<\varepsilon\}\).
Sequential compactness is another useful test: \(X\) is sequentially compact if every sequence in \(X\) has a convergent subsequence whose limit belongs to \(X\). For metric spaces, the earlier equivalence of compactness tests identifies compactness with sequential compactness. In particular, the earlier result that compact metric spaces are totally bounded lets us certify noncompactness by proving failure of total boundedness.
Separated Points and Total Boundedness
A collection of points is uniformly separated if there is one positive distance that every distinct pair meets or exceeds. Such a collection cannot fit inside finitely many sufficiently small balls. It also provides sequences that cannot have convergent subsequences.
Proof. Consider the radius \(\delta/3\). If one open ball \(B(z,\delta/3)\), with center \(z\in X\), contained two distinct points \(x,y\in E\), the triangle inequality would give \(d(x,y)\le d(x,z)+d(z,y)<2\delta/3<\delta\), contradicting the separation condition. Thus each such ball contains at most one point of \(E\). A finite collection of these balls contains only finitely many points of \(E\), so it cannot cover \(X\). Hence there is no finite \(\delta/3\)-net and \(X\) is not totally bounded.
Now let \((x_{n_j})\) be any subsequence of a sequence of distinct points of \(E\). Its terms are still distinct, so any two terms have distance at least \(\delta\). If this subsequence converged, it would be Cauchy: given \(\varepsilon>0\), sufficiently late terms would each be within \(\varepsilon/3\) of the limit, and the triangle inequality would put their pairwise distance below \(2\varepsilon/3\). Taking \(\varepsilon=\delta\) contradicts the lower bound \(\delta\). Therefore no subsequence converges. \(\square\)
This proposition isolates the obstruction: the points do not become closer together as the sequence proceeds. Completeness cannot create a limit for a sequence that is not Cauchy.
Worked Example: The Natural Numbers with the Discrete Metric
Let \(X=\mathbb{N}=\{1,2,\dots\}\), with the discrete metric \(d(m,n)=0\) if \(m=n\) and \(d(m,n)=1\) if \(m\ne n\). Verify completeness, failure of total boundedness, and failure of sequential compactness.
Completeness. Let \((x_n)\) be Cauchy in this metric. Use the Cauchy condition with \(\varepsilon=1\). There is \(N\) such that \(d(x_m,x_n)<1\) whenever \(m,n\ge N\). The only possible distance strictly less than \(1\) is \(0\), so \(x_m=x_n\) for all \(m,n\ge N\). The sequence is eventually constant, and therefore converges to that constant value in \(\mathbb{N}\). Every Cauchy sequence converges, so \(X\) is complete.
Failure of total boundedness. Take radius \(\varepsilon=1/2\). For each \(z\in\mathbb{N}\), the ball \(B(z,1/2)\) equals \(\{z\}\): its center has distance \(0<1/2\), while every other point has distance \(1\), which is not less than \(1/2\). A finite collection of these balls covers only finitely many natural numbers. It cannot cover \(X\), so \(X\) is not totally bounded.
Failure of sequential compactness. The sequence \(x_n=n\) consists of distinct points, and \(d(x_m,x_n)=1\) whenever \(m\ne n\). The proposition, with \(\delta=1\), shows that it has no convergent subsequence. Thus this complete metric space is not sequentially compact, and consequently is not compact.
Check. The same distance calculation confirms the Cauchy argument: a sequence with two unequal terms arbitrarily far out has a pairwise distance \(1\), so it cannot satisfy the Cauchy condition for \(\varepsilon=1\).
Worked Example: The Real Line with Its Usual Metric
Equip \(\mathbb{R}\) with the usual metric \(d(x,y)=|x-y|\). The completeness of \(\mathbb{R}\) is the standard completeness property of the real numbers. To see why this complete space is not totally bounded, use the subset \(E=\mathbb{Z}\).
For distinct integers \(m,n\), the nonzero integer \(m-n\) has absolute value at least \(1\). Hence \(|m-n|\ge1\). Apply the proposition with \(\delta=1\): each open ball of radius \(1/3\), regardless of its center in \(\mathbb{R}\), contains at most one integer. Indeed, if it contained integers \(m\ne n\), then \[ |m-n|\le |m-z|+|z-n|<1/3+1/3=2/3<1, \] a contradiction. Finitely many such balls therefore cover only finitely many integers and cannot cover \(\mathbb{R}\). Thus \(\mathbb{R}\) is not totally bounded.
The sequence of integers \(1,2,3,\dots\) has pairwise distances at least \(1\), so it has no convergent subsequence. This gives an explicit failure of sequential compactness as well. Notice that completeness is not in question: its Cauchy sequences do converge in \(\mathbb{R}\). The example fails compactness because it has points arbitrarily far apart, not because it is missing limits of Cauchy sequences.
Check. The numerical radius was \(1/3\), and two points in one ball would have distance strictly less than \(2/3\), while distinct integers have distance at least \(1\). The strict inequality is what rules out two integers in the same ball.
A Complete Space of Bounded Sequences
Failure of total boundedness is not limited to spaces whose points move indefinitely along a line. It can also arise from infinitely many directions, even when the vectors in question all have norm \(1\). We give a complete metric space where this happens.
Worked Example: Bounded Real Sequences with the Supremum Metric
Let \(X\) be the set of all bounded real sequences \(x=(x_1,x_2,\dots)\), and define \(d(x,y)=\sup_{k\in\mathbb{N}}|x_k-y_k|\). This supremum is finite because both sequences are bounded. We verify that this metric space is complete and not totally bounded.
Completeness. Let \((x^{(m)})_{m\in\mathbb{N}}\) be a Cauchy sequence in this metric, where \(x^{(m)}=(x^{(m)}_1,x^{(m)}_2,\dots)\). For each fixed coordinate \(k\), \[ |x^{(m)}_k-x^{(n)}_k|\le d(x^{(m)},x^{(n)}). \] Thus \((x^{(m)}_k)_{m\in\mathbb{N}}\) is Cauchy in \(\mathbb{R}\). By completeness of \(\mathbb{R}\), it has a limit \(x_k\in\mathbb{R}\). These coordinate limits define a sequence \(x=(x_1,x_2,\dots)\).
It remains to check both that \(x\) is bounded and that convergence holds in the supremum metric. Given \(\varepsilon>0\), the Cauchy property supplies \(M\) such that \(d(x^{(m)},x^{(n)})<\varepsilon/2\) for all \(m,n\ge M\). Fix \(m\ge M\) and \(k\). As \(n\) tends to infinity, \(x^{(n)}_k\) tends to \(x_k\); therefore \[ |x^{(m)}_k-x_k|\le\varepsilon/2. \] This holds for every \(k\). In particular, using \(m=M\), \[ |x_k|\le |x^{(M)}_k|+\varepsilon/2 \] for every \(k\). Since \(x^{(M)}\) is bounded, this inequality shows that \(x\) is bounded, so \(x\in X\). Also, for every \(m\ge M\), taking the supremum over \(k\) gives \(d(x^{(m)},x)\le\varepsilon/2<\varepsilon\). Hence \(x^{(m)}\to x\) in \(X\), proving completeness.
Failure of total boundedness. For each \(n\in\mathbb{N}\), let \(e_n\) be the sequence whose \(n\)-th coordinate is \(1\) and whose other coordinates are \(0\). Each \(e_n\) is bounded. If \(m\ne n\), the \(m\)-th coordinate of \(e_m-e_n\) is \(1\), and every coordinate difference has absolute value at most \(1\). Consequently \(d(e_m,e_n)=1\). The set \(\{e_n:n\in\mathbb{N}\}\) is infinite and its distinct points are separated by \(\delta=1\). The proposition shows that \(X\) is not totally bounded and that the sequence \((e_n)\) has no convergent subsequence.
Check. The supremum of the coordinate differences is exactly \(1\): one coordinate difference equals \(1\), and none is larger. Thus the separation used above holds for every distinct pair, not merely for consecutive terms.
A Common Mistake: Bounded Does Not Mean Totally Bounded
Common pitfall. In the discrete metric on an infinite set, it is tempting to reason: “All distances are at most \(1\), so the space is totally bounded.” This confuses a bound on distances with the existence of finite nets. In the natural numbers example, at radius \(1/2\), each ball is a singleton, so no finite family of these balls covers the infinite space. The space has diameter \(1\), but it is not totally bounded.
The failed reasoning would apply a single bound to all pairwise distances and treat that as if it produced finitely many small balls. The correction is to test the definition at a specific positive radius: total boundedness requires a finite cover at every positive radius. At radius \(1/2\), the singleton balls expose the failure immediately.
The examples share a useful pattern. To disprove total boundedness, find infinitely many points separated by at least one fixed positive distance. To disprove sequential compactness, list distinct points from that collection: the resulting sequence cannot have a Cauchy, and hence cannot have a convergent, subsequence. Completeness answers a different question. It guarantees limits for Cauchy sequences that occur in the space; it does not force arbitrary sequences to contain Cauchy subsequences.
Check Your Understanding
For each question, use the definitions of completeness, total boundedness, and sequential compactness rather than relying on a picture of the space.
- In an infinite set with the discrete metric \(d(x,y)=0\) for \(x=y\) and \(d(x,y)=1\) for \(x\ne y\), describe all Cauchy sequences and explain why the space is complete.
- Why do finitely many open balls of radius \(1/2\) fail to cover an infinite discrete metric space?
- Suppose \(E\subseteq X\) is infinite and distinct points of \(E\) have distance at least \(\delta>0\). Give a radius at which no finite collection of balls can cover \(E\).
- Why does the sequence \(e_n\) in the bounded-sequence example have no convergent subsequence, even though every \(e_n\) belongs to the same metric space?
- In the real line example, what strict inequality follows if two integers lie in the same open ball of radius \(1/3\), and why is it a contradiction?