Complete and Totally Bounded Spaces

Two Different Forms of Control

Completeness and total boundedness impose different kinds of control on a metric space. Completeness concerns sequences: every Cauchy sequence must converge to a point of the space. Total boundedness concerns finite approximation: at each positive scale, finitely many balls of that radius cover the space. Neither condition should be mistaken for the other. The useful situation is when both hold.

Recall that a metric space is complete if every Cauchy sequence in it converges in the space, and is totally bounded if, for every \(\varepsilon>0\), it has a finite \(\varepsilon\)-net. As in the previous tutorial, a finite \(\varepsilon\)-net consists of centers in the space such that every point is at distance strictly less than \(\varepsilon\) from at least one center.

Earlier in this course, the Sequential Characterization of Total Boundedness established that every sequence in a totally bounded metric space has a Cauchy subsequence. If the space is also complete, that Cauchy subsequence converges in the space. We will use this sequence-extraction mechanism to prove a useful consequence for continuous functions.

Worked Examples: Checking the Two Conditions

Continuous Maps Are Uniformly Continuous

The combination of the two conditions gives a global conclusion from pointwise continuity. Total boundedness supplies a Cauchy subsequence from any sequence, and completeness turns that subsequence into a convergent one. If a continuous map failed to be uniformly continuous, points could be chosen arbitrarily close together while their images stayed a fixed positive distance apart. The subsequence mechanism rules out that possibility.

Proof. If \(X\) is empty, both conclusions hold trivially. Suppose \(X\) is nonempty. We first prove uniform continuity. Assume, to the contrary, that \(f\) is not uniformly continuous. By the negation of the definition of uniform continuity, there is an \(\eta>0\) such that for every \(\delta>0\) there are \(x,y\in X\) satisfying

For each positive integer \(n\), apply this statement with \(\delta=1/n\), obtaining \(x_n,y_n\in X\) such that

By the Sequential Characterization of Total Boundedness, \((x_n)\) has a Cauchy subsequence \((x_{n_k})\). Since \(X\) is complete, there is an \(x\in X\) such that \(x_{n_k}\to x\). The triangle inequality gives

Both terms on the right tend to zero, so \(y_{n_k}\to x\) as well. Continuity of \(f\) at \(x\), or equivalently the Sequential Characterization of Continuity, now gives \(f(x_{n_k})\to f(x)\) and \(f(y_{n_k})\to f(x)\). Another application of the triangle inequality yields

This contradicts \(d_Y(f(x_{n_k}),f(y_{n_k}))\geq\eta\) for every \(k\). Therefore \(f\) is uniformly continuous.

To prove boundedness of the image, use uniform continuity with the target tolerance \(1\). There is a \(\delta>0\) such that \(d_X(u,v)<\delta\) implies \(d_Y(f(u),f(v))<1\). Total boundedness gives a finite \(\delta\)-net \(\{a_1,\ldots,a_m\}\) for \(X\). For every \(x\in X\), some \(a_j\) satisfies \(d_X(x,a_j)<\delta\), and therefore \(d_Y(f(x),f(a_j))<1\). Thus

A finite union of balls of radius \(1\) is bounded: for example, fix \(f(a_1)\); the triangle inequality bounds the distance from any point in any one of these balls to \(f(a_1)\) by \(1+\max_j d_Y(f(a_j),f(a_1))\). Hence \(f[X]\) is bounded. \(\square\)

Finite Products Preserve Both Conditions

The two properties also combine well under a basic construction. For two metric spaces \(X\) and \(Y\), equip \(X\times Y\) with the maximum metric \(d((x_1,y_1),(x_2,y_2))=\max\{d_X(x_1,x_2),d_Y(y_1,y_2)\}\). Total boundedness of this product follows from the earlier theorem that finite products preserve total boundedness. Completeness can be checked coordinate by coordinate.

Proof. Total boundedness follows from the Finite Products Preserve Total Boundedness theorem. To prove completeness, let \(((x_n,y_n))\) be a Cauchy sequence in \(X\times Y\). For every \(\varepsilon>0\), there is an \(N\) such that for \(m,n\geq N\),

Each coordinate distance is bounded above by this maximum, so \((x_n)\) is Cauchy in \(X\) and \((y_n)\) is Cauchy in \(Y\). Completeness gives points \(x\in X\) and \(y\in Y\) such that \(x_n\to x\) and \(y_n\to y\). Given \(\varepsilon>0\), for all sufficiently large \(n\) both \(d_X(x_n,x)<\varepsilon\) and \(d_Y(y_n,y)<\varepsilon\). Therefore

Thus \((x_n,y_n)\to(x,y)\) in the maximum metric, proving that the product is complete. \(\square\)

Why the Pairing Matters

The examples distinguish the jobs done by the two hypotheses. Total boundedness lets us find Cauchy subsequences at any chosen scale; it does not ensure that their limits belong to the space, as the rational example shows. Completeness ensures that Cauchy sequences converge in the space; it does not provide finite nets, as the real line shows. In the continuous-map theorem, both steps are needed: first extract a Cauchy subsequence, then use completeness to obtain a limit at which continuity can be applied.

A common pitfall is to replace total boundedness with boundedness. A single large ball gives no finite cover by small balls, and therefore does not provide the subsequence control used in the proof. Another is to use pointwise continuity where a global estimate is required. The contradiction argument explains why the stronger conclusion of uniform continuity follows here: a failure of uniformity would produce two sequences approaching each other while their images remain separated.

These examples and results prepare the way for the next step: understanding what the combination of completeness and total boundedness implies for compactness in a metric space.