Compactness Produces Finite Nets
A metric space is totally bounded when, at every positive scale, finitely many points suffice to lie within that scale of every point in the space. Compactness supplies exactly the finiteness needed: the open balls of a chosen radius cover the space, and compactness reduces that cover to a finite subcover. The centers of the selected balls form the desired finite set.
Throughout, a metric space \((X,d)\) has the topology generated by its open balls \(B_d(x,r)=\{y\in X:d(x,y)<r\}\), where \(x\in X\) and \(r>0\). Compactness means that every open cover has a finite subcover. The empty space is compact, since every open cover of it has the empty finite subcover.
The centers in a net must belong to \(X\). The net condition can also be written \(X\subseteq\bigcup_{f\in F}B_d(f,\varepsilon)\), so a net is a finite collection of centers whose open \(\varepsilon\)-balls cover the space. We now show that compactness always provides such a collection. The radius of the balls in the proof will be exactly the desired net scale \(\varepsilon\).
The Compactness Argument
Proof. Fix \(\varepsilon>0\). If \(X=\varnothing\), take \(F=\varnothing\). It is finite, and the requirement that every \(x\in X\) have a center is vacuous. Now suppose \(X\ne\varnothing\). For each \(x\in X\), the set \(B_d(x,\varepsilon)\) is open in the metric topology. The family \(\{B_d(x,\varepsilon):x\in X\}\) covers \(X\), because \(x\in B_d(x,\varepsilon)\) follows from \(d(x,x)=0<\varepsilon\). By compactness, finitely many members cover \(X\); write them as \(B_d(x_1,\varepsilon),\ldots,B_d(x_m,\varepsilon)\), where each center \(x_i\) belongs to \(X\). Set \(F=\{x_1,\ldots,x_m\}\), removing repetitions if any. For every \(x\in X\), the finite balls cover \(X\), so \(x\in B_d(x_i,\varepsilon)\) for some \(i\). By the definition of an open ball, this means \(d(x,x_i)<\varepsilon\). Thus \(F\) is a finite \(\varepsilon\)-net. Since \(\varepsilon>0\) was arbitrary, \(X\) is totally bounded. \(\square\)
The radius choice matters. Membership in an open ball of radius \(\varepsilon\) gives the strict inequality \(d(x,x_i)<\varepsilon\) directly, with no boundary case to resolve. Compactness does not tell us in advance which centers will suffice; it guarantees that some finite selection from the entire ball cover does suffice. In specific spaces we may be able to exhibit a net explicitly, as in the next examples.
Worked Example: A Finite Discrete Metric Space
Let \(X=\{0,1,2,3\}\) with the discrete metric \(d(x,y)=0\) if \(x=y\), and \(d(x,y)=1\) if \(x\ne y\). This metric gives \(X\) its metric topology. First, \(X\) is compact: given any open cover, choose one covering set for each of the four points. The chosen collection has at most four members and covers \(X\).
Take \(\varepsilon=1/2\). If a proposed set \(F\subseteq X\) omits a point \(x\), then every \(f\in F\) differs from \(x\), so \(d(x,f)=1\). Since \(1\) is not less than \(1/2\), no such \(F\) is a \(1/2\)-net. Thus a \(1/2\)-net must contain all four points, and \(F=X\) works: for each \(x\in X\), the center \(f=x\) gives \(d(x,f)=0<1/2\).
Now take \(\varepsilon=2\). The single center \(F=\{0\}\) works. The distances from \(0\) to \(0,1,2,3\), respectively, are \(0,1,1,1\), and each is strictly less than \(2\). In particular, \(d(0,2)=1\), not \(2\): the metric is discrete, not the usual distance between these labels.
Check: At scale \(1/2\), omitting any point leaves it at distance \(1\) from every selected center, so the strict net condition fails. At scale \(2\), every point has distance at most \(1<2\) from \(0\), so the one-point set really is a net.
Worked Example: A Net in the Unit Interval
Let \(X=[0,1]\) with the usual metric \(d(x,y)=|x-y|\) and its metric topology. By the Heine–Borel theorem, the closed bounded interval \([0,1]\) is compact. For \(\varepsilon=1/4\), consider \(F=\{0,1/4,1/2,3/4,1\}\).
To check the net condition, divide the interval into the four subintervals \([k/4,(k+1)/4]\) for \(k=0,1,2,3\). For any \(x\) in one of these subintervals, at least one endpoint is at distance at most half the subinterval’s length. The length is \(1/4\), so one of the two endpoints in \(F\) is at distance at most \(1/8\). Since \(1/8<1/4\), this distance is strictly less than \(\varepsilon\), as required.
For example, on \([1/4,1/2]\), the endpoints \(1/4\) and \(1/2\) are \(1/4\) apart. If \(x\leq 3/8\), then \(0\leq x-1/4\leq 1/8\); if \(x\geq 3/8\), then \(0\leq 1/2-x\leq 1/8\). The same midpoint argument applies on each of the other three subintervals. At the outer endpoints \(0\) and \(1\), the distance to a center is \(0\).
Check: The largest distance needed by this construction is at most \(1/8\), and \(1/8<1/4\). Thus every point of \([0,1]\) lies in an open ball of radius \(1/4\) around one of the five listed centers.
What Fails Without Compactness
The theorem is a sufficient-condition result: compactness guarantees total boundedness. The next example shows why the compactness assumption cannot simply be discarded. It also illustrates a direct way to test whether a proposed finite net can exist.
Worked Example: An Infinite Discrete Metric Space
Let \(X=\mathbb{N}=\{1,2,\ldots\}\) with the discrete metric \(d(m,n)=0\) when \(m=n\) and \(d(m,n)=1\) when \(m\ne n\); use its metric topology. This space is not compact: the open sets \(\{\{n\}:n\in\mathbb{N}\}\) cover \(X\), and no finite selection of these singletons covers all of \(\mathbb{N}\).
At scale \(\varepsilon=1/2\), no finite set \(F\subseteq X\) is a net. If \(F\) is finite, choose \(n\in\mathbb{N}\setminus F\), which is possible because \(\mathbb{N}\) is infinite. For every \(f\in F\), we have \(f\ne n\), and hence \(d(n,f)=1\). This is not less than \(1/2\), so \(n\) is not within the required distance of any center in \(F\).
Check: The same singleton cover verifies failure of compactness, while the calculation \(d(n,f)=1\) for every \(f\in F\) verifies failure of a finite \(1/2\)-net. Both failures use the stated discrete metric and the strict inequality in the definition.
A Consequence: Compact Metric Spaces Are Bounded
Proof. Apply the theorem with \(\varepsilon=1\), obtaining a finite \(1\)-net \(F\). Since \(X\ne\varnothing\), the net condition implies \(F\ne\varnothing\); choose \(p\in X\). The finite set of real numbers \(\{d(f,p):f\in F\}\) has a maximum \(M\). For any \(x\in X\), choose \(f\in F\) with \(d(x,f)<1\). The triangle inequality gives \(d(x,p)\leq d(x,f)+d(f,p)<1+M\). Consequently \(d(x,p)\leq 1+M\) for every \(x\in X\), so \(R=1+M>0\) is a bound. The empty space is bounded vacuously if boundedness is expressed as a condition on all its points. \(\square\)
This argument also indicates why total boundedness is stronger than merely being bounded: a single finite net controls distances at a specified scale, and total boundedness requires such finite control at every positive scale. The theorem obtains that control from open-cover compactness, not from a prior assumption that the space is bounded or complete.
A Common Pitfall: Confusing the Metric with the Labels
For the finite discrete space \(\{0,1,2,3\}\), it is incorrect to compute \(d(0,2)\) by subtracting the labels and obtaining \(2\). That subtraction would be relevant for the usual metric inherited from \(\mathbb{R}\), but the metric in the first example was defined to assign distance \(1\) to every pair of distinct points. Therefore \(d(0,2)=1<2\), and \(\{0\}\) is a \(2\)-net. The correction is to return to the given metric formula before testing any net condition.
A related error is to replace the strict condition \(d(x,f)<\varepsilon\) by \(d(x,f)\leq\varepsilon\). In the discrete metric example at \(\varepsilon=1\), a point omitted from \(F\) is at distance exactly \(1\) from every chosen center. It therefore prevents \(F\) from being a \(1\)-net under the definition used here. The theorem’s open balls avoid this ambiguity: membership in \(B_d(f,\varepsilon)\) means strictly less than \(\varepsilon\).
The practical point is that compactness turns an infinite family of possible centers into a finite selection while preserving the chosen radius. In applications, one can first choose the scale required by the problem, cover the space by balls of that radius, and then invoke compactness. The resulting finite subcover is the finite net; the selected ball centers are its points.
Check Your Understanding
For each question, use the strict inequality in the definition of an \(\varepsilon\)-net.
- In the discrete metric on \(\{a,b,c\}\), determine whether \(\{a\}\) is a \(1\)-net and whether it is a \(3/2\)-net.
- Explain why the family \(\{B_d(x,\varepsilon):x\in X\}\) covers a metric space for every \(\varepsilon>0\).
- In \([0,1]\) with the usual metric, verify that \(\{0,1/2,1\}\) is a \(1/2\)-net, or identify precisely why the strict condition fails.
- For the infinite discrete metric space \(\mathbb{N}\), explain why no finite set is a \(1/2\)-net.
- In the boundedness corollary, identify where finiteness of the net is used to obtain a uniform bound.