Bounded at One Scale, Finite at Every Scale
A bounded set fits inside one ball of some finite radius. Total boundedness asks for more: no matter how small a positive scale is chosen, finitely many balls of that radius must cover the set. The number of balls may depend on the scale. This distinction is central in the study of compact metric spaces.
We will use the notation \(B_\varepsilon(a)\) for the open ball of radius \(\varepsilon\) about \(a\). When considering a subset \(E\) of a metric space \(X\), distances are measured using the restricted metric. Thus, if \(a\in E\), the ball in the subspace \(E\) is \(B_\varepsilon(a)\cap E\). For convenience, a finite cover of \(E\) by such balls can also be written using balls in \(X\).
Equivalently, total boundedness means that for every \(\varepsilon>0\), finitely many open balls of radius \(\varepsilon\), with centers in \(E\), cover \(E\). The strict inequality in the definition comes from using open balls. The choice of radius matters: a finite cover at one scale does not automatically give a finite cover at every smaller scale.
For a fixed \(\varepsilon\), a finite \(\varepsilon\)-net records only that the set can be approximated to that precision using finitely many reference points. As \(\varepsilon\) decreases, the net may need more points. Total boundedness requires a finite net at each scale, but it does not require one fixed finite set of centers to work for all scales.
Finite Sets, Subsets, and Finite Unions
Worked Example: Every Finite Metric Space Is Totally Bounded
Let \(E=\{p,q,r,s\}\) be a finite subset of any metric space, with the restricted metric. Fix \(\varepsilon>0\). Take \(F=E\). Each point is within distance \(\varepsilon\) of itself, since \(d(p,p)=d(q,q)=d(r,r)=d(s,s)=0<\varepsilon\). Thus every point of \(E\) belongs to the ball of radius \(\varepsilon\) centered at itself, and
So \(F\) is a finite \(\varepsilon\)-net. Since the choice of \(\varepsilon>0\) was arbitrary, \(E\) is totally bounded. No estimate on the distances between distinct points is needed.
Total boundedness is preserved when we pass to a subset, although a small adjustment may be needed because the old net’s centers might not lie in the subset. It is also preserved when we combine finitely many sets.
Proof. Let \(A\) be totally bounded and \(C\subseteq A\). If \(C\) is empty, it is totally bounded by definition. Suppose \(C\) is nonempty, and fix \(\varepsilon>0\). Choose a finite \(\varepsilon/2\)-net \(F=\{a_1,\ldots,a_m\}\subseteq A\) for \(A\). Keep only those indices \(i\) for which \(B_{\varepsilon/2}(a_i)\cap C\) is nonempty. For each such index, choose a point \(c_i\in B_{\varepsilon/2}(a_i)\cap C\). These finitely many chosen points belong to \(C\).
For any \(c\in C\), the net property gives an index \(i\) with \(d(c,a_i)<\varepsilon/2\). This index is among those kept, since \(c\in B_{\varepsilon/2}(a_i)\cap C\). By the choice of \(c_i\), \(d(a_i,c_i)<\varepsilon/2\). The triangle inequality now gives
Thus the chosen points form a finite \(\varepsilon\)-net for \(C\). Since \(\varepsilon>0\) was arbitrary, \(C\) is totally bounded.
Now let \(A_1,\ldots,A_k\) be totally bounded sets, where \(k\) is a positive integer. Fix \(\varepsilon>0\). For each nonempty \(A_j\), choose a finite \(\varepsilon\)-net \(F_j\subseteq A_j\). The union \(F_1\cup\cdots\cup F_k\) is finite. Every point in \(A_1\cup\cdots\cup A_k\) belongs to some \(A_j\), and so lies within distance \(\varepsilon\) of a point in \(F_j\). The union of the nets is therefore an \(\varepsilon\)-net for the union. Empty sets contribute no points and cause no change. This proves the finite-union assertion. \(\square\)
Constructing Nets in Familiar Spaces
Worked Example: A Closed Interval in the Real Line
Consider \(I=[-2,3]\) with the usual metric. Fix \(\varepsilon>0\), and choose a positive integer \(N\) so large that \(5/(2N)<\varepsilon\). Divide the interval into \(N\) equal subintervals, each of length \(h=5/N\), and let
All \(N+1\) points \(c_j\) belong to \(I\). Every \(x\in I\) lies in at least one subinterval \([c_j,c_{j+1}]\), where \(0\leq j<N\). Of the two endpoints, at least one is at distance at most \(h/2\) from \(x\): if both distances were greater than \(h/2\), their sum would be greater than \(h\), even though \(c_{j+1}-c_j=h\). Therefore some center \(c_j\) from the finite set of grid points satisfies
The grid points form a finite \(\varepsilon\)-net. Since this construction works for every \(\varepsilon>0\), \([-2,3]\) is totally bounded. Smaller values of \(\varepsilon\) may require larger \(N\), which is allowed.
Worked Example: A Rectangle with the Maximum Metric
Let \(R=[0,2]\times[-1,1]\), and use the maximum metric \(d_\infty((x,y),(u,v))=\max\{|x-u|,|y-v|\}\). Fix \(\varepsilon>0\). Choose positive integers \(N\) and \(M\) such that \(1/N<\varepsilon\) and \(1/M<\varepsilon\). Use the horizontal grid points \(x_i=2i/N\), for \(0\leq i\leq N\), and the vertical grid points \(y_j=-1+2j/M\), for \(0\leq j\leq M\). The finite set of grid points in the rectangle is
For any \((x,y)\in R\), one can choose a horizontal grid point \(x_i\) within distance \(1/N\) of \(x\), and a vertical grid point \(y_j\) within distance \(1/M\) of \(y\). Hence
Thus \(F\) is a finite \(\varepsilon\)-net. Since this holds at every positive scale, the rectangle is totally bounded in the maximum metric.
Total Boundedness Implies Boundedness
Although total boundedness requires a cover at every scale, even one of those covers is enough to prove ordinary boundedness. This gives a new implication in addition to the result from the previous tutorial that compact sets are bounded.
Proof. The empty space is bounded by convention, so let \(X\) be a nonempty totally bounded metric space. By total boundedness, there is a finite \(1\)-net \(F=\{a_1,\ldots,a_m\}\subseteq X\). Since \(X\) is nonempty, the net cannot be empty, so \(m\geq1\). The finitely many distances \(d(a_i,a_1)\) have a finite maximum; write
For each \(x\in X\), the net property supplies an index \(i\) such that \(d(x,a_i)<1\). The triangle inequality then gives
Consequently \(X\subseteq B_{M+1}(a_1)\), a ball with finite positive radius. Thus \(X\) is bounded. \(\square\)
In particular, the proposition from the previous tutorial characterizing boundedness by finite diameter shows that every totally bounded nonempty metric space has finite diameter. The reverse implication fails: a single finite distance bound says nothing about whether infinitely many points can be covered by finitely many balls of a much smaller radius.
Worked Example: A Bounded Space That Is Not Totally Bounded
Let \(X\) be any infinite set with the discrete metric \(\delta\), which assigns distance \(0\) to equal points and distance \(1\) to distinct points. Every pair of points in \(X\) is at distance at most \(1\), so \(X\subseteq B_2(a)\) for any chosen \(a\in X\). Thus \(X\) is bounded.
Now take \(\varepsilon=1/2\). By the previously established description of balls in the discrete metric, each ball \(B_{1/2}(a)\) contains only \(a\). A finite collection of these balls therefore contains only finitely many points. Because \(X\) is infinite, no finite collection covers \(X\). So \(X\) has no finite \(1/2\)-net and is not totally bounded.
This example shows why the quantifier “for every \(\varepsilon>0\)” matters. A bounded set can fit inside a large ball and still have infinitely many points separated at a smaller scale.
Why Total Boundedness Matters
Total boundedness captures a form of finite approximation. At any prescribed accuracy, a finite list of points suffices to represent the whole space to within that accuracy. The finite list may change and grow as the accuracy increases. In analysis, this makes it possible to reduce questions about an entire space to finitely many local regions at a chosen scale.
Earlier in this course, the Sequential Characterization of Total Boundedness connected this covering condition with sequences: a metric space is totally bounded exactly when every sequence in it has a Cauchy subsequence. This is one reason total boundedness appears alongside completeness. The theorem Complete and Totally Bounded Metric Spaces Are Compact, established earlier, states that the two properties together characterize compactness for metric spaces.
A common pitfall is to treat boundedness as if it already supplied finite approximation. The infinite discrete example rules this out: its diameter is finite, but at radius \(1/2\) each ball captures only one point. In the next step of the compactness argument, compactness will supply the stronger, scale-by-scale finite-cover property.
Check Your Understanding
Use the definition of an \(\varepsilon\)-net and the results proved above to answer these questions.
- What must be true about the centers and the covering distances for a finite set to be an \(\varepsilon\)-net?
- Why might the proof that a subset of a totally bounded set is totally bounded use nets at radius \(\varepsilon/2\) rather than \(\varepsilon\)?
- In the proof that total boundedness implies boundedness, why is it useful to compare every point with the same net center \(a_1\)?
- Why is an infinite discrete metric space bounded but not totally bounded?
- Does a finite net at one radius establish total boundedness? Explain what the definition requires instead.