Finite Nets as a Measure of Metric Size
A metric space is totally bounded when, no matter how small a positive scale is chosen, finitely many balls of that radius cover the space. This requirement is stronger than simply having finite diameter: a space may have all its points within a fixed bounded distance of one another and still require infinitely many small balls to cover it. We will construct nets in Euclidean spaces and use separated points to identify failures of total boundedness.
For a subset \(A\) of a metric space, the same definition applies with \(F\subseteq A\) and every point of \(A\) within distance \(\varepsilon\) of a point of \(F\). The centers are required to belong to the set being covered. When \(A\) is empty, the empty set is a finite \(\varepsilon\)-net for \(A\).
By contrast, a metric space is bounded if there is some finite \(D\) such that \(d(x,y)\leq D\) for every \(x,y\in X\). Boundedness concerns one fixed distance scale. Total boundedness asks for finite covers at every positive scale, including arbitrarily small ones.
Constructing Nets in Euclidean Space
Proof. Let \(A\subseteq\mathbb{R}^n\) be bounded, and fix \(\varepsilon>0\). If \(A=\varnothing\), the empty set is an \(\varepsilon\)-net, so suppose \(A\ne\varnothing\). Boundedness gives a number \(M\geq0\) such that every \(x=(x_1,\dots,x_n)\in A\) satisfies \(|x_i|\leq M\) for each coordinate \(i\). Set \(h=\varepsilon/(2\sqrt n)>0\). Partition \(\mathbb{R}^n\) into half-open cubes of side length \(h\): the cube indexed by \((k_1,\dots,k_n)\in\mathbb{Z}^n\) consists of the points whose \(i\)-th coordinate lies in \([k_i h,(k_i+1)h)\) for every \(i\).
Only finitely many of these cubes can meet \(A\). Indeed, if the \(i\)-th coordinate of a point in a cube lies between \(-M\) and \(M\), then its index \(k_i\) lies between the fixed finite bounds \(\lfloor -M/h\rfloor\) and \(\lfloor M/h\rfloor\). There are finitely many possible indices in each of the \(n\) coordinates, and hence finitely many possible cubes. From each cube that meets \(A\), choose one point of \(A\) in that cube, and let \(F\) be the set of chosen points. This is a finite subset of \(A\).
For any \(x\in A\), let \(f\in F\) be the chosen point from the same cube as \(x\). In each coordinate, \(|x_i-f_i|<h\), so
Thus \(\|x-f\|<\varepsilon/2<\varepsilon\). Every point of \(A\) is within \(\varepsilon\) of a point of the finite set \(F\), so \(F\) is an \(\varepsilon\)-net. Since \(\varepsilon>0\) was arbitrary, \(A\) is totally bounded. \(\square\)
Worked Example: A Net for a Closed Interval
Consider \(A=[0,1]\subseteq\mathbb{R}\) with the standard metric, and take \(\varepsilon=1/4\). Use the finite set
The intervals of radius \(1/4\) about these centers are open. Consecutive centers are \(1/4\) apart, so a point between them is at distance at most \(1/8\) from one of the two centers. More explicitly, for \(x\in[0,1/4]\), the center \(0\) is at distance \(x\leq1/4\), but at the endpoint \(x=1/4\) we instead use the center \(1/4\), which is at distance \(0\). For \(x\in[0,1/8]\), \(d(x,0)=x\leq1/8<1/4\); for \(x\in[1/8,1/4]\), \(d(x,1/4)=1/4-x\leq1/8<1/4\). The same calculation applies to each of the other three subintervals between consecutive centers.
Check: the largest distance needed to the nearer center is \(1/8\), and \(1/8<1/4=\varepsilon\). The intervals of radius \(1/4\) about the five listed centers therefore cover \([0,1]\), so \(F\) is an \(\varepsilon\)-net.
Worked Example: A Net for a Rectangle
Let \(A=[-1,1]\times[0,1]\subseteq\mathbb{R}^2\), with the Euclidean metric, and take \(\varepsilon=1/2\). Use the finite set
There are \(9\) choices for \(j\) and \(5\) choices for \(k\), so \(F\) has \(9\cdot5=45\) points, all in \(A\). For each \(x\)-coordinate in \([-1,1]\), some grid coordinate \(-1+j/4\) is at distance at most \(1/8\). This follows because consecutive grid coordinates are \(1/4\) apart, including the endpoints. The same reasoning shows that each \(y\)-coordinate in \([0,1]\) is within \(1/8\) of some \(k/4\). Therefore, for every \(a\in A\), there is a point \(f\in F\) with
The strict inequality holds because \(\sqrt2<4\), as \(2<16\). Thus each point of \(A\) lies in an open ball of radius \(1/2\) about one of the \(45\) grid points, and \(F\) is a finite \(1/2\)-net.
Separated Points Obstruct Finite Nets
Proof. Take \(\varepsilon=\delta/2>0\). Any open ball \(B_d(x,\varepsilon)\) contains at most one point of \(S\). If it contained distinct \(s,t\in S\), the triangle inequality would give \(d(s,t)\leq d(s,x)+d(x,t)<2\varepsilon=\delta\), contradicting \(d(s,t)\geq\delta\). If a finite \(\varepsilon\)-net for \(X\) existed, its finitely many balls would therefore cover at most finitely many points of \(S\). That cannot cover the infinite set \(S\). So \(X\) has no finite \(\varepsilon\)-net and is not totally bounded. \(\square\)
Worked Example: An Infinite Discrete Metric Space
Let \(X\) be an infinite set with the discrete metric \(d(x,y)=0\) when \(x=y\) and \(d(x,y)=1\) when \(x\ne y\). Take \(\varepsilon=1/2\). For any center \(x\in X\), the open ball \(B_d(x,1/2)\) is exactly \(\{x\}\): the distance from \(x\) to itself is \(0<1/2\), while the distance from \(x\) to every other point is \(1\), which is not less than \(1/2\).
A finite collection of these balls is a finite collection of singletons and cannot cover the infinite set \(X\). Thus \(X\) is not totally bounded. The metric is nevertheless bounded: for all \(x,y\in X\), \(d(x,y)\leq1\). This gives an explicit example where boundedness does not imply total boundedness.
For comparison, if \(X\) is finite, then \(F=X\) is an \(\varepsilon\)-net for every \(\varepsilon>0\), because each \(x\in X\) has distance \(0<\varepsilon\) from itself. Hence every finite metric space is totally bounded.
Worked Example: A Bounded Metric on the Real Line
Define a metric on \(\mathbb{R}\) by \(d_b(x,y)=\min\{|x-y|,1\}\). Its values are at most \(1\), so this metric is bounded. Nevertheless, it is not totally bounded. Consider the infinite set \(S=\mathbb{Z}\). For distinct integers \(m,n\), \(|m-n|\geq1\), and consequently
Thus distinct points of \(S\) are separated by at least \(\delta=1\). The proposition, with \(\varepsilon=1/2\), shows that no finite collection of open balls of radius \(1/2\) can cover \(\mathbb{R}\). Directly, each such ball contains at most one integer: if it contained two integers \(m\ne n\), the triangle inequality would give \(d_b(m,n)<1\), contrary to the calculation above.
Check: the metric has diameter at most \(1\), but a ball of radius \(1/2\) cannot contain two integers. A finite set of such balls covers only finitely many integers, so it cannot cover the real line.
What the Examples Tell Us
The Euclidean theorem and the discrete examples distinguish several useful situations. In \(\mathbb{R}^n\), boundedness lets us place a bounded set inside finitely many small cubes and choose a representative from each cube that meets the set. The construction works at every scale because the number of relevant cubes is finite for each fixed positive side length. In an infinite discrete metric space, however, choosing a scale smaller than the distance between distinct points makes each ball cover at most one point. An infinite separated set provides the same obstruction even when the metric space is not discrete.
A common mistake is to infer total boundedness merely from a uniform bound on distances. For the bounded metric \(d_b\) above, one might reason: “All distances are at most \(1\), so one ball covers the whole space.” That conclusion is valid for a ball of radius \(2\), since \(d_b(x,0)\leq1<2\) for every \(x\in\mathbb{R}\). But total boundedness requires a finite net for every positive radius. At radius \(1/2\), each ball contains at most one integer, and finitely many such balls cannot cover all integers. The one-ball calculation at radius \(2\) does not meet the requirement at radius \(1/2\).
Total boundedness is therefore a small-scale finiteness condition, not simply another name for finite diameter. In the next stage of metric-space study, this condition will be considered alongside completeness; here the key tests are explicit finite nets for bounded Euclidean sets and explicit separated subsets when total boundedness fails.
Check Your Understanding
Use the definition with open balls and strict inequalities when checking whether a set is an \(\varepsilon\)-net.
- For \(A=[0,2]\subseteq\mathbb{R}\) and \(\varepsilon=1/2\), give a finite \(\varepsilon\)-net whose centers lie in \(A\), and verify the largest distance to a center.
- Why does the proof for bounded subsets of \(\mathbb{R}^n\) require only finitely many cubes to meet the set at each fixed side length?
- Show that an infinite set with the discrete metric is not totally bounded by choosing a positive radius and describing every ball of that radius.
- Suppose an infinite subset \(S\) of a metric space has pairwise distances at least \(3\). Which radius can be used in the separated-set argument to rule out a finite net?
- Explain why the fact that a metric space has diameter at most \(1\) gives a finite net at radius \(2\), but does not by itself establish total boundedness.