Finite Nets: A Stronger Form of Boundedness
Metric arguments often ask whether a space can be controlled using finitely many points at a prescribed accuracy. This is stronger than asking whether the space is bounded. A bounded set may fit inside one large ball while still requiring infinitely many small balls to cover it. The finite-covering property that rules out this behavior is called total boundedness.
For a subset \(A\) of a metric space \(X\), a finite set of points can serve as approximate representatives for every point of \(A\). The allowed approximation error is a positive radius \(\varepsilon\). The centers need not belong to \(A\), although in many examples they can be chosen there. Thinking in terms of such finite nets gives a useful bridge between geometric covering arguments and sequence arguments.
Equivalently, total boundedness means that for every positive radius, \(A\) is covered by finitely many open balls of that radius. Indeed, if \(F=\{z_1,\ldots,z_m\}\) is an \(\varepsilon\)-net, then \(A\subseteq\bigcup_{j=1}^m B_\varepsilon(z_j)\). Conversely, the centers of any such finite ball cover form an \(\varepsilon\)-net.
Total Boundedness Gives Cauchy Subsequences
The finite-net condition has a precise sequential consequence. At each successively smaller scale, a finite cover must place infinitely many terms of a sequence into at least one ball. Repeating this selection produces nested infinite collections of indices. A diagonal choice from those collections yields a subsequence whose terms become arbitrarily close to one another.
Proof. First suppose \(X\) is totally bounded, and let \((x_n)\) be any sequence in \(X\). Cover \(X\) by finitely many open balls of radius \(1/2\). At least one of these balls contains \(x_n\) for infinitely many indices, since a finite union of sets containing only finitely many terms could contain only finitely many terms in total. Let \(I_1\) be an infinite set of indices for which \(x_n\) lies in that ball.
Next cover \(X\) by finitely many open balls of radius \(1/4\). One of them contains \(x_n\) for infinitely many indices in \(I_1\); let \(I_2\subseteq I_1\) be an infinite set of such indices. Continue inductively: at stage \(k\), use a finite cover by balls of radius \(2^{-k}\) and choose an infinite set \(I_k\subseteq I_{k-1}\) whose terms all lie in one of those balls. For the first stage use the cover of radius \(2^{-1}\).
Choose indices \(n_k\in I_k\) so that \(n_1<n_2<\cdots\). This is possible because each \(I_k\) is infinite and hence unbounded in the positive integers. If \(j,\ell\geq k\), then \(n_j,n_\ell\in I_k\), because the sets \(I_m\) are nested. Thus \(x_{n_j}\) and \(x_{n_\ell}\) lie in the same ball of radius \(2^{-k}\). The triangle inequality gives $$ d(x_{n_j},x_{n_\ell})<2^{-k}+2^{-k}=2^{1-k}. $$ Given \(\varepsilon>0\), choose \(k\) so large that \(2^{1-k}<\varepsilon\). Then all terms of the subsequence with indices at least \(k\) are within \(\varepsilon\) of one another. The subsequence is Cauchy.
For the reverse implication, suppose \(X\) is not totally bounded. Then there is some \(\varepsilon_0>0\) for which no finite collection of open balls of radius \(\varepsilon_0\) covers \(X\). Choose \(x_1\in X\). Having chosen \(x_1,\ldots,x_n\), the balls \(B_{\varepsilon_0}(x_1),\ldots,B_{\varepsilon_0}(x_n)\) do not cover \(X\). Choose \(x_{n+1}\) outside their union. This gives $$ d(x_{n+1},x_j)\geq\varepsilon_0\qquad (1\leq j\leq n). $$ Consequently, any two distinct terms of the sequence are at least \(\varepsilon_0\) apart. No subsequence can be Cauchy: the Cauchy condition with tolerance \(\varepsilon_0\) would require all sufficiently late pairs of its terms to have distance less than \(\varepsilon_0\). This contradicts the separation just established. Hence, if every sequence has a Cauchy subsequence, \(X\) must be totally bounded. \(\square\)
The argument also provides a practical diagnostic. Failure of total boundedness gives a fixed scale at which points can be chosen one after another, always outside the balls around earlier choices. The result is a sequence with a uniform separation, which cannot contain a Cauchy subsequence.
Worked Examples: Nets at Different Scales
Worked Example: A Bounded Discrete Space That Is Not Totally Bounded
Let \(X\) be an infinite set with the discrete metric \(\delta(x,y)=0\) when \(x=y\) and \(\delta(x,y)=1\) when \(x\ne y\). This space is bounded: for any \(x,y\in X\), \(\delta(x,y)\leq1\), so one ball of radius \(2\) covers \(X\). But it is not totally bounded. For \(\varepsilon=1/2\), each ball \(B_{1/2}(x)\) is the singleton \(\{x\}\). A finite collection of such balls covers only finitely many points, so it cannot cover the infinite set \(X\).
The sequence characterization gives another verification. Choose distinct points \(x_1,x_2,\ldots\) in \(X\). Every pair of distinct terms has distance \(1\), so no subsequence is Cauchy. This example shows why boundedness alone is too weak: a single large ball can cover the space, even though arbitrarily fine covers cannot be made finite.
Worked Example: A Finite Net for a Bounded Interval
Consider \(A=(2,5)\) with the usual metric and let \(\varepsilon>0\). Choose a positive integer \(N\) such that \(3/N<\varepsilon\), and set $$ z_j=2+\frac{3j}{N},\qquad j=0,1,\ldots,N. $$ These points form a finite \(\varepsilon\)-net for \(A\). To check this, take any \(x\in(2,5)\). The mesh points divide \([2,5]\) into intervals of length \(3/N\). Thus \(x\) lies in some interval \([z_j,z_{j+1}]\), and its distance to \(z_j\) is at most \(3/N<\varepsilon\). The finite set \(\{z_0,\ldots,z_N\}\) therefore covers \(A\) by open \(\varepsilon\)-balls.
The centers at the endpoints need not belong to \(A\), and the definition permits that. If centers inside \(A\) are preferred, choose points in \(A\) sufficiently close to the finitely many endpoints; alternatively, the definition as stated already proves total boundedness of the interval. The key feature is that the number of centers may depend on \(\varepsilon\), but it is finite for each chosen \(\varepsilon\).
Worked Example: A Convergent Set Has Finite Nets
Let \(A=\{0\}\cup\{1/n:n\in\mathbb{N},\,n\geq1\}\subseteq\mathbb{R}\). Fix \(\varepsilon>0\), and choose \(N\) such that \(1/(N+1)<\varepsilon\). Use the finite set $$ F=\{0,1,1/2,\ldots,1/N\}. $$ Every point \(1/n\) with \(n\leq N\) is itself a center. If \(n>N\), then $$ \left|\frac1n-0\right|=\frac1n\leq\frac1{N+1}<\varepsilon. $$ The point \(0\) is also a center. Hence \(F\) is an \(\varepsilon\)-net for \(A\). Since this construction works for every \(\varepsilon>0\), \(A\) is totally bounded.
Consequences and a Product Construction
Total boundedness implies ordinary boundedness, but the converse can fail, as the discrete example shows. For completeness, the implication follows directly from a single finite net. Choose a \(1\)-net \(F=\{z_1,\ldots,z_m\}\) for a nonempty totally bounded set \(A\). Fix \(a_0\in A\). For each \(a\in A\), choose \(z_j\in F\) with \(d(a,z_j)<1\). Then $$ d(a,a_0)\leq d(a,z_j)+d(z_j,a_0)<1+\max_{1\leq i\leq m}d(z_i,a_0). $$ The maximum is finite, so all points of \(A\) lie within one fixed finite distance of \(a_0\). This proves boundedness. The empty set is bounded as well.
Finite products retain the finite-net property when equipped with the maximum metric. This is useful because the maximum metric measures the largest coordinate error, so a net in each coordinate immediately gives a net for pairs.
Proof. Fix \(\varepsilon>0\). Choose a finite \(\varepsilon\)-net \(F\) for \(X\) and a finite \(\varepsilon\)-net \(G\) for \(Y\). The set \(F\times G\) is finite. For any \((x,y)\in X\times Y\), choose \(u\in F\) with \(d_X(x,u)<\varepsilon\) and \(v\in G\) with \(d_Y(y,v)<\varepsilon\). Then $$ d_{\max}((x,y),(u,v)) =\max\{d_X(x,u),d_Y(y,v)\}<\varepsilon. $$ Thus \(F\times G\) is an \(\varepsilon\)-net for \(X\times Y\). Since this works for every \(\varepsilon>0\), the product is totally bounded. \(\square\)
Worked Example: A Square Has a Finite Net
Give \([0,1]\times[0,1]\) the maximum metric. For any \(\varepsilon>0\), choose a positive integer \(N\) with \(1/N<\varepsilon\), and define $$ F=\left\{\left(\frac{i}{N},\frac{j}{N}\right):0\leq i,j\leq N\right\}. $$ This is finite. Given \((x,y)\in[0,1]^2\), choose grid coordinates \(i/N\) and \(j/N\) within \(1/N\) of \(x\) and \(y\), respectively. Such choices exist because consecutive grid points are \(1/N\) apart and the grid includes both endpoints. Then $$ d_{\max}\left((x,y),\left(\frac{i}{N},\frac{j}{N}\right)\right) =\max\left\{\left|x-\frac{i}{N}\right|,\left|y-\frac{j}{N}\right|\right\} \leq\frac1N<\varepsilon. $$ So the grid is a finite \(\varepsilon\)-net. The coordinate construction is the concrete version of the finite-product theorem.
What the Finite-Net Viewpoint Does—and Does Not—Say
Total boundedness is a scale-by-scale condition: for each requested accuracy there must be a finite cover, but the number of centers can grow as the accuracy becomes finer. It is not enough to find one finite cover at one radius. Nor is it enough to know that every point lies within some finite distance of a fixed center. The infinite discrete example satisfies the latter boundedness condition but fails the former.
The sequential characterization is especially useful when the object under study is a sequence. In a totally bounded space, it produces a Cauchy subsequence, not necessarily a convergent one. To conclude convergence inside the space, an additional property may be needed. Conversely, a separated sequence is an efficient way to disprove total boundedness. These distinctions will matter when finite coverings are combined with other metric-space properties in the study of compactness.
To prove total boundedness, begin with an arbitrary \(\varepsilon>0\) and construct a finite \(\varepsilon\)-net.
In intervals and products, divide each coordinate into finitely many pieces whose diameters are smaller than the desired accuracy.
If no finite cover works at some radius, choose each new point outside the balls around all earlier points.
Total boundedness guarantees a Cauchy subsequence; convergence in the space is a separate conclusion.
Check Your Understanding
Use the definitions, constructions, and proofs in this tutorial to answer the following questions.
- What does it mean for a finite set to be an \(\varepsilon\)-net for \(A\)?
- In the sequence characterization, why can one select infinitely many indices in one ball at each stage?
- Why does the sequence constructed when total boundedness fails have no Cauchy subsequence?
- How does the infinite discrete metric space show that boundedness does not imply total boundedness?
- Why does taking products of finite \(\varepsilon\)-nets work for the maximum metric?
- Does total boundedness alone guarantee that every sequence has a convergent subsequence in the space? Explain which conclusion the theorem actually provides.