From Convergent Subsequences to Finite Nets
Sequential compactness says that every sequence has a convergent subsequence. Total boundedness, by contrast, asks for a finite collection of small balls that covers the whole space. The link between these ideas is separation: if no finite collection of balls of a fixed radius covers the space, one can keep choosing points outside the balls already chosen. The resulting sequence stays uniformly separated, so it cannot have a convergent subsequence.
Throughout, \((X,d)\) is a metric space with its metric topology. A sequence \((x_n)\) converges to \(x\in X\) if, for every \(r>0\), there is \(N\in\mathbb{N}\) such that \(d(x_n,x)<r\) whenever \(n\geq N\). A subsequence has the form \((x_{n_k})\), where \(n_1<n_2<\cdots\).
The strict inequality in the definition matters: a point at distance exactly \(\varepsilon\) from a center is not in that center’s open \(\varepsilon\)-ball. The proof below uses this convention and obtains distances greater than or equal to \(\varepsilon\) between the selected points.
The Main Result
Proof. If \(X=\varnothing\), the empty set is a finite \(\varepsilon\)-net for every \(\varepsilon>0\), so the conclusion holds. Now suppose \(X\ne\varnothing\), and fix an arbitrary \(\varepsilon>0\). We show that \(X\) has a finite \(\varepsilon\)-net. Suppose, to the contrary, that no finite set of centers gives such a net.
Choose \(x_1\in X\). Since the open ball \(B(x_1,\varepsilon)\) does not cover \(X\), choose \(x_2\in X\setminus B(x_1,\varepsilon)\). More generally, after choosing \(x_1,\ldots,x_n\), their finitely many open \(\varepsilon\)-balls do not cover \(X\), by our supposition. Choose \[ x_{n+1}\in X\setminus\bigcup_{i=1}^{n}B(x_i,\varepsilon). \] This recursive construction gives a sequence in \(X\). For every pair of indices \(i<j\), the choice of \(x_j\) places it outside \(B(x_i,\varepsilon)\). Hence \(d(x_i,x_j)\geq\varepsilon\).
Sequential compactness gives a convergent subsequence \((x_{n_k})\), with limit \(x\in X\). By convergence, there is \(K\) such that \(d(x_{n_k},x)<\varepsilon/3\) whenever \(k\geq K\). In particular, this holds for \(k=K\) and \(k=K+1\). The triangle inequality then gives \[ d(x_{n_K},x_{n_{K+1}}) \leq d(x_{n_K},x)+d(x,x_{n_{K+1}}) <\frac{\varepsilon}{3}+\frac{\varepsilon}{3} =\frac{2\varepsilon}{3} <\varepsilon. \] But \(n_K\ne n_{K+1}\), so the construction gave \(d(x_{n_K},x_{n_{K+1}})\geq\varepsilon\). This contradiction shows that a finite \(\varepsilon\)-net exists. Since \(\varepsilon>0\) was arbitrary, \(X\) is totally bounded. \(\square\)
The argument uses only the assumed existence of a convergent subsequence and the triangle inequality. In particular, it does not require a formula for the number of net points; it proves that some finite net exists at each chosen scale.
Separated Sets Must Be Finite
Proof. Suppose an \(\varepsilon\)-separated subset \(A\) were infinite. Choose distinct points \(a_1,a_2,\ldots\) from \(A\), one at a time; this is a recursive selection from the infinite set after finitely many points have been removed. For any distinct indices \(i,j\), the selected points are distinct elements of \(A\), so \(d(a_i,a_j)\geq\varepsilon\). Sequential compactness would give a convergent subsequence. As in the theorem’s proof, two sufficiently late, distinct terms of that subsequence would have distance less than \(2\varepsilon/3<\varepsilon\), a contradiction. Therefore \(A\) is finite. \(\square\)
The condition is about distinct elements of a subset, or, equivalently, about all distinct indices of a sequence. Merely saying that the distinct values of a sequence are separated is not enough if repeated terms are allowed: repetition can make consecutive terms have distance zero.
Worked Example: A Finite Net for the Unit Interval
Give an explicit \(2/5\)-net for \(I=[0,1]\), with the usual metric inherited from \(\mathbb{R}\).
Take \(F=\{0,1/2,1\}\subseteq I\). The open balls in \(\mathbb{R}\) of radius \(2/5\) about these centers are \[ B(0,2/5)=(-2/5,2/5),\qquad B(1/2,2/5)=(1/10,9/10),\qquad B(1,2/5)=(3/5,7/5). \] For any \(x\in[0,1/2]\), at least one of \(0\) and \(1/2\) is within distance \(1/4\): if \(x\leq 1/4\), use \(0\), since \(d(x,0)=x\leq1/4\); if \(x>1/4\), use \(1/2\), since \(d(x,1/2)=1/2-x<1/4\). For \(x\in[1/2,1]\), use \(1/2\) when \(x\leq3/4\), and use \(1\) when \(x>3/4\); in either case the distance is at most \(1/4\). Since \(1/4<2/5\), each \(x\in I\) lies in an open ball of radius \(2/5\) about a center in \(F\). Thus \(F\) is a \(2/5\)-net.
Check: \(F\) has three points, all in \(I\). The largest distance to a center needed in the preceding cases is at most \(1/4\), and \(1/4<2/5\), so the open-ball condition is satisfied, including at the endpoints \(0\) and \(1\).
Worked Example: How Large Can a Separated Subset Be?
In \(I=[0,1]\) with its usual metric, find the largest possible size of a \(1/4\)-separated subset.
Consider any six distinct points of \(I\), and list them in increasing order as \(y_1<y_2<y_3<y_4<y_5<y_6\). If they belong to a \(1/4\)-separated subset, then each consecutive pair has distance at least \(1/4\). Therefore \[ y_6-y_1 =(y_2-y_1)+(y_3-y_2)+(y_4-y_3)+(y_5-y_4)+(y_6-y_5) \geq 5\cdot\frac14=\frac54. \] But \(0\leq y_1<y_6\leq1\), so \(y_6-y_1\leq1\), contradicting \(5/4>1\). Thus such a subset has at most five points.
The set \(\{0,1/4,1/2,3/4,1\}\) has five points. For any two distinct points in it, their distance is \(k/4\) for some \(k\in\{1,2,3,4\}\), so the distance is at least \(1/4\). It is therefore \(1/4\)-separated, and the largest possible size is exactly five.
Check: The four consecutive gaps are each \(1/4\), and their sum is \(4\cdot(1/4)=1\), the length of \(I\). Six separated points would require five gaps with total length at least \(5/4\), which cannot fit in an interval of length \(1\).
Worked Example: An Infinite Discrete Metric Space
Let \(X=\mathbb{N}=\{1,2,\ldots\}\) with the discrete metric \(d(m,n)=0\) if \(m=n\), and \(d(m,n)=1\) if \(m\ne n\). Its metric topology is the discrete topology. Show that it has no finite \(1/2\)-net and is not sequentially compact.
For each \(m\in\mathbb{N}\), the open ball \(B(m,1/2)\) contains exactly \(m\): the distance from \(m\) to itself is \(0<1/2\), while the distance from \(m\) to any \(n\ne m\) is \(1\), which is not less than \(1/2\). Thus a finite collection of such balls covers only the finitely many centers in that collection. It cannot cover all of \(\mathbb{N}\), so no finite \(1/2\)-net exists.
Now consider the sequence \(x_n=n\). For any distinct indices \(i,j\), the terms \(i\) and \(j\) are distinct, so \(d(x_i,x_j)=1\). Every subsequence also has pairwise distances equal to \(1\). If a subsequence converged to some \(m\in\mathbb{N}\), then eventually its terms would lie in \(B(m,1/2)=\{m\}\), making all sufficiently late terms equal to \(m\). That contradicts the fact that the subsequence has distinct terms. Therefore this sequence has no convergent subsequence.
Check: The radius test uses the strict inequality \(d(m,n)<1/2\): it includes \(m\), since \(0<1/2\), and excludes every \(n\ne m\), since \(1\not<1/2\). This confirms both the singleton-ball calculation and the failure of a finite net.
A Common Mistake About Repeated Terms
Common pitfall. One might state that “a sequence whose distinct terms are at least \(\varepsilon\) apart must be finite” and apply it to any sequence. As written, that condition can be vacuous for a constant sequence: if \(x_n=p\) for every \(n\), there are no two distinct values to compare. The sequence is infinite, and the distance between terms at distinct indices is \(d(p,p)=0\), not at least \(\varepsilon\).
The correction is to make the separation condition apply to all distinct indices, \(d(x_i,x_j)\geq\varepsilon\) whenever \(i\ne j\), or to state the result for an \(\varepsilon\)-separated subset, whose elements are distinct by definition. The proof then has exactly the pairwise lower bound it needs. In the construction in the main theorem, each newly chosen point lies outside every earlier open \(\varepsilon\)-ball, which is why the required inequality holds for every pair of distinct indices.
Total boundedness is useful because it expresses a global finiteness property at every chosen scale: finitely many centers suffice to approximate all points within that scale. The theorem shows that sequential compactness prevents a space from containing indefinitely many points that remain a fixed positive distance apart. It establishes this implication directly, without asserting its converse.
Check Your Understanding
For each question, keep track of whether the distance condition is strict or non-strict.
- In the main proof, why does choosing a point outside \(\bigcup_{i=1}^{n}B(x_i,\varepsilon)\) give \(d(x_{n+1},x_i)\geq\varepsilon\) rather than \(d(x_{n+1},x_i)>\varepsilon\)?
- Why does a convergent subsequence have two distinct terms at distance less than \(2\varepsilon/3\) when its terms are eventually within \(\varepsilon/3\) of its limit?
- In \([0,1]\) with the usual metric, is \(\{0,1/2,1\}\) a \(1/4\)-net under the strict open-ball definition? Explain your answer.
- In the discrete metric, what is \(B(m,r)\) when \(0<r\leq1\), and why can finitely many such balls not cover \(\mathbb{N}\)?
- State the correct pairwise condition for a sequence to be \(\varepsilon\)-separated, and explain why a constant sequence does not satisfy it for \(\varepsilon>0\).