From Finite Nets to a Convergent Subsequence
A sequence in a metric space may move around indefinitely, so finding a convergent subsequence requires more than knowing that the space is bounded. Total boundedness supplies finite approximations at every positive scale. By successively keeping infinitely many terms inside smaller and smaller balls, we can arrange that the retained terms form a Cauchy sequence. Completeness then supplies a limit in the space.
Throughout, \((X,d)\) is a metric space. A sequence \((x_n)\) converges to \(x\in X\) if for every \(\varepsilon>0\) there is \(N\in\mathbb{N}\) such that \(d(x_n,x)<\varepsilon\) whenever \(n\ge N\). It is Cauchy if for every \(\varepsilon>0\) there is \(N\in\mathbb{N}\) such that \(d(x_m,x_n)<\varepsilon\) whenever \(m,n\ge N\). The space is complete if every Cauchy sequence in \(X\) converges to a point of \(X\).
Total boundedness is a requirement at every scale: a net that is adequate for one radius need not be adequate for a smaller radius. The proof uses radii that decrease to zero. At each scale, a finite cover ensures that at least one ball contains terms from infinitely many of the indices retained so far.
The Finite Selection Step
Proof. Suppose instead that every \(J_i\) is finite. Then their finite union \(J_1\cup\cdots\cup J_r\) is finite. Since the sets cover \(J\), we would have \(J\subseteq J_1\cup\cdots\cup J_r\), making \(J\) finite. This contradicts the assumption that \(J\) is infinite. Therefore at least one \(J_i\) is infinite. \(\square\)
In the application, a finite net gives a finite cover of \(X\) by balls. The lemma says that among infinitely many retained terms, at least one of those balls contains terms with infinitely many retained indices. Repeating this selection at each finer scale is what produces the Cauchy estimate.
Completeness and Total Boundedness
Proof. If \(X=\varnothing\), there is no sequence in \(X\), so the assertion is vacuously true. Otherwise, let \((x_n)\) be any sequence in \(X\). We construct nested infinite sets of indices \(J_0\supseteq J_1\supseteq J_2\supseteq\cdots\), beginning with \(J_0=\mathbb{N}\). For each \(k\in\mathbb{N}\), total boundedness gives a finite cover of \(X\) by open balls of radius \(2^{-k}\). The indices in \(J_{k-1}\) are covered by the finitely many corresponding sets of indices. By the lemma, one ball contains \(x_n\) for infinitely many \(n\in J_{k-1}\). Let \(J_k\) be the set of those indices. Thus \(J_k\) is infinite, \(J_k\subseteq J_{k-1}\), and all terms with indices in \(J_k\) lie in one ball of radius \(2^{-k}\).
Choose indices \(n_k\in J_k\) successively so that \(n_1<n_2<\cdots\). This is possible because each \(J_k\) is infinite and therefore contains an index larger than the previously chosen one. The nested property gives \(n_j\in J_k\) whenever \(j\ge k\). Hence, for \(p,q\ge k\), both \(x_{n_p}\) and \(x_{n_q}\) lie in the same open ball of radius \(2^{-k}\). If its center is \(c\), the triangle inequality gives
To verify the Cauchy condition, let \(\varepsilon>0\). Choose \(k\) so large that \(2^{1-k}<\varepsilon\). For all \(p,q\ge k\), the displayed estimate then gives \(d(x_{n_p},x_{n_q})<\varepsilon\). Thus \((x_{n_k})\) is Cauchy. Since \(X\) is complete, there is \(x\in X\) such that \(x_{n_k}\to x\). The selected terms form a subsequence of the original sequence, so every sequence in \(X\) has a convergent subsequence in \(X\). \(\square\)
The two hypotheses do distinct jobs. Total boundedness provides the finite choices at each scale and thereby yields a Cauchy subsequence. Completeness turns that Cauchy subsequence into a convergent one whose limit belongs to \(X\). The proof does not require that the original sequence converge, nor does it identify a unique subsequence or limit.
Worked Examples
Worked Example: A Finite Metric Space
Let \(X=\{a,b,c\}\) with the discrete metric \(d(x,y)=0\) when \(x=y\) and \(d(x,y)=1\) when \(x\ne y\). Consider the sequence defined by \(x_{3k-2}=a\), \(x_{3k-1}=b\), and \(x_{3k}=c\) for \(k\in\mathbb{N}\).
First, \(X\) is totally bounded. For any \(\varepsilon>0\), the three points \(a,b,c\) form a finite \(\varepsilon\)-net: each point is at distance \(0<\varepsilon\) from itself. Next, \(X\) is complete. If \((y_n)\) is Cauchy, apply the Cauchy condition with \(\varepsilon=1\). There is \(N\) such that \(d(y_m,y_n)<1\) for all \(m,n\ge N\). Distinct points in this metric have distance \(1\), so \(y_m=y_n\) for all \(m,n\ge N\). The sequence is eventually constant and converges to that constant value in \(X\).
For the specified sequence, choose \(n_k=3k-2\). Then \(n_1=1\), \(n_2=4\), and \(n_3=7\), and the selected terms are \(x_1=a\), \(x_4=a\), and \(x_7=a\); in fact \(x_{n_k}=a\) for every \(k\). This subsequence converges to \(a\), since \(d(x_{n_k},a)=0\) for every \(k\).
Check: \(n_k=3k-2\) is strictly increasing because \(n_{k+1}-n_k=(3(k+1)-2)-(3k-2)=3>0\), so the selected terms really do form a subsequence.
Worked Example: The Unit Interval and a Sequence Approaching Zero
Give \(I=[0,1]\) the metric inherited from \(\mathbb{R}\). This space is complete: a Cauchy sequence in \(I\) converges in \(\mathbb{R}\), by completeness of the real numbers, and its limit remains in \(I\). Indeed, if \(x_n\in[0,1]\) and \(x_n\to x\in\mathbb{R}\), then \(x\ge0\) and \(x\le1\); either inequality failing would contradict convergence by taking a sufficiently small positive tolerance.
It is also totally bounded. Given \(\varepsilon>0\), choose \(m\in\mathbb{N}\) with \(1/m<\varepsilon\), and use the finite set of centers \(j/m\) for \(j=0,1,\ldots,m\). For \(x\in[0,1)\), let \(j\) be the greatest integer not exceeding \(mx\). Then \(0\le j\le m-1\) and \(0\le x-j/m<1/m<\varepsilon\). For \(x=1\), choose \(j=m\), giving distance \(0<\varepsilon\). Thus these centers form a finite \(\varepsilon\)-net.
Now set \(x_n=1/(n+1)\). One subsequence is \(x_{2k-1}=1/(2k)\), for \(k\in\mathbb{N}\). Its first three terms are \(x_1=1/2\), \(x_3=1/4\), and \(x_5=1/6\), as follows by substituting \(n=1,3,5\) into \(1/(n+1)\). Given \(\varepsilon>0\), choose \(K\in\mathbb{N}\) with \(1/(2K)<\varepsilon\). For every \(k\ge K\), \(0\le x_{2k-1}=1/(2k)\le1/(2K)<\varepsilon\). Hence this subsequence converges to \(0\in I\).
Check: Substitution of the proposed limit gives \(d(x_{2k-1},0)=|1/(2k)-0|=1/(2k)\), exactly the quantity bounded above by \(\varepsilon\).
Worked Example: A Sequence with Two Different Convergent Subsequences
Again use \(I=[0,1]\), which is complete and totally bounded as verified above. Define \(x_n=1/2+(-1)^n/(n+2)\). The terms lie in \(I\): for even \(n\), the added fraction is positive and at most \(1/4\), so \(1/2<x_n\le3/4\); for odd \(n\), it is negative with magnitude at most \(1/3\), so \(1/6\le x_n<1/2\).
Take the even indices \(n=2k\). Direct substitution gives \(x_{2k}=1/2+1/(2k+2)\). For example, \(x_2=1/2+1/4=3/4\), and \(x_4=1/2+1/6=2/3\). Given \(\varepsilon>0\), choose \(K\) with \(1/(2K+2)<\varepsilon\). For \(k\ge K\), \(d(x_{2k},1/2)=1/(2k+2)\le1/(2K+2)<\varepsilon\), so the even-indexed subsequence converges to \(1/2\).
The odd indices also give a convergent subsequence: substitution of \(n=2k-1\) gives \(x_{2k-1}=1/2-1/(2k+1)\). For \(k=1\), this is \(1/2-1/3=1/6\); for \(k=2\), it is \(1/2-1/5=3/10\). Its distance from \(1/2\) is \(1/(2k+1)\), which tends to zero by the same \(\varepsilon\)-criterion. Thus this sequence has two subsequences with the same limit, though the theorem only promises at least one convergent subsequence.
Check: The distance calculations use the actual formulas: \(|x_{2k}-1/2|=|1/(2k+2)|=1/(2k+2)\), while \(|x_{2k-1}-1/2|=|-1/(2k+1)|=1/(2k+1)\).
A Common Pitfall: One Scale Is Not Enough
A single finite net can locate infinitely many sequence terms in one ball, but that alone does not establish a Cauchy subsequence. For example, in \(I=[0,1]\), the ball of radius \(1/2\) centered at \(0\) contains both \(0\) and \(1/4\), whose distance is \(1/4\). Therefore, knowing that all selected terms lie in that ball only gives a distance bound of \(1\), not an arbitrarily small bound.
The correction is to repeat the selection at every scale. At stage \(k\), use radius \(2^{-k}\), and retain infinitely many indices inside one ball at that scale. The estimate \(d(x_{n_p},x_{n_q})<2^{1-k}\) then tends to zero as \(k\) increases. This is why the proof needs a nested sequence of infinite index sets rather than just one application of a finite net.
The theorem is useful because it converts a geometric condition on a metric space—finite nets at every scale—into a conclusion about arbitrary sequences. Its proof also separates two ideas that should not be conflated: the net construction gives a Cauchy subsequence, while completeness guarantees convergence inside the space.
Check Your Understanding
Use the finite-net selection argument and the definitions of completeness and total boundedness in your reasoning.
- In the theorem’s proof, why must one of finitely many balls contain terms at infinitely many indices from the current infinite set?
- Explain why \(n_j\in J_k\) whenever \(j\ge k\), and use this to obtain the distance estimate for two selected terms.
- Where does completeness enter the proof, and what conclusion has already been obtained before it is used?
- For the discrete metric on a finite set, explain why applying the Cauchy condition with \(\varepsilon=1\) forces a sequence to be eventually constant.
- Why does finding infinitely many sequence terms in one ball of a fixed radius not, by itself, prove that those terms form a Cauchy subsequence?