The Completeness Problem
Starting from a metric space \(X\), the Cauchy-sequence construction adds points represented by Cauchy sequences in \(X\). The metric between two such points is the limit of the distances between corresponding terms. The remaining task is to prove that this new space has no missing limits: every Cauchy sequence of its points must converge to a point in the constructed space.
We recall the construction to fix notation. Two Cauchy sequences \((x_j)\) and \((y_j)\) in a metric space \((X,d)\) are equivalent when \(d(x_j,y_j)\to0\). Write \([x_j]\) for the equivalence class of \((x_j)\), and write \(\widehat X\) for the set of these classes. The metric verified for this construction is \(\widehat d([x_j],[y_j])=\lim_{j\to\infty}d(x_j,y_j)\). The embedding \(i:X\to\widehat X\) sends \(x\) to the class of the constant sequence \(x,x,\ldots\). We use that this formula gives a well-defined metric, as established in the verification of the completion metric.
If \(X=\varnothing\), then there are no sequences in \(X\), so \(\widehat X=\varnothing\), and completeness holds vacuously. For the proof below, assume \(X\ne\varnothing\).
The Embedded Space Is Dense
The proof will choose points of \(X\) that approximate selected classes. Here “dense” means that every open ball in \(\widehat X\) around a class contains an embedded point \(i(x)\).
Proof. Write \(\xi=[x_j]\). Since \((x_j)\) is Cauchy, there is \(N\in\mathbb{N}\) such that \(d(x_j,x_N)<\varepsilon/2\) for every \(j\ge N\). By the metric formula, \(\widehat d(\xi,i(x_N))=\lim_{j\to\infty}d(x_j,x_N)\). Every term with \(j\ge N\) is less than \(\varepsilon/2\), so the limit is at most \(\varepsilon/2\). Since \(\varepsilon/2<\varepsilon\), the required point is \(x_N\). \(\square\)
This lemma is the density of the embedded copy \(i(X)\) in \(\widehat X\). Notice what it does and does not say: the approximating point may depend on both the class and the requested error, and the error can be made arbitrarily small. It does not say that an arbitrary term of an arbitrary representative is a good approximation.
Completeness of the Constructed Space
Proof. Let \((\xi_n)\) be a Cauchy sequence in \(\widehat X\). We first choose a subsequence whose tail distances have explicit bounds. For each \(k\in\mathbb{N}\), the Cauchy property gives an index \(N_k\) such that \(\widehat d(\xi_n,\xi_m)<2^{-k}\) whenever \(n,m\ge N_k\). Choose indices \(n_k\) strictly increasing with \(n_k\ge N_k\). Then, whenever \(l\ge k\), \(\widehat d(\xi_{n_k},\xi_{n_l})<2^{-k}\).
By the density lemma, choose \(a_k\in X\) with \(\widehat d(\xi_{n_k},i(a_k))<2^{-k}\). These choices approximate the selected classes by errors that tend to zero. For \(l\ge k\), the triangle inequality gives \[ d(a_k,a_l) =\widehat d(i(a_k),i(a_l)) \le \widehat d(i(a_k),\xi_{n_k}) +\widehat d(\xi_{n_k},\xi_{n_l}) +\widehat d(\xi_{n_l},i(a_l)) <2^{-k}+2^{-k}+2^{-l} \le 3\cdot2^{-k}. \] Here \(i\) preserves distances, since the distance between the two constant sequences \(i(a_k)\) and \(i(a_l)\) is \(d(a_k,a_l)\). Given any \(\varepsilon>0\), choose \(k\) large enough that \(3\cdot2^{-k}<\varepsilon\). The displayed bound then applies to every \(l\ge k\); reversing the order of the indices handles any pair of indices at least \(k\). Thus \((a_k)\) is Cauchy in \(X\).
It therefore defines a point \(\xi=[a_k]\in\widehat X\). We show first that the selected subsequence \((\xi_{n_k})\) converges to \(\xi\). Fix \(k\) and let \(m\ge k\). By the triangle inequality, \[ \widehat d(\xi_{n_k},\xi) \le \widehat d(\xi_{n_k},\xi_{n_m}) +\widehat d(\xi_{n_m},i(a_m)) +\widehat d(i(a_m),\xi). \] The first term is less than \(2^{-k}\), and the second is less than \(2^{-m}\). For the third term, its value is \(\widehat d(i(a_m),\xi)=\lim_{l\to\infty}d(a_m,a_l)\). For every \(l\ge m\), the bound already proved gives \(d(a_m,a_l)<3\cdot2^{-m}\), so the limit is at most \(3\cdot2^{-m}\). Hence \[ \widehat d(\xi_{n_k},\xi)<2^{-k}+4\cdot2^{-m} \qquad(m\ge k). \] The left side does not depend on \(m\). Letting \(m\to\infty\) yields \(\widehat d(\xi_{n_k},\xi)\le2^{-k}\). As \(k\to\infty\), this proves \(\xi_{n_k}\to\xi\).
Finally, the entire sequence converges to \(\xi\). Let \(\varepsilon>0\). Choose \(K\) with \(2^{-K}<\varepsilon/2\), and choose \(N\) such that \(\widehat d(\xi_n,\xi_m)<\varepsilon/2\) whenever \(n,m\ge N\). Take \(k\ge K\) with \(n_k\ge N\). For every \(n\ge N\), both \(n\) and \(n_k\) are at least \(N\), and the subsequence estimate gives \[ \widehat d(\xi_n,\xi) \le \widehat d(\xi_n,\xi_{n_k})+\widehat d(\xi_{n_k},\xi) <\varepsilon/2+2^{-k} \le\varepsilon/2+2^{-K} <\varepsilon. \] Thus \(\xi_n\to\xi\) in \(\widehat X\), proving completeness. \(\square\)
Worked Examples of the Construction
Worked Example: Approximating a Class in the Rational Metric Space
Take \(X=\mathbb{Q}\) with the usual metric \(d(x,y)=|x-y|\), and let \(\xi=[x_j]\), where \(x_j=1/(j+1)\). This sequence is Cauchy: for \(j,l\ge N\), both terms lie between \(0\) and \(1/(N+1)\), so \(|x_j-x_l|\le1/(N+1)\), which tends to zero as \(N\to\infty\).
The embedded point \(i(x_N)\) has distance \[ \widehat d(\xi,i(x_N)) =\lim_{j\to\infty}\left|\frac1{j+1}-\frac1{N+1}\right| =\frac1{N+1}. \] For example, choosing \(N=9\) gives \(x_N=1/10\) and distance \(1/10\). Since \(1/10<1/8\), this is an approximation within error \(1/8\). Direct substitution in the distance formula gives the same value: the expression inside the limit tends to \(|0-1/10|=1/10\).
Worked Example: A Sequence of Classes in the Open Interval
Let \(X=(0,1)\) with the usual metric. For \(n,j\in\mathbb{N}\), set \(x^{(n)}_j=1/(n+2)+1/(j+2)\), and let \(\xi_n=[x^{(n)}_j]\). Every term belongs to \(X\): its smallest possible value is positive, and its largest is \(1/3+1/3=2/3<1\). For fixed \(n\), the sequence is Cauchy because \[ |x^{(n)}_j-x^{(n)}_l| =\left|\frac1{j+2}-\frac1{l+2}\right| \le\frac1{j+2}+\frac1{l+2}, \] whose right side tends to zero as \(j,l\to\infty\).
For \(n,m\in\mathbb{N}\), corresponding terms differ by \[ |x^{(n)}_j-x^{(m)}_j| =\left|\frac1{n+2}-\frac1{m+2}\right|. \] This expression is independent of \(j\), so \(\widehat d(\xi_n,\xi_m)=|1/(n+2)-1/(m+2)|\). The sequence of classes is Cauchy: for \(n,m\ge N\), this distance is at most \(1/(n+2)+1/(m+2)\le2/(N+2)\), which tends to zero.
To see the representative choice in the proof explicitly, use the subsequence \(\xi_{2^k}\) and choose \(a_k=2/(2^k+2)\in X\). The representative term \(x^{(2^k)}_{2^k}\) is exactly \(a_k\). Its approximation error is \[ \widehat d(\xi_{2^k},i(a_k)) =\lim_{j\to\infty}\left|\frac1{2^k+2}+\frac1{j+2}-\frac2{2^k+2}\right| =\left|0-\frac1{2^k+2}\right| =\frac1{2^k+2}. \] This is less than \(2^{-k}\), since \(2^k+2>2^k\). The diagonal sequence \((a_k)\) is Cauchy: for \(l\ge k\), \[ |a_k-a_l| =\left|\frac2{2^k+2}-\frac2{2^l+2}\right| \le\frac2{2^k+2}+\frac2{2^l+2} \le\frac4{2^k+2}, \] and the last bound tends to zero. Thus \([a_k]\) is a class in \(\widehat X\). For a direct convergence check, the metric formula gives \[ \widehat d(\xi_n,[a_k]) =\lim_{j\to\infty}\left|\frac1{n+2}+\frac1{j+2}-\frac2{2^j+2}\right| =\frac1{n+2}. \] This tends to zero as \(n\to\infty\), so the sequence of classes converges to \([a_k]\).
Worked Example: The Two-Point Discrete Metric
Let \(X=\{p,q\}\) with \(d(p,q)=d(q,p)=1\) and \(d(p,p)=d(q,q)=0\). Every Cauchy sequence in \(X\) is eventually constant: apply the Cauchy condition with \(\varepsilon=1/2\). Beyond some index, any two terms have distance less than \(1/2\); because distinct points have distance \(1\), all those terms must be equal.
Two eventually constant sequences are equivalent exactly when their eventual values agree. Indeed, if the eventual values agree, corresponding terms eventually have distance zero; if they differ, corresponding terms eventually have distance \(1\), which does not tend to zero. Thus \(\widehat X\) has exactly the two classes \(i(p)\) and \(i(q)\), whose distance is \(1\). A Cauchy sequence of classes is eventually constant by the same \(\varepsilon=1/2\) argument, and therefore converges to its eventual value. This checks completeness directly, including the case where the original space is already complete.
A Common Pitfall: Choosing Terms Without Control
In the proof, it matters that \(a_k\) approximates \(\xi_{n_k}\) with a specified small error. Choosing the first term of each representative instead can produce a sequence that is not Cauchy, even if the classes are constant.
For a concrete example, use \(X=\mathbb{R}\) with the usual metric, and let every \(\xi_n\) be the class of the constant zero sequence. Represent \(\xi_n\) by \(x^{(n)}_1=0\) when \(n\) is even and \(x^{(n)}_1=1\) when \(n\) is odd, with \(x^{(n)}_j=0\) for every \(j\ge2\). Each representative is eventually zero, so it is equivalent to the constant zero sequence. The class sequence is therefore constant and Cauchy. But the careless choice \(a_n=x^{(n)}_1\) alternates between \(0\) and \(1\). For every \(n\), \(|a_n-a_{n+1}|=1\), so \((a_n)\) is not Cauchy. The correction is to choose terms far enough along the representatives to control the error. Here \(a_n=x^{(n)}_{n+1}=0\) works for every \(n\), and the approximation error is exactly zero.
The completeness argument has two distinct choices with two distinct purposes: first pass to a subsequence of classes with uniformly small tail distances; then approximate each selected class closely enough that the chosen points form a Cauchy sequence in \(X\). The resulting class is the limit of the subsequence, and the original Cauchy condition then gives convergence of the full sequence.
Check Your Understanding
Use the metric formula for \(\widehat X\) and the controlled-approximation argument to answer the following.
- In the density lemma, why does the Cauchy property of \((x_j)\) imply that \(\widehat d([x_j],i(x_N))\le\varepsilon/2\) for the chosen \(N\)?
- Why is it useful to choose \(n_k\) so that the bound \(2^{-k}\) holds for every later subsequence index \(l\ge k\), rather than only for the adjacent pair \(n_k,n_{k+1}\)?
- In the theorem proof, which two facts are used to prove that \((a_k)\) is Cauchy in \(X\)?
- For \(X=(0,1)\) and the classes in the second worked example, compute \(\widehat d(\xi_2,\xi_5)\) exactly.
- In the two-point discrete metric example, explain why the Cauchy condition with \(\varepsilon=1/2\) forces eventual constancy.