From Cauchy Sequences to Candidate Points
A complete metric space contains the limits of all its Cauchy sequences. To construct a complete space from one that may not be complete, we can use the Cauchy sequences themselves as candidate points. Some sequences should represent the same candidate point: the intended test is that the distance between their corresponding terms tends to zero. We will make this construction precise and define a distance between the resulting classes.
Let \((X,d)\) be a metric space, with its metric topology understood. A sequence \((x_n)\) in \(X\) is Cauchy if for every \(\varepsilon>0\) there is \(N\in\mathbb{N}\) such that \(d(x_n,x_m)<\varepsilon\) whenever \(n,m\ge N\). Write \(\mathcal{C}(X)\) for the set of all Cauchy sequences in \(X\). If \(X=\varnothing\), there are no sequences in \(X\), so \(\mathcal{C}(X)=\varnothing\); the construction below then gives an empty set of classes. Unless stated otherwise, examples use nonempty spaces.
The condition is about matching terms with the same index. It does not require either sequence to converge in \(X\). That distinction matters: the purpose of the construction is precisely to allow Cauchy sequences that have no limit in \(X\) to determine candidate points.
Why the Sequences Form Equivalence Classes
Proof. Let \((x_n),(y_n),(z_n)\in\mathcal{C}(X)\). For reflexivity, \(d(x_n,x_n)=0\) for every \(n\), so \(d(x_n,x_n)\to0\) and \((x_n)\sim(x_n)\). For symmetry, \(d(x_n,y_n)=d(y_n,x_n)\) for every \(n\), so if \(d(x_n,y_n)\to0\), then \(d(y_n,x_n)\to0\). For transitivity, suppose \((x_n)\sim(y_n)\) and \((y_n)\sim(z_n)\). The triangle inequality gives \(0\le d(x_n,z_n)\le d(x_n,y_n)+d(y_n,z_n)\) for every \(n\). Both terms on the right tend to zero. Given \(\varepsilon>0\), for all sufficiently large \(n\) each is less than \(\varepsilon/2\), so \(d(x_n,z_n)<\varepsilon\). Thus \(d(x_n,z_n)\to0\), which means \((x_n)\sim(z_n)\). \(\square\)
Equivalence classes collect precisely those sequences whose terms become arbitrarily close to one another at matching indices. The equivalence relation is the reason we can refer to a class without depending on which sequence in that class is chosen.
Worked Example: Cauchy Sequences in the Rationals
Give \(\mathbb{Q}\) the metric \(d(x,y)=|x-y|\), with its metric topology. Consider \(x_n=1/n\), \(u_n=2/n\), and \(y_n=1+1/n\), all of which are rational for \(n\in\mathbb{N}\). Each is Cauchy: for \(n,m\ge N\), both \(1/n\) and \(1/m\) lie between \(0\) and \(1/N\), so \(|1/n-1/m|\le1/N\); the same estimate applies to \(2/n,2/m\) with upper bound \(2/N\), and to \(1+1/n,1+1/m\) with upper bound \(1/N\). Choosing \(N\) large enough makes each bound smaller than any prescribed \(\varepsilon>0\).
Compare \(x_n\) and \(u_n\). Their termwise distance is \(|x_n-u_n|=|1/n-2/n|=1/n\), which tends to zero. Hence \((x_n)\sim(u_n)\), so they represent the same class. In contrast, \(|x_n-y_n|=|1/n-(1+1/n)|=1\) for every \(n\), so \((x_n)\not\sim(y_n)\).
As a check on the distinction, replace \(x_n\) by \(u_n\) when comparing with \(y_n\). For every \(n\ge2\), \(|u_n-y_n|=|2/n-(1+1/n)|=|1/n-1|=1-1/n\), which tends to \(1\), not to \(0\). Thus the class represented by \(u_n\) is still different from the class represented by \(y_n\), as it must be since \(u_n\sim x_n\) and \(x_n\not\sim y_n\).
Defining the Candidate Distance
For two classes, the natural candidate distance is the limit of the distances between corresponding terms. Before using that formula, we must establish that the limit exists and does not depend on the chosen representatives. The key estimate comes directly from the triangle inequality.
Proof. For any \(n,m\in\mathbb{N}\), the triangle inequality gives \(d(x_n,y_n)\le d(x_n,x_m)+d(x_m,y_m)+d(y_m,y_n)\). Reversing the roles of \(n\) and \(m\), and combining the two inequalities, yields \[ \bigl|d(x_n,y_n)-d(x_m,y_m)\bigr| \le d(x_n,x_m)+d(y_n,y_m). \] Since \((x_n)\) and \((y_n)\) are Cauchy, for any \(\varepsilon>0\) there is \(N\) such that both terms on the right are less than \(\varepsilon/2\) when \(n,m\ge N\). The left side is then less than \(\varepsilon\). Thus \(\bigl(d(x_n,y_n)\bigr)\) is a Cauchy sequence of real numbers. Completeness of \(\mathbb{R}\) implies that it has a real limit.
Now suppose \((x_n)\sim(x'_n)\) and \((y_n)\sim(y'_n)\). Applying the triangle inequality twice gives \[ \bigl|d(x_n,y_n)-d(x'_n,y'_n)\bigr| \le d(x_n,x'_n)+d(y_n,y'_n). \] The two terms on the right tend to zero by the assumed equivalences. Therefore the difference on the left tends to zero. Both distance sequences have limits by the first part of the proof, so those limits are equal. The proposed value is independent of the representatives. \(\square\)
At this stage, \(\widehat{X}\) is a set equipped with a well-defined candidate distance. The formula is the construction; the further verification that it satisfies all metric axioms is a separate step.
Worked Example: Candidate Points at the Ends of an Open Interval
Let \(X=(0,1)\) with the metric \(d(x,y)=|x-y|\) and its metric topology. Define \(x_n=1/(n+1)\) and \(y_n=2/(n+2)\). Both sequences lie in \(X\): their terms are positive, and \(1/(n+1)<1\) and \(2/(n+2)<1\) for every \(n\ge1\).
For \(n,m\ge N\), each \(x_n\) and \(x_m\) lies between \(0\) and \(1/(N+1)\), so \(|x_n-x_m|\le1/(N+1)\). Each \(y_n\) and \(y_m\) lies between \(0\) and \(2/(N+2)\), so \(|y_n-y_m|\le2/(N+2)\). These bounds tend to zero as \(N\) increases, proving that both sequences are Cauchy.
Their termwise distance is \[ |x_n-y_n| =\left|\frac{1}{n+1}-\frac{2}{n+2}\right| =\left|\frac{n+2-2n-2}{(n+1)(n+2)}\right| =\frac{n}{(n+1)(n+2)}. \] For \(n\ge1\), this is positive and is at most \(1/(n+1)\), since \(n/(n+2)\le1\). Therefore it tends to zero, and the sequences determine the same class.
Now let \(z_n=1/2\) for every \(n\). It is Cauchy because \(|z_n-z_m|=0\) for every \(n,m\). For the class represented by \((x_n)\) and the class represented by \((z_n)\), the candidate distance is \[ \widehat{d}([(x_n)],[(z_n)]) =\lim_{n\to\infty}\left|\frac{1}{n+1}-\frac12\right| =\lim_{n\to\infty}\frac{n-1}{2(n+1)} =\frac12. \] Here \(n\ge1\), so the absolute value was evaluated using \(1/(n+1)\le1/2\). The result is consistent with the sequence \(x_n\) approaching the missing endpoint \(0\) in the ambient real line, while \(z_n\) stays at \(1/2\); the construction itself only uses points and distances in \(X\).
Worked Example: A Discrete Metric Space
Let \(X=\{a,b\}\) with the discrete metric \(d(x,y)=0\) if \(x=y\) and \(d(x,y)=1\) if \(x\ne y\). Its metric topology is the discrete topology: for each point, the open ball of radius \(1/2\) contains that point alone. Consider any Cauchy sequence \((x_n)\) in \(X\). Apply the Cauchy condition with \(\varepsilon=1/2\). There is \(N\) such that \(d(x_n,x_m)<1/2\) for all \(n,m\ge N\). Since distances are either \(0\) or \(1\), this forces \(d(x_n,x_m)=0\), hence \(x_n=x_m\), for all \(n,m\ge N\). Every Cauchy sequence is therefore eventually constant.
The sequence that is constantly \(a\) and any sequence eventually equal to \(a\) are equivalent: their termwise distance is eventually \(0\), so its limit is \(0\). The same reasoning applies to \(b\). A sequence eventually equal to \(a\) cannot be equivalent to one eventually equal to \(b\), because their termwise distance is eventually \(1\), whose limit is not \(0\). Thus there are exactly two classes, represented by the constant sequences at \(a\) and \(b\).
The candidate distance between those classes is \(\lim_{n\to\infty}d(a,b)=\lim_{n\to\infty}1=1\). The distance from either class to itself is \(\lim_{n\to\infty}0=0\). These calculations check the construction directly in this finite example.
A Common Mistake: Requiring Eventual Equality
A tempting but incorrect replacement for \(\sim\) is to declare two sequences equivalent only when their terms are eventually equal. That condition is too restrictive: two sequences can represent the same candidate point even if their terms differ at every index.
For a concrete calculation, return to \(\mathbb{Q}\) with its usual metric and take \(x_n=1/n\) and \(u_n=2/n\). Both are Cauchy, as shown above, and \(x_n\ne u_n\) for every \(n\in\mathbb{N}\). Thus they are never equal, even eventually. But the correct test gives \(d(x_n,u_n)=|1/n-2/n|=1/n\to0\), so \((x_n)\sim(u_n)\). Declaring them different classes would split sequences that become arbitrarily close term by term, defeating the intended construction.
The construction has two essential safeguards: equivalence is based on the limit of the matching-term distances, and the distance between classes is defined only after proving that the corresponding real limit exists and is independent of representatives. These steps produce the raw material for a completion without assuming that the original space already contains the relevant limits.
Check Your Understanding
For each question, use the definition of Cauchy sequence, equivalence, or candidate distance as appropriate.
- In \(\mathbb{Q}\) with its usual metric, let \(x_n=3+1/n\) and \(y_n=3+2/n\). Are the sequences Cauchy, and are they equivalent?
- In \((0,1)\) with the usual metric, compute the candidate distance between the class of the constant sequence \(x_n=1/3\) and the class of \(y_n=1/(n+1)\).
- Explain where the triangle inequality is used to prove that \(d(x_n,y_n)\) has a limit.
- Why does the formula for \(\widehat{d}\) need a proof of independence from representatives?
- For the discrete metric on \(\{a,b\}\), can a Cauchy sequence contain infinitely many occurrences of both \(a\) and \(b\)? Justify your answer using \(\varepsilon=1/2\).