From Cauchy Sequences to a Metric
A metric space can have Cauchy sequences that do not converge in the space. The Cauchy-sequence construction addresses this by treating such a sequence as a new point. For that idea to work, two details must be checked: different sequences that ought to represent the same point must give the same distances, and the resulting distances must satisfy all the metric axioms. We will also verify that the original space sits inside the construction without changing any distances, and that every constructed point can be approximated by points from the original space.
Let \((X,d)\) be a metric space. 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 \(n,m\geq N\). Let \(C(X)\) denote the set of all Cauchy sequences in \(X\). For \((x_n),(y_n)\in C(X)\), write \((x_n)\sim(y_n)\) when \(d(x_n,y_n)\to0\). The construction uses the set \(\widehat X=C(X)/{\sim}\) of equivalence classes, with the class of \((x_n)\) denoted by \([(x_n)]\).
Why the Formula Is Well-Defined
First, \(\sim\) really is an equivalence relation. It is reflexive because \(d(x_n,x_n)=0\) for every \(n\), and symmetric because \(d(x_n,y_n)=d(y_n,x_n)\). If \((x_n)\sim(y_n)\) and \((y_n)\sim(z_n)\), then \(d(x_n,z_n)\leq d(x_n,y_n)+d(y_n,z_n)\). Both terms on the right tend to zero, so \(d(x_n,z_n)\to0\), proving transitivity. Thus the classes in \(\widehat X\) are defined.
Proof. For any four points in a metric space, the triangle inequality gives \(d(x_n,y_n)\leq d(x_n,x_m)+d(x_m,y_m)+d(y_m,y_n)\). Interchanging the pairs \((x_n,y_n)\) and \((x_m,y_m)\) gives the reverse comparison. Together these inequalities imply \(|d(x_n,y_n)-d(x_m,y_m)|\leq d(x_n,x_m)+d(y_n,y_m)\). Given \(\varepsilon>0\), the Cauchy property gives an \(N\) such that each term on the right is less than \(\varepsilon/2\) whenever \(n,m\geq N\). The difference on the left is then less than \(\varepsilon\). Hence \((d(x_n,y_n))\) is a Cauchy sequence of real numbers. Completeness of \(\mathbb{R}\) implies it has a finite real limit.
Now suppose \((x_n)\sim(x'_n)\) and \((y_n)\sim(y'_n)\). The same triangle-inequality comparison gives \(|d(x_n,y_n)-d(x'_n,y'_n)|\leq d(x_n,x'_n)+d(y_n,y'_n)\). The right side tends to zero. Since both distance sequences converge, their limits are equal. Therefore the formula assigns the same value to any representatives of the same two classes. \(\square\)
This proves the two necessary facts: every proposed distance is a real number, and changing representatives does not change it. In particular, it is now legitimate to speak of \(\widehat d\) as a function on pairs of classes, rather than on pairs of sequences.
The Metric Axioms
Proof. Let \(\alpha=[(x_n)]\), \(\beta=[(y_n)]\), and \(\gamma=[(z_n)]\) be any three classes. Theorem 1 ensures all distances below are defined independently of representatives.
For each \(n\), \(d(x_n,y_n)\geq0\); taking limits gives \(\widehat d(\alpha,\beta)\geq0\). Symmetry follows from \(d(x_n,y_n)=d(y_n,x_n)\) for every \(n\), so \(\widehat d(\alpha,\beta)=\widehat d(\beta,\alpha)\).
For separation, \(\widehat d(\alpha,\beta)=0\) means exactly that \(\lim_n d(x_n,y_n)=0\). By the definition of \(\sim\), this is equivalent to \((x_n)\sim(y_n)\), which is equivalent to \(\alpha=\beta\). Conversely, if \(\alpha=\beta\), their representatives are equivalent, so that same limit is zero. Thus \(\widehat d(\alpha,\beta)=0\) if and only if \(\alpha=\beta\).
Finally, \(d(x_n,z_n)\leq d(x_n,y_n)+d(y_n,z_n)\) for every \(n\). Each of the three real sequences of distances converges by Theorem 1. Taking limits preserves this inequality and yields \(\widehat d(\alpha,\gamma)\leq\widehat d(\alpha,\beta)+\widehat d(\beta,\gamma)\). These are all the metric axioms, so \(\widehat d\) is a metric. If \(X\) is empty, then \(C(X)\) and \(\widehat X\) are empty, and the metric axioms hold vacuously. \(\square\)
Worked Example: Equivalent Sequences in \(\mathbb{Q}\)
Give \(\mathbb{Q}\) its usual metric \(d(p,q)=|p-q|\). Consider the three sequences \(u_n=0\), \(v_n=1/n\), and \(w_n=1\). The constant sequences \(u\) and \(w\) are Cauchy. To check that \(v\) is Cauchy, take \(m,n\geq N\). Since \(1/m\) and \(1/n\) both lie in \((0,1/N]\), \(|1/n-1/m|\leq1/N\). Given \(\varepsilon>0\), choose \(N\) with \(1/N<\varepsilon\); this proves \(v\) is Cauchy.
For every \(n\), \(d(u_n,v_n)=|0-1/n|=1/n\), which tends to zero. Thus \(u\sim v\), so they represent the same class. The class of \(w\) is at distance \(1\) from that class: using \(u\) gives \(\lim_n|0-1|=1\), while using the equivalent representative \(v\) gives \(\lim_n|1/n-1|=1\). For the second limit, \(n=1\) gives \(0\), and for every \(n\geq2\) the value is \(1-1/n\), which tends to \(1\).
Check: the distance computed from either representative is \(1\), as representative-independence requires. The example also shows why individual terms need not agree: \(d(u_1,w_1)=1\), whereas \(d(v_1,w_1)=0\). It is the limit of the distances, not a distance at one selected index, that defines the distance between classes.
Worked Example: A Discrete Metric Space
Let \(X=\{a,b\}\) and give it the discrete metric: \(d(x,y)=0\) when \(x=y\), and \(d(x,y)=1\) when \(x\ne y\). Every Cauchy sequence in \(X\) is eventually constant. Indeed, apply the Cauchy condition with \(\varepsilon=1/2\). Beyond some index \(N\), every pair of terms has distance less than \(1/2\). Since the only possible distances are \(0\) and \(1\), those terms must all be equal.
Consequently, every class has a constant sequence as a representative. The constant sequences at \(a\) and \(b\) are not equivalent, since their termwise distance is always \(1\). The distance in \(\widehat X\) between their classes is \(\lim_n d(a,b)=1\), while the distance from either class to itself is \(0\). Thus the construction has exactly two classes, with the same distances as \(X\).
Check: if a Cauchy sequence is eventually equal to \(a\), its distance from the constant sequence at \(a\) is eventually \(0\), so the sequences are equivalent. If it is eventually equal to \(b\), its distance from the constant sequence at \(b\) is eventually \(0\). This verifies that no additional classes occur.
The Original Space Inside the Construction
Proof. A constant sequence is Cauchy, so \(\iota\) is defined for each \(x\in X\). For \(x,y\in X\), the distance between their constant sequences is constant at every index: \(\widehat d(\iota(x),\iota(y))=\lim_n d(x,y)=d(x,y)\). Thus \(\iota\) preserves distances. If \(\iota(x)=\iota(y)\), distance preservation gives \(d(x,y)=0\), and the metric axiom on \(X\) gives \(x=y\); hence \(\iota\) is injective. A distance-preserving injective map is an isometric embedding.
To prove density, let \(\alpha=[(x_k)]\) be any point of \(\widehat X\), and let \(\varepsilon>0\). Since \((x_k)\) is Cauchy, choose \(N\) such that \(d(x_n,x_k)<\varepsilon/2\) whenever \(n,k\geq N\). Fix any \(n\geq N\). For every \(k\geq N\), \(d(x_n,x_k)<\varepsilon/2\), so taking the limit as \(k\to\infty\) gives \(\widehat d(\iota(x_n),\alpha)=\lim_{k\to\infty}d(x_n,x_k)\leq\varepsilon/2<\varepsilon\). Therefore the open ball of radius \(\varepsilon\) about \(\alpha\) contains \(\iota(x_n)\), a point of \(\iota(X)\). Every open ball about every point of \(\widehat X\) meets \(\iota(X)\), which is the neighborhood criterion for density. If \(X\) is empty, the density assertion is vacuous. \(\square\)
Worked Example: A Boundary Point from \((0,1)\)
Give \(X=(0,1)\) the usual metric \(d(x,y)=|x-y|\), and take \(x_n=1/(n+1)\). For \(m,n\geq N\), both terms lie between \(0\) and \(1/(N+1)\), so \(|x_n-x_m|\leq1/(N+1)\). Given \(\varepsilon>0\), choose \(N\) with \(1/(N+1)<\varepsilon\); hence \((x_n)\) is Cauchy in \(X\).
Let \(\alpha=[(x_n)]\). For any constant point \(c\in(0,1)\), \(\widehat d(\alpha,\iota(c))=\lim_n|1/(n+1)-c|=c\), since \(1/(n+1)\to0\). The value \(c\) is positive, so \(\alpha\ne\iota(c)\). This holds for every \(c\in X\); hence \(\alpha\) is not in the embedded image.
Nevertheless, \(\iota(x_n)\) approaches \(\alpha\). More explicitly, for \(n\geq N\), the Cauchy condition gives \(d(x_n,x_k)<\varepsilon/2\) for all \(k\geq N\), so \(\widehat d(\iota(x_n),\alpha)\leq\varepsilon/2<\varepsilon\). This illustrates density: the constructed point need not come from \(X\), but every one of its neighborhoods contains an embedded point. The example makes no claim about completeness of \(\widehat X\); that requires a separate argument.
Check: for \(c=1/3\), the distance is \(\lim_n|1/(n+1)-1/3|=1/3\), so this particular constant point does not represent \(\alpha\). The general calculation gives the same conclusion for every allowed \(c>0\).
A Common Mistake: Choosing One Index
A tempting shortcut is to define the distance between classes using the distance between representative terms at a fixed index, such as \(d(x_1,y_1)\). That number can change when a representative changes, even though the class stays the same. In the rational example, \(u\sim v\), but with \(w_n=1\) we computed \(d(u_1,w_1)=1\) and \(d(v_1,w_1)=0\). Thus the fixed-index rule would assign two different distances to the same pair of classes. The correction is to use the limit \(\lim_n d(x_n,y_n)\), and to prove both its existence and its independence from representatives before treating it as a metric.
The construction succeeds because the Cauchy condition controls how distances change at large indices, while the equivalence relation identifies sequences whose termwise distances vanish in the limit. The resulting metric separates precisely the classes that are not equivalent. Constant sequences preserve the original distances, and the Cauchy property ensures that the terms of every representing sequence give embedded points arbitrarily close to its class.
Check Your Understanding
Use the definitions and proofs above to answer the following questions.
- Why does the inequality \(|d(x_n,y_n)-d(x_m,y_m)|\leq d(x_n,x_m)+d(y_n,y_m)\) imply that the proposed distance limit exists?
- If \((x_n)\sim(x'_n)\) and \((y_n)\sim(y'_n)\), which inequality proves that the two representative choices give the same distance?
- Explain why \(\widehat d([(x_n)],[(y_n)])=0\) implies equality of the two equivalence classes.
- For the discrete metric on \(\{a,b\}\), why must every Cauchy sequence eventually be constant?
- In the proof of density, why is it enough to find, for each \(\varepsilon>0\), an embedded point at distance less than \(\varepsilon\) from an arbitrary class?