What Completeness Requires
A convergent sequence is always Cauchy, but the converse need not hold in a given metric space. Completeness names the property that makes the converse true: a Cauchy sequence must have a limit that belongs to the space. The location of the limit matters. A sequence can converge in a larger space without converging in the metric space under consideration.
Throughout, \((X,d)\) denotes a metric space, equipped with the topology generated by its open balls. A sequence \((x_n)\) in \(X\) 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 distinction is that convergence compares terms with a point of \(X\), while the Cauchy condition compares terms with each other.
The definition includes the requirement that the limit belongs to \(X\). The empty metric space is complete vacuously: it has no sequences indexed by \(\mathbb{N}\), so it has no Cauchy sequence that fails to converge. In the examples below, \(\mathbb{Q}\) and intervals are given the metrics obtained by restricting the standard metric \(d(x,y)=|x-y|\) on \(\mathbb{R}\).
A Limit in the Ambient Space
We will compare convergence inside a subset of \(\mathbb{R}\) with convergence in \(\mathbb{R}\). The following basic fact explains why an ambient limit outside the subset rules out convergence within it.
Proof. Suppose \(x_n\to u\) and \(x_n\to v\). For every \(n\), the triangle inequality gives \(|u-v|\le |u-x_n|+|x_n-v|\). Given any \(\varepsilon>0\), convergence gives an index after which both terms on the right are less than \(\varepsilon/2\). Thus \(|u-v|<\varepsilon\) for every \(\varepsilon>0\). A nonnegative real number smaller than every positive real number must be zero, so \(|u-v|=0\) and \(u=v\). \(\square\)
If \(A\subseteq\mathbb{R}\) has the restricted metric, convergence of a sequence in \(A\) to \(c\in A\) is exactly convergence to \(c\) in \(\mathbb{R}\): both statements use the same distances \(|x_n-c|\). Consequently, if the sequence converges in \(\mathbb{R}\) to \(u\notin A\), it cannot also converge to a point of \(A\), by the proposition.
The Real Line and the Rationals
The real line with its standard metric is complete. This is the standard completeness theorem for \(\mathbb{R}\), a consequence of the least-upper-bound property of the real numbers; its proof is given in the later treatment of completeness of the real line. We use the theorem here without reproving it. Its conclusion concerns every Cauchy sequence in \(\mathbb{R}\), not just sequences with an obvious formula for a limit.
Worked Example: A Cauchy Sequence in the Real Line
Consider \(x_n=\frac{n}{n+1}\) in \(\mathbb{R}\). For every \(n\ge1\), \(1-x_n=\frac{1}{n+1}\), so \(x_n\to1\), and \(1\in\mathbb{R}\). To check the Cauchy condition directly, if \(m,n\ge N\), then
Given \(\varepsilon>0\), choose \(N\) so that \(2/(N+1)<\varepsilon\). Then \(|x_m-x_n|<\varepsilon\) for all \(m,n\ge N\). For a numerical check, \(x_1=1/2\) and \(x_2=2/3\); the formula for the distance to the limit gives \(1-x_1=1/2\) and \(1-x_2=1/3\), as required. This sequence illustrates completeness, while the theorem above establishes completeness for every Cauchy sequence in \(\mathbb{R}\).
The rationals behave differently. The metric on \(\mathbb{Q}\) uses the same distance formula, but a real limit of a rational sequence need not be rational.
Worked Example: A Cauchy Sequence in the Rationals
Let \(\alpha=\sqrt{2}\), and for each \(n\in\mathbb{N}\) define the rational number \(x_n=\lfloor 10^n\alpha\rfloor/10^n\), where \(\lfloor t\rfloor\) is the greatest integer not exceeding \(t\). The defining property of the floor gives \(0\le\alpha-x_n<10^{-n}\). In particular, \(x_n\to\alpha\) in \(\mathbb{R}\).
Each \(x_n\) is rational. For \(m,n\ge N\), the triangle inequality and the floor bounds give \(|x_m-x_n|\le|\alpha-x_m|+|\alpha-x_n|<10^{-m}+10^{-n}\le2\cdot10^{-N}\). Given \(\varepsilon>0\), choose \(N\) with \(2\cdot10^{-N}<\varepsilon\). This proves that \((x_n)\) is Cauchy, also when regarded as a sequence in \(\mathbb{Q}\), because the restricted metric has the same distances.
The first terms can be checked from the bounds. Since \(1.4^2=1.96<2<2.25=1.5^2\), \(x_1=1.4\). Since \(1.41^2=1.9881<2<2.0164=1.42^2\), \(x_2=1.41\). Also \(1.414^2=1.999396<2<2.002225=1.415^2\), so \(x_3=1.414\). These values agree with the definition by decimal truncation.
To see that \(\alpha\notin\mathbb{Q}\), suppose \(\sqrt{2}=p/q\) for integers \(p,q\) with \(q\ne0\), written in lowest terms. Then \(p^2=2q^2\), so \(p^2\) is even and hence \(p\) is even. Write \(p=2k\). Substitution gives \(q^2=2k^2\), so \(q\) is even as well, contradicting that \(p/q\) was in lowest terms. Thus \(\alpha\) is irrational.
If \((x_n)\) converged to a rational number \(r\) in \(\mathbb{Q}\), it would also converge to \(r\) in \(\mathbb{R}\). But it converges in \(\mathbb{R}\) to \(\sqrt{2}\), so uniqueness of real limits would give \(r=\sqrt{2}\), contradicting irrationality. Therefore this Cauchy sequence has no limit in \(\mathbb{Q}\), and \(\mathbb{Q}\) is incomplete.
Proper Open Intervals
Every nonempty proper open interval in \(\mathbb{R}\), with its restricted Euclidean metric, is incomplete. A sequence can approach an excluded finite endpoint while remaining inside the interval. The sequence is Cauchy because its terms become arbitrarily close to one another, but its only real limit is not in the interval.
Proof. First consider \((a,b)\), where \(a,b\in\mathbb{R}\) and \(a<b\). Define \(x_n=a+(b-a)/(n+1)\). Since \(n\ge1\), \(0<(b-a)/(n+1)\le(b-a)/2<b-a\), so \(a<x_n<b\). For \(m,n\ge N\), \(|x_m-x_n|\le(b-a)/(N+1)\), because both fractions \(1/(m+1)\) and \(1/(n+1)\) lie between \(0\) and \(1/(N+1)\). Choosing \(N\) large enough that \((b-a)/(N+1)<\varepsilon\) proves that the sequence is Cauchy. Also \(0<x_n-a=(b-a)/(n+1)\to0\), so \(x_n\to a\) in \(\mathbb{R}\). Since \(a\notin(a,b)\), uniqueness of limits shows that \((x_n)\) has no limit in \((a,b)\).
For \((a,\infty)\), set \(x_n=a+1/n\). Every term satisfies \(x_n>a\). If \(m,n\ge N\), then \(|x_m-x_n|\le1/N\), so the sequence is Cauchy. Since \(0<x_n-a=1/n\to0\), its real limit is \(a\notin(a,\infty)\); uniqueness again rules out convergence within the interval.
For \((-\infty,b)\), set \(x_n=b-1/n\). Every term satisfies \(x_n<b\), and for \(m,n\ge N\), \(|x_m-x_n|\le1/N\), so it is Cauchy. Since \(0<b-x_n=1/n\to0\), it converges in \(\mathbb{R}\) to \(b\notin(-\infty,b)\). It therefore has no limit in that interval. Each of the three spaces has a Cauchy sequence without a limit in the space, proving the theorem. \(\square\)
Worked Example: A Bounded Open Interval
In \((0,1)\), take \(x_n=1/(n+1)\). For \(n\ge1\), \(0<1/(n+1)\le1/2<1\), so every term belongs to the interval. If \(m,n\ge N\), then \(|x_m-x_n|\le1/(N+1)\); choosing \(N\) with \(1/(N+1)<\varepsilon\) proves the Cauchy condition. In \(\mathbb{R}\), \(x_n\to0\), since \(|x_n-0|=1/(n+1)\to0\). The point \(0\) is not in \((0,1)\), and uniqueness of limits means there is no limit in \((0,1)\). For a direct numerical check, \(x_1=1/2\), \(x_2=1/3\), and their distance is \(|1/2-1/3|=1/6\); their distances from \(0\) are \(1/2\) and \(1/3\).
Worked Example: An Unbounded Open Interval
In \((2,\infty)\), let \(x_n=2+1/n\). For every \(n\ge1\), \(x_n>2\), so the terms are in the interval. For \(m,n\ge N\), \(|x_m-x_n|=|1/m-1/n|\le1/N\), which proves the Cauchy condition by taking \(N\) with \(1/N<\varepsilon\). In \(\mathbb{R}\), \(|x_n-2|=1/n\to0\), but \(2\notin(2,\infty)\). Thus the sequence does not converge in the interval. As a check, \(x_1=3\), \(x_2=2.5\), and \(|x_1-x_2|=0.5\); their distances from the excluded endpoint are \(1\) and \(0.5\).
The classifications are therefore different: \(\mathbb{R}\) is complete by the standard completeness theorem for the real numbers; \(\mathbb{Q}\) is incomplete; and every proper open interval in \(\mathbb{R}\) is incomplete with its restricted Euclidean metric. The examples use sequences approaching missing real numbers, but the defining test is always internal: does every Cauchy sequence converge to a point of the space?
A Common Pitfall
Common mistake. One might write, “In \((0,1)\), \(1/(n+1)\to0\), so the sequence converges in \((0,1)\).” This confuses convergence in the ambient real line with convergence in the interval. The calculation \(|1/(n+1)-0|=1/(n+1)\to0\) proves convergence to \(0\) in \(\mathbb{R}\), but \(0\notin(0,1)\). Convergence in \((0,1)\) would require a limit \(c\in(0,1)\); uniqueness of limits in \(\mathbb{R}\) would force \(c=0\), which is impossible. The sequence is Cauchy in \((0,1)\) and has no limit there.
Check Your Understanding
For each question, keep track of which metric space contains the sequence and whether the proposed limit belongs to that space.
- State the definition of a complete metric space, including where the limit must lie.
- Why does the decimal-truncation sequence for \(\sqrt{2}\) show that \(\mathbb{Q}\) is incomplete even though it converges in \(\mathbb{R}\)?
- For \((3,7)\), give a Cauchy sequence in the interval whose real limit is outside it, and verify its terms lie strictly between the endpoints.
- Does the sequence \(2+1/n\) converge in \((2,\infty)\)? Distinguish its real limit from convergence in the interval.
- Why does completeness of \(\mathbb{R}\) not imply completeness of every subset with the restricted metric?