From Cauchy Behavior to a Limit
A Cauchy sequence has terms that eventually lie arbitrarily close to one another. Convergence asks for more: those terms must approach a point that belongs to the space. The previous tutorial showed what can happen when that point is missing. Completeness rules out precisely this failure: every Cauchy sequence in a complete metric space converges to a point of the space.
That definition gives a broad guarantee, but it does not always make it easy to recognize a Cauchy sequence. One useful approach is to control the distance between consecutive terms. If the total of these successive distances is finite, then the triangle inequality controls the distance between any two later terms. This gives a practical route from a sequence’s construction to its limit.
The key distinction is that completeness does not say every sequence converges, nor does it say every sequence with small consecutive steps converges. It applies to Cauchy sequences. We begin with a useful sufficient condition for being Cauchy.
Summable Successive Distances
Let \((x_n)\) be a sequence in a metric space. Write \(a_n=d(x_n,x_{n+1})\) for the distance traveled at the \(n\)th step. If the series of these nonnegative distances has finite sum, then the distance between \(x_n\) and any later term is bounded by a tail of that series.
Proof. For integers \(m>n\), repeated application of the triangle inequality gives $$ d(x_n,x_m)\leq\sum_{j=n}^{m-1}d(x_j,x_{j+1}). $$ Because the series of successive distances converges, its tails tend to zero. Given \(\varepsilon>0\), choose \(N\) so that $$ \sum_{j=N}^{\infty}d(x_j,x_{j+1})<\varepsilon. $$ If \(m>n\geq N\), then $$ d(x_n,x_m)\leq\sum_{j=n}^{m-1}d(x_j,x_{j+1}) \leq\sum_{j=N}^{\infty}d(x_j,x_{j+1})<\varepsilon. $$ If \(n=m\), the distance is zero; if \(n>m\), interchange the two indices because \(d(x_n,x_m)=d(x_m,x_n)\). Thus \((x_n)\) is Cauchy. If \(X\) is complete, the definition of completeness now gives convergence in \(X\). \(\square\)
The estimate also records how much distance remains after any particular step. When \(X\) is complete and \(x_n\to x\), the triangle inequality gives, for \(m>n\), $$ d(x_n,x)\leq d(x_n,x_m)+d(x_m,x). $$ Letting \(m\) tend to infinity and using continuity of distance to a fixed point, established in Tutorial 660, yields $$ d(x_n,x)\leq\sum_{j=n}^{\infty}d(x_j,x_{j+1}). $$ Thus the same tail that proves the Cauchy property also bounds the error from the limit.
Worked Example: Finite Geometric Sums in the Real Line
For \(n\geq1\), define \(x_n=\sum_{k=1}^{n}2^{-k}\), a real number. The difference between consecutive terms is $$ d(x_n,x_{n+1})=|x_{n+1}-x_n|=2^{-(n+1)}. $$ The sum of all these step distances is finite: $$ \sum_{n=1}^{\infty}2^{-(n+1)}=\frac12. $$ The Summable Steps Give a Cauchy Sequence Theorem therefore shows that \((x_n)\) is Cauchy. Since \(\mathbb{R}\) with its usual metric is complete, it converges in \(\mathbb{R}\).
In this example, the terms can also be calculated directly. The finite geometric-sum identity gives \(x_n=1-2^{-n}\); indeed, multiplying \(\sum_{k=1}^n2^{-k}\) by \(2\) and subtracting the original sum leaves \(1-2^{-n}\). Hence \(x_n\to1\), and the remaining error is exactly $$ |x_n-1|=2^{-n}=\sum_{j=n}^{\infty}2^{-(j+1)}. $$ This agrees with the tail estimate from the theorem.
Worked Example: Geometrically Shrinking Steps in a Complete Space
Let \((X,d)\) be complete, and suppose a sequence \((x_n)\) satisfies $$ d(x_n,x_{n+1})\leq 4^{-n}\qquad(n\geq1). $$ For \(m>n\), the triangle inequality gives $$ d(x_n,x_m)\leq\sum_{j=n}^{m-1}4^{-j} \leq\sum_{j=n}^{\infty}4^{-j} =\frac{4}{3}\,4^{-n}. $$ The last expression tends to zero as \(n\to\infty\). Thus \((x_n)\) is Cauchy, and completeness gives a limit \(x\in X\).
The tail estimate gives a quantitative bound as well. For any \(m>n\), the finite-sum bound above applies. Letting \(m\to\infty\), using \(x_m\to x\) and continuity of distance to the fixed point \(x_n\), gives $$ d(x_n,x)\leq\frac{4}{3}\,4^{-n}. $$ For instance, after step \(n=3\), the bound is \(\frac{4}{3}\cdot4^{-3}=\frac{1}{48}\). No coordinates or algebraic structure on \(X\) are needed: the argument uses only the metric and completeness.
A Sequence-Based Characterization of Completeness
The summable-steps theorem has a converse when it is expressed as a property of the whole space. If every sequence whose successive distances have finite total must converge, then every Cauchy sequence must converge too. The connection is that a Cauchy sequence contains a subsequence whose successive distances decrease fast enough to be summable.
Proof. First suppose that \(X\) is complete. The Summable Steps Give a Cauchy Sequence Theorem shows that every sequence with finite total successive distance is Cauchy. Completeness then implies that it converges in \(X\).
Conversely, suppose that every sequence in \(X\) with summable successive distances converges in \(X\). Let \((x_n)\) be any Cauchy sequence in \(X\). We will select a subsequence whose successive distances have finite sum. For each positive integer \(k\), the Cauchy property gives an integer \(N_k\) such that $$ p,q\geq N_k\quad\Longrightarrow\quad d(x_p,x_q)<2^{-k}. $$ Set \(M_k=\max\{N_1,\ldots,N_k\}\). Choose indices \(n_k\) recursively so that \(n_1\geq M_1\) and, for every \(k\geq1\), $$ n_{k+1}\geq\max\{M_{k+1},n_k+1\}. $$ These indices are strictly increasing. Since \(n_k\geq M_k\geq N_k\) and \(n_{k+1}\geq n_k\geq N_k\), the choice of \(N_k\) ensures that $$ d(x_{n_k},x_{n_{k+1}})<2^{-k}. $$ Consequently, $$ \sum_{k=1}^{\infty}d(x_{n_k},x_{n_{k+1}}) \leq\sum_{k=1}^{\infty}2^{-k}=1. $$ By the assumed property of \(X\), the subsequence \((x_{n_k})\) converges to some point of \(X\). The A Cauchy Sequence with a Convergent Subsequence Converges Theorem from Tutorial 662 implies that the original Cauchy sequence converges to that same point. Since the original Cauchy sequence was arbitrary, \(X\) is complete. \(\square\)
This characterization is useful when a sequence is built step by step. Rather than estimate every pair of distant terms directly, one can bound the total distance traveled. The converse proof shows why the condition tests completeness: any Cauchy sequence, even one with no apparent summable sequence of steps, has a subsequence that does have summable steps.
Worked Example: Summable Steps Can Reveal Incompleteness
Consider \(X=(0,1)\) with the usual metric, and set \(x_n=2^{-n}\) for \(n\geq1\). Every term belongs to \(X\), and $$ d(x_n,x_{n+1})=2^{-n}-2^{-(n+1)}=2^{-(n+1)}. $$ Therefore $$ \sum_{n=1}^{\infty}d(x_n,x_{n+1}) =\sum_{n=1}^{\infty}2^{-(n+1)} =\frac12. $$ But \(x_n\to0\) in \(\mathbb{R}\), and \(0\notin X\). If \((x_n)\) converged to some \(x\in(0,1)\), it would also converge to \(x\) as a real sequence. Uniqueness of metric limits, established in Tutorial 661, would force \(x=0\), a contradiction. Hence the sequence does not converge in \(X\).
This sequence has summable successive distances but no limit in the space. It therefore witnesses the failure of the summable-step property in an incomplete space, consistent with the Summable-Step Characterization of Completeness.
What the Criterion Does—and Does Not—Say
Summable successive distances are a sufficient condition for convergence in a complete space, not a necessary condition. A convergent sequence can move back and forth enough that the total distance traveled is infinite. For example, in \(\mathbb{R}\), let \(x_n=(-1)^n/n\). Since \(|x_n|=1/n\to0\), this sequence converges to \(0\). But its successive distances satisfy $$ |x_{n+1}-x_n|=\frac{1}{n+1}+\frac{1}{n}. $$ The sum of these distances diverges: it is at least \(\sum_{n=1}^{\infty}1/n\), the divergent harmonic series. So convergence alone does not imply a finite total of successive distances.
A second point is that small individual steps, by themselves, are not enough. The condition \(d(x_n,x_{n+1})\to0\) does not guarantee that the sequence is Cauchy: it gives no control over the total distance across many steps. Summability supplies that missing control through tail estimates. Finally, convergence must always be checked in the stated space. The sequence in the open interval has a perfectly good real limit, but that limit is not a point of the interval.
Check Your Understanding
Use the definitions and results in this tutorial to answer the following questions.
- How does the triangle inequality bound \(d(x_n,x_m)\) using the successive distances when \(m>n\)?
- Why does convergence of \(\sum_{n=1}^{\infty}d(x_n,x_{n+1})\) imply that \((x_n)\) is Cauchy?
- In the summable-step characterization, why is a subsequence selected from an arbitrary Cauchy sequence?
- Why does the sequence \(x_n=2^{-n}\) fail to converge in \((0,1)\), even though it converges in \(\mathbb{R}\)?
- Give one reason that finite total successive distance is sufficient for convergence in a complete space but not necessary.