---
title: Cauchy Sequences and the Completeness of the Reals
module: Sequences and Series
moduleNumber: 2
lessonNumber: 4
order: 204
summary: >
  The Cauchy criterion tests convergence without knowing the limit: a sequence
  converges exactly when its terms eventually all lie within any tolerance of
  one another. Cauchy sequences are bounded, in the reals Cauchy and convergent
  are equivalent, and this completeness property is interchangeable with the
  least-upper-bound axiom — the single feature that separates the real line
  from the rationals.
topics: [Sequences and Series]
sources:
  - book: Lebl
    ref: "§2.4 Cauchy sequences"
  - book: Rosenlicht
    ref: "Ch. 2 — The real number system"
draft: false
---

Often a number is described by a sequence that approaches it, and one wants to
know the sequence converges before the limit is available to name — a
decimal expansion, a Newton iteration, or the partial sums of a series. The
$\epsilon$-$M$ definition of convergence names the limit and so cannot be used.
The **Cauchy criterion** removes it: it certifies convergence using only the
terms of the sequence and their distances from each other. That the criterion
is equivalent to convergence is the completeness of $\mathbb{R}$, the property
that distinguishes the real line from the rationals.

## Cauchy sequences

> **Definition (Cauchy sequence).** A sequence $\{x_n\}$ is a **Cauchy
> sequence** if for every $\epsilon > 0$ there exists $M \in \mathbb{N}$ such
> that
> $$
> |x_n - x_k| < \epsilon \qquad \text{for all } n \ge M \text{ and } k \ge M.
> $$

Informally, the terms are eventually all within $\epsilon$ of one another. The
critical feature is that $n$ and $k$ range independently past $M$: no matter how
far apart the two indices are chosen, the corresponding terms stay close. This
is strictly stronger than requiring consecutive terms to bunch.

$$
% caption: The Cauchy condition. Past index M every pair of terms differs by less
% than epsilon, so all later terms fit inside one band of width epsilon.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black] (0,0) -- (10.2,0) node[right, black] {$n$};
\draw[black] (0,0) -- (0,4.2);
% band of width epsilon covering the tail
\fill[acc!10] (4.6,2.05) rectangle (10.2,2.75);
\draw[acc!70, dashed] (4.6,2.05) -- (10.2,2.05);
\draw[acc!70, dashed] (4.6,2.75) -- (10.2,2.75);
\draw[black, dashed] (4.6,0) -- (4.6,4.2);
\node[black, anchor=south] at (4.6,4.2) {index $M$};
\node[acc, anchor=west] at (10.0,3.1) {band};
\draw[acc, <->] (10.05,2.05) -- (10.05,2.75);
% early scattered points
\foreach \n/\y in {1/0.8, 2/3.6, 3/1.5, 4/3.0}
  \fill[black] (\n*1.0,\y) circle (2.0pt);
% tail points inside band
\foreach \n/\y in {5/2.5, 6/2.2, 7/2.55, 8/2.35, 9/2.45, 10/2.4}
  \fill[black] (\n*1.0,\y) circle (2.2pt);
\end{tikzpicture}
$$

> **Worked example.** The sequence $\{1/n\}$ is Cauchy. Given $\epsilon > 0$,
> take $M > 2/\epsilon$. For $n, k \ge M$, each of $1/n$ and $1/k$ is below
> $\epsilon/2$, so
> $$
> \left| \tfrac{1}{n} - \tfrac{1}{k} \right| \le \tfrac{1}{n} + \tfrac{1}{k} < \tfrac{\epsilon}{2} + \tfrac{\epsilon}{2} = \epsilon.
> $$

> **Worked example.** The sequence $\left\{\frac{n^2-1}{n^2}\right\}$ is Cauchy,
> directly from the definition. Since $\frac{n^2-1}{n^2} = 1 - \frac{1}{n^2}$,
> for any $n, k$,
> $$
> \left| \frac{n^2-1}{n^2} - \frac{k^2-1}{k^2} \right| = \left| \frac{1}{k^2} - \frac{1}{n^2} \right| \le \frac{1}{k^2} + \frac{1}{n^2}.
> $$
> Given $\epsilon > 0$, take $M > \sqrt{2/\epsilon}$; then $n, k \ge M$ gives
> $\frac{1}{n^2} \le \frac{1}{M^2} < \frac{\epsilon}{2}$ and likewise for $k$,
> so the difference is below $\epsilon$.[^lebl-ex241]

> **Worked example.** The sequence $\{(-1)^n\}$ is not Cauchy. For any $M$, take
> an even $n \ge M$ and $k = n + 1$; then $|(-1)^n - (-1)^{n+1}| = |1 - (-1)| =
> 2$. No $\epsilon \le 2$ can be satisfied, so the sequence is not Cauchy.

The first step toward the main theorem is that the Cauchy property, like
convergence, forces boundedness.

> **Proposition.** Every Cauchy sequence is bounded.

> **Proof.** Apply the definition with $\epsilon = 1$: there is $M$ with
> $|x_n - x_M| < 1$ for all $n \ge M$. By the reverse triangle inequality,
> $|x_n| < 1 + |x_M|$ for those $n$. Bounding the finitely many earlier terms by
> their maximum,
> $$
> B = \max\bigl\{ |x_1|, \dots, |x_{M-1}|,\ 1 + |x_M| \bigr\}
> $$
> is a bound for the whole sequence.[^lebl-cbd] $\square$

## Cauchy equals convergent in the reals

> **Theorem (Completeness of $\mathbb{R}$).** A sequence of real numbers is
> Cauchy if and only if it converges.

> **Proof.** One direction is quick. If $x_n \to x$, then given $\epsilon > 0$
> choose $M$ with $|x_n - x| < \epsilon/2$ for $n \ge M$; for $n, k \ge M$,
> $$
> |x_n - x_k| \le |x_n - x| + |x - x_k| < \tfrac{\epsilon}{2} + \tfrac{\epsilon}{2} = \epsilon,
> $$
> so the sequence is Cauchy. This half holds in any ordered field and uses
> nothing special about $\mathbb{R}$.
>
> For the converse, suppose $\{x_n\}$ is Cauchy. It is bounded, so its limit
> superior and inferior exist; their existence uses the
> [least-upper-bound property](/real-analysis/foundations/ordered-fields-completeness),
> since the tail suprema and infima require it. Set $a = \limsup x_n$ and
> $b = \liminf x_n$. By the
> [subsequence-extraction theorem](/real-analysis/sequences-series/limsup-bolzano-weierstrass)
> there are subsequences with $x_{n_i} \to a$ and $x_{m_i} \to b$. Given
> $\epsilon > 0$, take $M_1, M_2, M_3$ so that $|x_{n_i} - a| < \epsilon/3$,
> $|x_{m_i} - b| < \epsilon/3$, and $|x_n - x_k| < \epsilon/3$ past each, and
> set $M = \max\{M_1, M_2, M_3\}$. For $i \ge M$ both $n_i, m_i \ge M$, so
> $$
> |a - b| \le |a - x_{n_i}| + |x_{n_i} - x_{m_i}| + |x_{m_i} - b| < \tfrac{\epsilon}{3} + \tfrac{\epsilon}{3} + \tfrac{\epsilon}{3} = \epsilon.
> $$
> Since $|a - b| < \epsilon$ for all $\epsilon > 0$, we get $a = b$, and a
> bounded sequence with $\liminf = \limsup$ converges.[^lebl-comp] $\square$

The theorem gives a limit-free test: to prove a sequence converges, show its
terms bunch, and the limit is guaranteed to exist even before it is identified.

> **Worked example.** Suppose successive gaps contract by a fixed factor: there
> is $C$ with $0 < C < 1$ and $|x_{n+1} - x_n| \le C\,|x_n - x_{n-1}|$ for all
> $n \ge 2$. Then $\{x_n\}$ is Cauchy, hence convergent. Induction gives
> $|x_{n+1} - x_n| \le C^{n-1}\,|x_2 - x_1|$. For $k > n$, chain the gaps and
> sum the geometric bound:
> $$
> |x_k - x_n| \le \sum_{i=n}^{k-1} |x_{i+1} - x_i| \le |x_2 - x_1| \sum_{i=n}^{k-1} C^{i-1} \le |x_2 - x_1|\, \frac{C^{n-1}}{1 - C}.
> $$
> The right side tends to $0$ as $n \to \infty$, so given $\epsilon > 0$ there
> is $M$ with $|x_2 - x_1|\,C^{M-1}/(1-C) < \epsilon$, and all pairs past $M$
> differ by less than $\epsilon$.[^lebl-ex242] Every contraction iteration
> converges by this argument; no candidate limit is ever named.

## Completeness and the least-upper-bound property

The equivalence just proved is often taken as the definition of completeness.

> **Definition (Cauchy-complete).** An ordered set is **Cauchy-complete** (or
> just **complete**) if every Cauchy sequence in it converges to a point of the
> set.

The theorem says $\mathbb{R}$, equipped with the least-upper-bound property, is
Cauchy-complete. The converse holds too: an ordered field in which $\mathbb{Q}$
is dense and every Cauchy sequence converges necessarily has the
least-upper-bound property. So the two formulations of completeness are
interchangeable, and one can build $\mathbb{R}$ either way — by cutting
$\mathbb{Q}$ into pieces with a least upper bound, or by "completing"
$\mathbb{Q}$, adjoining just enough points to make every Cauchy sequence of
rationals converge. The resulting field is the same.[^lebl-remark]

The Cauchy formulation has one decisive advantage: it never mentions order
beyond the distance $|x_n - x_k|$. It therefore transplants verbatim to any
setting with a notion of distance, and this is the definition of completeness
adopted for [metric spaces](/real-analysis/metric-spaces/convergence-completeness) later
in the subject.

$$
% caption: The completeness statements are logically equivalent as axioms of the
% real line; each can be taken as primitive and the others derived.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize,
  box/.style={draw, minimum width=34mm, minimum height=11mm, align=center, font=\footnotesize}]
\definecolor{acc}{HTML}{4A6FA5}
\node[box, draw=acc, text=acc] (lub) at (0,1.6) {least upper bound};
\node[box] (nest) at (5.6,1.6) {nested intervals};
\node[box] (cau) at (0,-1.6) {Cauchy complete};
\node[box] (bw) at (5.6,-1.6) {Bolzano Weierstrass};
\draw[<->, black, thick] (lub) -- (nest);
\draw[<->, black, thick] (lub) -- (cau);
\draw[<->, black, thick] (nest) -- (bw);
\draw[<->, black, thick] (cau) -- (bw);
\draw[<->, black, thick] (lub) -- (bw);
\end{tikzpicture}
$$

## The rationals are incomplete

A Cauchy sequence of rationals can fail to have a rational limit. The Newton iteration for $\sqrt{2}$ from the
[monotone convergence lesson](/real-analysis/sequences-series/limit-laws-monotone),
started at $x_1 = 2$, produces rationals

$$
2,\ \tfrac{3}{2},\ \tfrac{17}{12},\ \tfrac{577}{408},\ \dots
$$

that form a Cauchy sequence, yet its limit $\sqrt{2}$ is
[not rational](/real-analysis/foundations/ordered-fields-completeness). Inside
$\mathbb{Q}$ the sequence bunches but converges to nothing; the point it
approaches is missing. Completing $\mathbb{Q}$ fills every such hole at once.

$$
% caption: A Cauchy sequence of rationals aiming at an irrational value. In the
% rationals the target is missing; the real line supplies it.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black] (0,0) -- (10,0);
% tick at 1 and 2
\draw[black] (1,0.12) -- (1,-0.12) node[below, black] {$1$};
\draw[black] (9,0.12) -- (9,-0.12) node[below, black] {$2$};
% rational approximants closing in from both sides on the hole at ~6.3
\foreach \x in {2.2,3.6,4.6,5.3,5.75,6.0}
  \fill[black] (\x,0) circle (2.0pt);
\foreach \x in {8.4,7.6,7.1,6.8,6.6,6.5}
  \fill[black] (\x,0) circle (2.0pt);
% the missing limit as a hollow circle
\draw[acc, thick, fill=white] (6.32,0) circle (2.6pt);
\draw[acc, dashed] (6.32,0.7) -- (6.32,-0.7);
\node[acc, anchor=south] at (6.32,0.75) {limit is irrational};
\end{tikzpicture}
$$

## Cauchy is stronger than consecutive closeness

A frequent error is to conclude a sequence is Cauchy from $|x_{n+1} - x_n| \to
0$. That is weaker and does not suffice. The partial sums of the harmonic series
$x_n = \sum_{i=1}^n \frac{1}{i}$ satisfy $x_{n+1} - x_n = \frac{1}{n+1} \to 0$,
and in fact $\lim_{n\to\infty} |x_{n+j} - x_n| = 0$ for every fixed $j$, yet its
[partial sums are unbounded](/real-analysis/sequences-series/series-convergence)
and the sequence diverges. The failure of the Cauchy condition is explicit at
the pair $(n, 2n)$: each of the $n$ summands is at least $\frac{1}{2n}$, so

$$
x_{2n} - x_n = \sum_{i=n+1}^{2n} \frac{1}{i} \ge n \cdot \frac{1}{2n} = \frac{1}{2}
$$

for every $n$, and no $M$ works for $\epsilon = \tfrac{1}{2}$. The Cauchy
condition requires that $n$ and $k$ be arbitrarily far apart, not merely
adjacent; over a long stretch the small gaps accumulate.

$$
% caption: Harmonic partial sums. Consecutive steps shrink to zero, yet the sums
% drift upward without bound: doubling the index always adds at least one half.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black] (0,0) -- (10.6,0) node[right, black] {$n$};
\draw[black] (0,0) -- (0,3.6);
\foreach \n/\y in {1/0.90, 2/1.35, 3/1.65, 4/1.88, 5/2.06, 6/2.21, 7/2.33, 8/2.45, 9/2.55, 10/2.64, 11/2.72, 12/2.79}
  \fill[black] (\n*0.8,\y) circle (2.0pt);
% guides from two distant partial sums to a drift arrow at the right margin
\draw[black, dashed] (2.4,1.65) -- (10.3,1.65);
\draw[black, dashed] (9.6,2.79) -- (10.3,2.79);
\draw[acc, <->] (10.15,1.65) -- (10.15,2.79);
\node[acc, anchor=east] at (10.0,2.22) {drift};
\node[black, anchor=north] at (5.6,1.45) {steps shrink toward zero};
\end{tikzpicture}
$$

> **Contrast.** $|x_{n+1} - x_n| \to 0$ controls only neighboring terms.
> Cauchy controls all pairs $x_n, x_k$ with $n, k \ge M$ simultaneously. The
> harmonic partial sums satisfy the first and fail the second.

## Summary

| Statement | Holds in $\mathbb{Q}$? | Holds in $\mathbb{R}$? |
| --- | --- | --- |
| convergent $\Rightarrow$ Cauchy | yes | yes |
| Cauchy $\Rightarrow$ convergent | no | yes |
| least-upper-bound property | no | yes |
| $\lvert x_{n+1} - x_n \rvert \to 0 \Rightarrow$ Cauchy | no | no |

The Cauchy criterion tests convergence with no reference to the limit, and its
equivalence with convergence is precisely the completeness of $\mathbb{R}$ — the
same fact as the least-upper-bound property, the nested-interval property, and
Bolzano–Weierstrass. This is the property $\mathbb{Q}$ lacks.

[^lebl-cbd]: Lebl, §2.4, Proposition 2.4.4.
[^lebl-ex241]: Lebl, §2.4, Exercise 2.4.1.
[^lebl-ex242]: Lebl, §2.4, Exercise 2.4.2.
[^lebl-comp]: Lebl, §2.4, Theorem 2.4.5.
[^lebl-remark]: Lebl, §2.4, Remark 2.4.6 (Cauchy-completeness and the construction of $\mathbb{R}$); the equivalence with the least-upper-bound property is Exercise 2.4.3.