Cauchy Sequences and the Completeness of the Reals

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 $ϵ$ - $M$ 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 $R$ , the property that distinguishes the real line from the rationals.

Cauchy sequences

Informally, the terms are eventually all within $ϵ$ 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.

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.

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

Cauchy equals convergent in the reals

Proof. One direction is quick. If $x_{n} \to x$ , then given $ϵ > 0$ choose $M$ with $∣ x_{n} - x ∣ < ϵ /2$ for $n \geq M$ ; for $n, k \geq M$ ,

∣ x_{n} - x_{k} ∣ \leq ∣ x_{n} - x ∣ + ∣ x - x_{k} ∣ < \frac{ϵ}{2} + \frac{ϵ}{2} = ϵ,

so the sequence is Cauchy. This half holds in any ordered field and uses nothing special about $R$ .

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, since the tail suprema and infima require it. Set $a = lim sup x_{n}$ and $b = lim inf x_{n}$ . By the subsequence-extraction theorem there are subsequences with $x_{n_{i}} \to a$ and $x_{m_{i}} \to b$ . Given $ϵ > 0$ , take $M_{1}, M_{2}, M_{3}$ so that $∣ x_{n_{i}} - a ∣ < ϵ /3$ , $∣ x_{m_{i}} - b ∣ < ϵ /3$ , and $∣ x_{n} - x_{k} ∣ < ϵ /3$ past each, and set $M = max {M_{1}, M_{2}, M_{3}}$ . For $i \geq M$ both $n_{i}, m_{i} \geq M$ , so

∣ a - b ∣ \leq ∣ a - x_{n_{i}} ∣ + ∣ x_{n_{i}} - x_{m_{i}} ∣ + ∣ x_{m_{i}} - b ∣ < \frac{ϵ}{3} + \frac{ϵ}{3} + \frac{ϵ}{3} = ϵ .

Since $∣ a - b ∣ < ϵ$ for all $ϵ > 0$ , we get $a = b$ , and a bounded sequence with $lim inf = lim sup$ converges.³ $□$

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.

Completeness and the least-upper-bound property

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

The theorem says $R$ , equipped with the least-upper-bound property, is Cauchy-complete. The converse holds too: an ordered field in which $Q$ 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 $R$ either way — by cutting $Q$ into pieces with a least upper bound, or by completing $Q$ , adjoining just enough points to make every Cauchy sequence of rationals converge. The resulting field is the same.⁵

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 later in the subject.

The completeness statements are logically equivalent as axioms of the real line; each can be taken as primitive and the others derived.

The rationals are incomplete

A Cauchy sequence of rationals can fail to have a rational limit. The Newton iteration for $2$ from the monotone convergence lesson, started at $x_{1} = 2$ , produces rationals

2, \frac{3}{2}, \frac{17}{12}, \frac{577}{408}, \dots

that form a Cauchy sequence, yet its limit $2$ is not rational. Inside $Q$ the sequence bunches but converges to nothing; the point it approaches is missing. Completing $Q$ fills every such hole at once.

A Cauchy sequence of rationals aiming at an irrational value. In the rationals the target is missing; the real line supplies it.

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 and the sequence diverges. The failure of the Cauchy condition is explicit at the pair $(n, 2 n)$ : each of the $n$ summands is at least $\frac{1}{2 n}$ , so

x_{2 n} - x_{n} = i = n + 1 \sum 2 n \frac{1}{i} \geq n \cdot \frac{1}{2 n} = \frac{1}{2}

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

Harmonic partial sums. Consecutive steps shrink to zero, yet the sums drift upward without bound: doubling the index always adds at least one half.

Summary

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

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

Lebl, §2.4, Exercise 2.4.1. ↩
Lebl, §2.4, Proposition 2.4.4. ↩
Lebl, §2.4, Theorem 2.4.5. ↩
Lebl, §2.4, Exercise 2.4.2. ↩
Lebl, §2.4, Remark 2.4.6 (Cauchy-completeness and the construction of $R$ ); the equivalence with the least-upper-bound property is Exercise 2.4.3. ↩