Cauchy sequence ($\forall \epsilon >0\exists N\forall m,n\ge N:d(x_m,x_n)<\epsilon$)

A Cauchy sequence is a sequence whose sufficiently late terms are all close to one another. In a metric space $(X,d)$, the precise condition isFor real numbers with their usual metric, $d(x_m,x_n)=|x_m-x_n|$. The useful feature is that the definition compares computed terms to other computed terms: it does not require an already known limit of a sequence. The one index $N$ must control every pair of later indices, not merely successive terms. ^[ref].

For an explicit example, take $q_n=1-2^{-n}$ for $n\ge 1$. Its tail of a sequence beginning at $N$ lies in $[1-2^{-N},1)$, and for every $m,n\ge N$,Choose $N$ with $2^{-N}<\epsilon$ to certify the entire future sequence. For $\epsilon =0.1$, $N=4$ works because $2^{-4}=0.0625$. The limit of a sequence is one, but the pairwise bound establishes the Cauchy sequence property without first needing that answer. By contrast, $a_n=(-1{)}^n$ is not a Cauchy sequence: arbitrarily late even integers and odd integers give values separated by two. ^[ref].

Tiny successive changes do not provide this guarantee. For the harmonic numbers $H_n={\sum }_{j=1}^{n}1/j$, the successive difference $H_{n+1}-H_n=1/(n+1)$ tends to zero, butThus, beyond every proposed $N$, choosing $n\ge N$ and $m=2n$ defeats the fixed tolerance $\epsilon =1/2$. Even every fixed positive lag $h$ passes the weaker test: $0<H_{n+h}-H_n\le h/(n+1)\to 0$. The failing lag grows with $n$, as explained more generally by convergence from arbitrary variable-lag increments. ^[ref].

Every convergent sequence in a metric space is a Cauchy sequence: apply the triangle inequality to two late terms and their common limit of a sequence. The reverse implication requires completeness. It holds in the real numbers by completeness of the real numbers, but can fail in the rational numbers: the rational endpoints from the bisection method approaching $\sqrt{2}$ are a Cauchy sequence with no rational limit of a sequence. The existing construction in completeness of the real numbers shows how this happens. A Cauchy sequence is always a bounded sequence; in $\mathbb{R}$, boundedness gives a convergent subsequence by the Bolzano-Weierstrass theorem, and the pairwise condition forces the whole sequence to share its limit. ^[ref].

The tail diameter of a sequence in a metric space is $D_N=\sup \{d(x_m,x_n):m,n\ge N\}$, allowing $+\infty$. The bounds are nonincreasing as $N$ grows because fewer pairs remain. A sequence is a Cauchy sequence exactly when $D_N\to 0$. The supremum need not be attained: for $q_n=1-2^{-n}$, $D_N=2^{-N}$ even though no two finite terms achieve that difference. For $(-1{)}^n$, $D_N=2$ for every $N$. For the harmonic numbers, $D_N=+\infty$ for every $N$, because each remaining tail is unbounded.

If a sequence in a metric space has ${\sum }_{j=1}^{\infty }d(x_{j+1},x_j)<\infty$, the triangle inequality gives $d(x_m,x_n)\le {\sum }_{j=n}^{m-1}d(x_{j+1},x_j)$ for $m>n$. The tail of this convergent series tends to zero and bounds every later pair, proving the Cauchy sequence condition. In particular, an increment bound $d(x_{j+1},x_j)\le C{r}^j$ with $C>0$ and $0<r<1$ gives $d(x_m,x_n)\le C{r}^n/(1-r)$ by the geometric series. Unlike checking a small latest step, this bounds all remaining movement.

This sufficient condition is not necessary. The real sequence $x_n=(-1{)}^n/n$ converges to zero and is therefore a Cauchy sequence, although $|x_{n+1}-x_n|=1/(n+1)+1/n$ has a divergent sum. Cancellation allows convergence despite infinite accumulated absolute values of the successive changes.

A real sequence converges if and only if the displayed condition holds. A Cauchy sequence makes all future differences uniformly small, so the forward implication is immediate. Conversely, failure of the Cauchy sequence criterion produces a fixed tolerance and, for each $n$, a future term differing from $a_n$ by at least half that tolerance. Choosing the least such positive lag defines one function that contradicts the displayed condition. The completeness of the real numbers then gives convergence. Testing only every fixed integer lag is weaker: $a_n=\log n$ passes those tests but diverges.

A Cauchy subsequence is a subsequence that is a Cauchy sequence. Every subsequence of a Cauchy sequence is a Cauchy sequence, because its increasing indices eventually lie beyond any index required by the original pairwise bound. Every convergent subsequence is also a Cauchy sequence.

The converse about the original sequence fails: $a_n=(-1{)}^n$ has a constant subsequence at its even integers, hence a Cauchy sequence, while $a_n$ itself is not a Cauchy sequence. If the original sequence is already a Cauchy sequence and one subsequence converges to $L$, however, the whole sequence converges to $L$. Choose a late subsequence term close to $L$ and use the triangle inequality to compare it with every sufficiently late original term. ^[ref].

Every Cauchy sequence of real numbers converges to a real number. Explicitly, the conditionimplies the existence of $L\in \mathbb{R}$ with $x_n\to L$. The hypothesis compares approximations with one another, without requiring a candidate answer. This makes completeness of the real numbers an existence principle: sufficiently consistent approximations cannot approach a missing destination. ^[ref].

The equivalent least-upper-bound property says that every nonempty set $A\subseteq \mathbb{R}$ bounded above has a supremum $s\in \mathbb{R}$. Precisely, every $x\in A$ satisfies $x\le s$, and for every $\epsilon >0$ some $x\in A$ satisfies $s-\epsilon <x$. The second condition excludes any smaller upper bound. No largest element of $A$ is required.

The rational numbers lack this property. The set $A=\{q\in \mathbb{Q}:q\ge 0,q^2<2\}$ contains one and has upper bound two, but has no supremum in $\mathbb{Q}$. A proposed rational supremum $s$ would have to satisfy $s^2=2$: if $s^2<2$, a sufficiently small rational increase still lies in $A$; if $s^2>2$, a sufficiently small rational decrease remains an upper bound. The irrationality of the square root of two excludes equality. In $\mathbb{R}$, its supremum is $\sqrt{2}$. ^[ref].

Construct this square root by the bisection method, starting with rational bounds $[l_0,u_0]=[1,2]$. At each step square the midpoint and retain the half whose endpoints bracket two after squaring. The first four resulting intervals are $[1,3/2]$, $[5/4,3/2]$, $[11/8,3/2]$, and $[11/8,23/16]$. After $k$ steps, $u_k-l_k=2^{-k}$ and $l_k^2<2<u_k^2$. Both endpoint sequences are Cauchy sequences because every later endpoint remains in the current interval. Completeness of the real numbers gives a common limit of a sequence $L$; continuity of squaring gives $L^2=2$. The midpoint approximates $L$ with error at most $2^{-(k+1)}$. Thus exact rational arithmetic produces certified accuracy while completeness of the real numbers supplies the exact answer.

Richard Dedekind approached completeness of the real numbers by filling gaps in the rational numbers with Dedekind cuts. In the preface to his 1872 Continuity and Irrational Numbers, he traced the problem to teaching differential calculus in 1858 and wanting arithmetic foundations for limits. His construction makes a missing boundary, such as $\sqrt{2}$, into a number defined by the separation it produces among rational numbers. ^[ref].

A Dedekind cut can be represented by a nonempty proper subset $A$ of the rational numbers satisfying two conditions: if $q\in A$ and $r<q$ is rational, then $r\in A$; and $A$ has no maximum. The complementary set contains all the rational values on the other side of the boundary. Using the convention that the lower set never includes its boundary avoids two representations of a rational boundary.

The set $\{q\in \mathbb{Q}:q<0\}$ represents zero. The set $\{q\in \mathbb{Q}:q<0\text{ or }{q}^2<2\}$ represents $\sqrt{2}$, even though no rational number occupies that boundary by the irrationality of the square root of two. Thus a Dedekind cut describes an irrational number entirely using rational comparisons. Ordering these cuts by the subset relation and defining their arithmetic gives a construction of the real numbers. ^[ref].

The least-upper-bound property states that every nonempty set of real numbers that has an upper bound has a supremum in the real numbers. It is an order-theoretic formulation of the completeness of the real numbers.