Cauchy sequence in a metric space

Let $(X,d)$ be a metric space (Metric space: $d(x,y) = 0$ iff $x = y$, symmetry, and the triangle inequality; pseudometric and ultrametric) and let $(x_k)$ be a sequence in $X$, that is a function $\mathbb{N} \to X$ written $x_k := x(k)$ (Convergence of a sequence in a metric space: $x_k \to x$ iff $d(x_k, x) \to 0$ in $\mathbb{R}$, Sequences of reals: bounded, eventually, frequently, tails, subsequences). As everywhere in this library $\mathbb{N}$ contains $0$, so a sequence is indexed from $0$.

$(x_k)$ is a Cauchy sequence in $(X,d)$ if for every rational $\varepsilon > 0$ there is $K \in \mathbb{N}$ such that

$d(x_m, x_n) < \varepsilon \qquad \text{for all } m, n \ge K .$

Rational and real $\varepsilon$ agree here. The test is written with a rational $\varepsilon$ to match Limits and Cauchy sequences of reals and Convergence of a sequence in a metric space: $x_k \to x$ iff $d(x_k, x) \to 0$ in $\mathbb{R}$, and nothing is lost by using a real one: below any real $\eta > 0$ lies a positive rational (The rationals embed densely in the reals), and the index belonging to that rational serves for $\eta$. So a proof may establish Cauchyness by producing an index for every real $\varepsilon > 0$, and may use a Cauchy hypothesis at a real $\varepsilon$ by first passing to a rational below it. Both moves are used on this page and are always cited.

The condition is exactly $d(x_m,x_n) \to 0$ as $m$ and $n$ grow independently. The distances $d(x_m,x_n)$ are nonnegative reals (Nonnegativity of a metric is a consequence of the other axioms, not an axiom), and the displayed condition asks them to be uniformly small on a tail of the doubly indexed family. It is not the same as $d(x_{k+1}, x_k) \to 0$, which is a strictly weaker condition and is a standing source of error. The partial sums $H_n$ of the harmonic series separate the two: consecutive ones differ by $1/(n+1)$, which tends to $0$, while the sequence is unbounded, and an unbounded sequence of reals is not Cauchy (Every Cauchy sequence in a metric space is bounded).

Consistency with the real line. For $X = \mathbb{R}$ with the usual metric $d(x,y) = |x-y|$ (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ is a metric, its open balls are the intervals $(x-r, x+r)$, and it is unbounded) the condition above reads $|x_m - x_n| < \varepsilon$ for $m,n \ge K$, which is verbatim the definition of a Cauchy sequence of reals (Limits and Cauchy sequences of reals). So the notion introduced here restricts on $\mathbb{R}$ to the one already in use, and no ambiguity is created.

• A Cauchy sequence need not converge. The definition mentions no candidate limit, and that is the whole point of it: it is a condition on the sequence alone, testable without knowing where the sequence is going. Whether every Cauchy sequence converges is a property of the space, namely completeness (Complete metric space: every Cauchy sequence converges in the space), and it genuinely fails in some spaces (FALSE: every Cauchy sequence in a metric space converges).
• Cauchyness is a property of the metric, not of the topology. Two metrics on the same set may have exactly the same open sets and different Cauchy sequences (FALSE: two metrics inducing the same topology have the same Cauchy sequences). What does preserve Cauchy sequences is uniform equivalence (Topologically, uniformly and Lipschitz equivalent metrics on a set), and the reason is A uniformly continuous map sends Cauchy sequences to Cauchy sequences.
• Every subsequence of a Cauchy sequence is Cauchy, since a strictly increasing index map satisfies $n_j \ge j$ (A strictly increasing index map satisfies $n_k \ge k$), so the same $K$ works for the subsequence. Conversely a Cauchy sequence with one convergent subsequence already converges (A Cauchy sequence in a metric space with a convergent subsequence converges to that subsequence’s limit).