Complete metric space: every Cauchy sequence converges in the 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).

$(X,d)$ is complete if every Cauchy sequence in $(X,d)$ (Cauchy sequence in a metric space) converges to a point of $X$ (Convergence of a sequence in a metric space: $x_k\to x$ iff $d(x_k,x)\to 0$ in $\mathbb{R}$).

A subset $A\subseteq X$ is called complete when the metric subspace $(A,d_A)$ is complete (Isometry, isometric embedding, and the subspace metric on a subset); as always, the metric is part of the data, and $d_A$ is the restriction of $d$ to $A\times A$.

The limit is unique when it exists, since limits in a metric space are unique (A sequence in a metric space has at most one limit), so a complete space assigns to each of its Cauchy sequences one point and not a set of points.

Completeness is a property of the pair $(X,d)$, not of $X$ and not of the topology of $d$. Both quantifiers in the definition are about the metric: the Cauchy condition is stated with distances, and so is convergence. Two metrics on the same set can have the same open sets while exactly one of them is complete, which is the content of FALSE: completeness of a metric space is determined by its topology and its witness. Read the word complete as an abbreviation for complete with respect to this metric, always.

• Do not confuse this with Dedekind completeness. The least-upper-bound property of Complete ordered field (least-upper-bound property) is an order condition on an ordered field and is what defines $\mathbb{R}$; the condition here is a metric condition and makes sense in any metric space, with no order in sight. On $\mathbb{R}$ the first implies the second (The Cauchy criterion from the least-upper-bound property: in a complete ordered field every Cauchy sequence converges, $\mathbb{R}$ and ${\mathbb{R}}^n$ for $n\ge 1$ with the Euclidean metric are complete, componentwise from the Cauchy criterion in $\mathbb{R}$) and the two are not the same statement: the rationals with the usual metric are an ordered field that fails both, while there are complete metric spaces with no field structure at all.
• Every convergent sequence is Cauchy (Every convergent sequence in a metric space is Cauchy), so completeness is exactly the assertion that the two classes of sequences coincide. It is the converse inclusion that carries all the content.
• Three sources of completeness are proved on this page. The real line and ${\mathbb{R}}^n$ are complete ($\mathbb{R}$ and ${\mathbb{R}}^n$ for $n\ge 1$ with the Euclidean metric are complete, componentwise from the Cauchy criterion in $\mathbb{R}$); a closed subset of a complete space is complete (Closed subspaces of complete metric spaces are complete; the converse under countable choice); and, under Countable Choice, every metric space sits densely and isometrically inside a complete one (Every metric space has a completion, constructed as the equivalence classes of its Cauchy sequences).