A completion of a metric space: a complete metric space together with an isometric embedding onto a dense subspace

Let $(X,d)$ be a metric space (Metric space: $d(x,y) = 0$ iff $x = y$, symmetry, and the triangle inequality; pseudometric and ultrametric). A completion of $(X,d)$ is a pair $\big((\widehat{X}, \widehat{d}),\, \iota\big)$ in which

• $(\widehat{X}, \widehat{d})$ is a complete metric space (Complete metric space: every Cauchy sequence converges in the space);
• $\iota : X \to \widehat{X}$ is an isometric embedding, that is $\widehat{d}(\iota(x), \iota(y)) = d(x,y)$ for all $x,y \in X$ (Isometry, isometric embedding, and the subspace metric on a subset);
• $\iota[X]$ is dense in $\widehat{X}$, that is $\overline{\iota[X]} = \widehat{X}$ (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space).

The embedding is part of the data, not an afterthought. A completion is a pair, and two completions of the same space are compared through their embeddings (A completion is unique up to a unique isometry fixing the original space, and uniformly continuous maps into complete spaces extend through it); the underlying complete space alone carries no information about where $X$ sits inside it. This is the same discipline as for the metric itself: a metric space is a pair, not a set.

$\iota$ identifies $X$ with a subspace of $\widehat{X}$, metric and topology included. An isometric embedding is injective, is an isometry onto its image, and carries the metric topology of $X$ onto the subspace topology of $\iota[X]$ (An isometric embedding is injective and carries the metric topology of the source onto the subspace topology of its image). So "$X$ is a dense subspace of a complete space" is an accurate reading of the definition, and the pedantic version with $\iota$ written out is used only where two completions have to be compared.

Existence and uniqueness are theorems, not part of the definition. That every metric space has a completion is Every metric space has a completion, constructed as the equivalence classes of its Cauchy sequences; that any two are isometric by a unique isometry commuting with the embeddings is A completion is unique up to a unique isometry fixing the original space, and uniformly continuous maps into complete spaces extend through it. Until the first of those is proved, the phrase the completion is not licensed, and it is not used here.

• A complete space is its own completion, with $\iota$ the identity: the identity is an isometric embedding and $X$ is dense in itself. Combined with uniqueness, this says that completing changes nothing when there was nothing to complete.
• Density is what pins the completion down. Without it, any complete space containing an isometric copy of $X$ would qualify, and $\mathbb{R}^2$ would be a "completion" of $\mathbb{Q}$. Density is exactly the demand that no room be added beyond what the missing limits require.
• What a completion adds is limits, not points of a different kind. Every point of $\widehat{X}$ is a limit of points of $\iota[X]$ (A point lies in the closure of $A$ iff some sequence in $A$ converges to it, and a set is closed iff it is sequentially closed), so the new points are precisely the destinations that the Cauchy sequences of $X$ were already aiming at (Cauchy sequence in a metric space). The construction in Every metric space has a completion, constructed as the equivalence classes of its Cauchy sequences makes that literal by taking the new points to be the Cauchy sequences themselves, up to the relation of having distance tending to $0$.