The completion of $\mathbb{Q}$ under the usual metric is $\mathbb{R}$

Assume the Axiom of Countable Choice (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)). Regard $\mathbb{Q}$ (The rationals as equivalence classes of pairs of integers) as the subset $\widehat{\mathbb{Q}}$ of $\mathbb{R}$ that is the image of the canonical embedding $q\mapsto \widehat{q}$ (The rationals embed densely in the reals), carrying the subspace metric $d_{\mathbb{Q}}(p,q)=\mathrm{\mid }\widehat{p}-\widehat{q}\mathrm{\mid }$ inherited from the usual metric of $\mathbb{R}$ (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ is a metric, its open balls are the intervals $(x-r,x+r)$, and it is unbounded, Isometry, isometric embedding, and the subspace metric on a subset). Let $\iota :\mathbb{Q}\to \mathbb{R}$ be that embedding.

Then $((\mathbb{R},d_{\mathbb{R}}),\iota )$ is a completion of $(\mathbb{Q},d_{\mathbb{Q}})$ (A completion of a metric space: a complete metric space together with an isometric embedding onto a dense subspace). Consequently, by uniqueness of completions (A completion is unique up to a unique isometry fixing the original space, and uniformly continuous maps into complete spaces extend through it), for every completion $((\widehat{Y},\widehat{d}),j)$ of $(\mathbb{Q},d_{\mathbb{Q}})$ there is exactly one continuous $\phi :\mathbb{R}\to \widehat{Y}$ with $\phi \circ \iota =j$, and it is an isometry. In this sense $\mathbb{R}$ is the completion of $\mathbb{Q}$.

• This is the metric statement of what the construction pages did by hand. This library builds $\mathbb{R}$ out of Cauchy sequences of rationals on its own page, and the general construction of Every metric space has a completion, constructed as the equivalence classes of its Cauchy sequences follows the same plan for an arbitrary metric space: classes of Cauchy sequences, with the distance read off as a limit. The present item is the observation that, run on $\mathbb{Q}$, that plan reaches a space isometric to the $\mathbb{R}$ already in hand, and that no separate verification of "which complete space it is" is needed once uniqueness is available.
• Density is the whole of the third condition, and it is exactly the Archimedean fact. That every interval of positive length contains a rational is The rationals embed densely in the reals; without it $\mathbb{Q}$ would sit inside $\mathbb{R}$ as a complete-looking but small subspace, and $\mathbb{R}$ would not be a completion of it but merely a complete space containing it.
• The metric is the one written above and no other. Completions are taken with respect to a named metric (A completion of a metric space: a complete metric space together with an isometric embedding onto a dense subspace), and a different metric on $\mathbb{Q}$ would in general have a different completion; nothing here is a statement about $\mathbb{Q}$ as a bare field or as a bare topological space.