Completion of an absolutely valued field

Given: The data and hypotheses of the statement.

Absolute values on a field: Let $F$ be a field. An absolute value on $F$ is a function $\mathrm{\mid }\cdot \mathrm{\mid }:F\to {\mathbb{R}}_{\ge 0}$ such that for all $x,y\in F$: $\mathrm{\mid }x\mathrm{\mid }=0⟺x=0,\mathrm{\mid }xy\mathrm{\mid }=\mathrm{\mid }x\mathrm{\mid }\mathrm{\mid }y\mathrm{\mid },\mathrm{\mid }x+y\mathrm{\mid }\le \mathrm{\mid }x\mathrm{\mid }+\mathrm{\mid }y\mathrm{\mid }\mathrm{.}$ It is nonarchimedean when it satisfies the stronger inequality $\mathrm{\mid }x+y\mathrm{\mid }\le \max \{\mathrm{\mid }x\mathrm{\mid },\mathrm{\mid }y\mathrm{\mid }\}$ for all $x,y\in F$. It is trivial when $\mathrm{\mid }x\mathrm{\mid }=1$ for every nonzero $x\in F$.

Every metric space has a completion, constructed as the equivalence classes of its Cauchy sequences: Let $(X,d)$ be a metric space (def-metric-space) and let $\mathcal{C}$ be the set of all Cauchy sequences in $X$ (def-cauchy-in-metric). Then: 1. For all $x=(x_n)$ and $y=(y_n)$ in $\mathcal{C}$ the real sequence $(d(x_n,y_n){)}_n$ converges, so $\rho (x,y):={\lim }_{n}d(x_n,y_n)$ is a single well-determined real (thm-cauchy-criterion-via-lub, lem-limit-unique). 2. The relation $x\sim y:⟺\rho (x,y)=0$ is an equivalence relation on $\mathcal{C}$. Write $\widehat{X}:=\mathcal{C}\mathrm{/}\text{\hspace{0.17em}⁣}\sim$ for the set of its classes and $[x]$ for the class of $x$. 3. $\widehat{d}([x],[y]):=\rho (x,y)$ does not depend on the chosen representatives, and $\widehat{d}$ is a metric on $\widehat{X}$. 4. The map $\iota :X\to \widehat{X}$ sending $p$ to the class of the constant sequence at $p$ is an isometric embedding with dense image (def-isometry-and-metric-embedding, def-metric-interior-closure-boundary). 5. $(\widehat{X},\widehat{d})$ is complete. Consequently $((\widehat{X},\widehat{d}),\iota )$ is a completion of $(X,d)$ (def-metric-completion), and every metric space has a completion. The notation is kept honest. A Cauchy sequence in $X$ need not converge in $X$, so no symbol ${\lim }_n{x}_n$ appears anywhere below; the only limits taken are limits of real sequences, and each is written only after its existence has been proved. The equivalence relation is defined and verified here rather than cited, as was done for def-integers, so that the construction is self-contained and its transitivity argument is visible at the point of use.

A completion is unique up to a unique isometry fixing the original space, and uniformly continuous maps into complete spaces extend through it: Let $(X,d)$ be a metric space (def-metric-space); completions of it exist (thm-metric-completion-exists, def-metric-completion). Then: 1. Universal property. Let $((\widehat{X},\widehat{d}),\iota )$ be a completion of $(X,d)$, let $(Z,d_Z)$ be a complete metric space (def-complete-metric-space) and let $f:X\to Z$ be uniformly continuous (def-metric-uniform-continuity). Then there is exactly one continuous $F:\widehat{X}\to Z$ with $F\circ \iota =f$, and that $F$ is uniformly continuous. 2. Uniqueness of the completion. Let $(({\widehat{X}}_1,{\widehat{d}}_1),{\iota }_1)$ and $(({\widehat{X}}_2,{\widehat{d}}_2),{\iota }_2)$ be completions of $(X,d)$. Then there is exactly one continuous $\phi :{\widehat{X}}_1\to {\widehat{X}}_2$ with $\phi \circ {\iota }_1={\iota }_2$, and that $\phi$ is an isometry (def-isometry-and-metric-embedding). So a completion is determined by $(X,d)$ up to a unique isometry compatible with the embeddings, which is what licenses the phrase the completion from here on.