# Leigh.Samphier/Sandbox/Complete Normed Division Ring is Completion of Dense Subring

## Theorem

Let $\struct {R, \norm {\, \cdot \,} }$ be a complete normed division ring.

Let $\struct {S, \norm {\, \cdot \,}}$ be a dense normed division subring of $\struct {R, \norm {\, \cdot \,}}$.

Then:

$\struct {R, \norm {\, \cdot \,} }$ is a completion of $\struct {S, \norm {\, \cdot \,}}$ where the inclusion mapping $i : S \to R$ is the required distance-preserving ring monomorphism.

## Proof

Let $d$ be the metric induced by the norm $\norm {\, \cdot \,}$.

By definition of a complete normed division ring, the metric space $\struct{R, d}$ is complete.

$\map {i^\to} S = S$

Thus $\map {i^\to} S$ is dense in $\struct {R, \norm {\, \cdot \,} }$.

It follows that $\struct {R, \norm {\, \cdot \,} }$ is a completion of $\struct {S, \norm {\, \cdot \,}}$ by definition.

$\blacksquare$