Normed Division Ring is Dense Subring of Completion

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

Let $\struct {R', \norm {\, \cdot \,}' }$ be a normed division ring completion of $\struct {R, \norm {\, \cdot \,} }$

Then:
 * $\struct {R, \norm {\, \cdot \,} }$ is isometrically isomorphic to a dense normed division subring of $\struct {R', \norm {\, \cdot \,}' }$.

Proof
By the definition of a normed division ring completion then:
 * $(1): \quad$ there exists a distance-preserving ring monomorphism $\phi: R \to R'$.
 * $(2): \quad \struct {R', \norm {\, \cdot \,}' }$ is a complete metric space.
 * $(3): \quad \phi \sqbrk R$ is a dense subspace in $\struct {R', \norm {\, \cdot \,}' }$.

By image of a ring homomorphism is a subring then $\phi \sqbrk R$ is a subring of $R'$ and $\phi: R \to \phi \sqbrk R$ is an isomorphism.

By Epimorphism from Division Ring to Ring then $\phi \sqbrk R$ is a division subring of $R'$.

By Division Subring of Normed Division Ring then $\struct {\phi \sqbrk R, \norm {\, \cdot \,}' }$ is a normed division subring of $\struct {R', \norm {\, \cdot \,}' }$.

By Distance-Preserving Surjection is Isometry of Metric Spaces then $\phi: R \to \phi \sqbrk R$ is an isometry.

The result follows.