Embedding Division Ring into Quotient Ring of Cauchy Sequences

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

Let $\mathcal C$ be the ring of Cauchy sequences over $R$

Let $\mathcal N = \set {\sequence {x_n}: \displaystyle \lim_{n \mathop \to \infty} x_n = 0}$

Let $\norm {\, \cdot \,}:\mathcal C \, \big / \mathcal N \to \R_{\ge 0}$ be the norm on the quotient ring $\mathcal C \, \big / \mathcal N$ defined by:
 * $\displaystyle \forall \sequence {x_n} + \mathcal N: \norm {\sequence {x_n} + \mathcal N} = \lim_{n \mathop \to \infty} \norm {x_n}$

Let $\phi: R \to \mathcal C \, \big / \mathcal N$ be the mapping from $R$ to the quotient ring $\mathcal C \, \big / \mathcal N$ defined by:
 * $\forall a \in R: \map \phi a = \sequence {a, a, a, \dotsc} + \mathcal N$

where $\sequence {a, a, a, \dotsc} + \mathcal N$ is the left coset in $\mathcal C \, \big / \mathcal N$ that contains the constant sequence $\sequence {a, a, a, \dotsc}$.

Then:
 * $\phi$ is a distance-preserving ring monomorphism.

Proof
By the definition of a distance-preserving mapping and a ring monomorphism it has to be shown that:
 * $(1): \quad \phi$ is a homomorphism.
 * $(2): \quad \phi$ is an injection.
 * $(3): \quad \phi$ is distance-preserving.

$(1): \quad \phi$ is a homomorphism
By definition, $\phi$ is the composition of two mappings:
 * $\phi = q \circ \phi'$

where:
 * $\text{(a)}: \quad \phi': R \to \mathcal C$, defined by: $\forall a \in R, \map {\phi'} a = \sequence {a, a, a, \dotsc}$
 * $\text{(b)}: \quad q$ is the quotient mapping $q: \mathcal C \to \mathcal C \, \big / \mathcal N$ defined by: $\map q {\sequence {x_n} } = \sequence {x_n} + \mathcal N$

By Embedding Normed Division Ring into Ring of Cauchy Sequences, $\phi'$ is a ring monomorphism.

By Quotient Ring Epimorphism is Epimorphism, then $q$ is a ring epimorphism.

By Composition of Ring Homomorphisms is Ring Homomorphism then the composition $\phi = q \circ \phi'$ is a ring homomorphism

$(2): \quad \phi$ is an injection
Let $a, b \in R$.

Suppose $\map \phi a = \map \phi b$.

Then:

By Constant Rule for Convergent Sequences then:
 * $\displaystyle \lim_{n \mathop \to \infty} {a - b} = a - b$

Hence $a-b = 0$.

The result follows.

$(3): \quad \phi$ is distance-preserving
Let $a, b \in R$.

Then: