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:
 * $\displaystyle \forall \sequence {x_n} + \mathcal {N}: \norm {\sequence {x_n} + \mathcal {N} } = \lim_{n \to \infty} \norm{x_n}$

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:
 * $\quad \quad \quad \forall a \in R: \phi \paren {a} = (a,a,a,\dots) + \mathcal {N}$

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

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 of $\phi$, it is the composition of two mappings:
 * $\phi = q \circ \phi'$

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

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

By Quotient Mapping 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 $\phi \paren {a} = \phi \paren {b}$:

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

Hence $a-b = 0$.

The result follows.

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

Then: