Embedding Division Ring into Quotient Ring of Cauchy Sequences

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

Let $\CC$ be the ring of Cauchy sequences over $R$

Let $\NN = \set {\sequence {x_n}: \ds \lim_{n \mathop \to \infty} x_n = 0}$

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

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

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

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 Sequence Converges to Constant in Normed Division Ring then:
 * $\ds \lim_{n \mathop \to \infty} {a - b} = a - b$

Hence $a-b = 0$.

The result follows.

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

Then: