Quotient of Cauchy Sequences is Metric Completion

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

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

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

Let $\mathcal {N}$ be the set of null sequences.

Let $\mathcal {C} \,\big / \mathcal {N}$ be the quotient ring of Cauchy sequences of $\mathcal {C}$ by the maximal ideal $\mathcal {N}$.

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 $d'$ be the metric induced by $\struct {\mathcal {C} \,\big / \mathcal {N}, \norm {\, \cdot \,} }$

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:
 * $\quad \quad \quad \forall a \in R: \map \phi a = \sequence {a, a, a, \ldots} + \mathcal N$

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

Then:
 * $\struct {\mathcal {C} \,\big / \mathcal {N}, d'}$ is the metric completion of $\struct {R,d}$

and:
 * $\map \phi R$ is a dense subset of $\mathcal {C} \,\big / \mathcal {N}$

Proof
By the definition of the metric induced by a norm then:
 * a sequence $\sequence {x_n}$ is a Cauchy sequence in $\struct {R, \norm {\, \cdot \,} }$ $\sequence {x_n}$ is a Cauchy sequence in $\struct {R, d}$.

So $\mathcal {C}$ is the set of Cauchy sequences in $\struct {R, d}$.

Let $\sim$ be the equivalence relation on $\mathcal C$ defined by:


 * $\displaystyle \sequence{x_n} \sim \sequence{y_n} \iff \lim_{n \mathop \to \infty} d \paren{x_n, y_n} = 0$

Let $\tilde {\mathcal C} = \mathcal C / \sim$ denote the set of equivalence classes under $\sim$.

For $\sequence {x_n} \in \mathcal C$, let $\eqclass {x_n}{}$ denote the equivalence class containing $\sequence {x_n}$.

Lemma 1
Let $\tilde d: \tilde {\mathcal C} \times \tilde {\mathcal C} \to \R_{\ge 0}$ be the metric defined by:
 * $\displaystyle \map {\tilde d} {\eqclass {x_n}{}, \eqclass {y_n}{}} = \lim_{n \mathop \to \infty} \map d {x_n, y_n}$

By Completion of a Metric Space then:
 * $\struct{\tilde {\mathcal C},\tilde d}$ is the metric completion of $\struct {R,d}$.

Lemma 2
Let $\tilde \phi: R \to \tilde {\mathcal C}$ be the mapping defined by:
 * $\map {\tilde \phi} a = \eqclass {a, a, a, \dotsc}{}$

where $\eqclass {a, a, a, \dotsc}{}$ denotes the equivalence class containing the constant sequence $\sequence {a, a, a, \dotsc}$.

By Completion of a Metric Space then:
 * $\map {\tilde \phi} R$ is a dense subset of $\tilde {\mathcal C}$.

By Lemma 1 then:
 * $\forall \sequence {x_n} \in \mathcal C: \sequence {x_n} + \mathcal N = \eqclass {x_n}{}$

In particular:
 * $\forall a \in R: \sequence {a, a, a, \dotsc} + \mathcal N = \eqclass {a, a, a, \dotsc}{}$

That is:
 * $\forall a \in R: \map \phi a = \map {\tilde \phi} a$

Hence $\phi = \tilde \phi$.

The result follows.