Quotient of Cauchy Sequences is Metric Completion/Lemma 2

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

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 $\sim$ 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 $\sequence {x_n}$.

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

Then:
 * $\quad d' = \tilde d$

Proof
By Lemma 1 of Quotient of Cauchy Sequences is Metric Completion then $\mathcal {C} \,\big / \mathcal {N} = \tilde {\mathcal C}$

Let $\eqclass{x_n}{}$ and $\eqclass{x_n}{}$ be equivalence classes in $\mathcal {C} \,\big / \mathcal {N} = \tilde {\mathcal C}$ then:

The result follows.