Quotient of Cauchy Sequences is Metric Completion/Lemma 1

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

Then:
 * $\quad \mathcal {C} \,\big / \mathcal {N} = \tilde {\mathcal C}$

Proof
Let $\sequence{x_n}$ and $\sequence{y_n}$ be Cauchy sequences in $\mathcal {C}$ then:

Hence: $\sequence{x_n}$ and $\sequence{y_n}$ belong to the same equvalence class in $\mathcal {C} \,\big / \mathcal {N}$ $\sequence{x_n}$ and $\sequence{y_n}$ belong to the same equivalence class in $\tilde {\mathcal C}$

The result follows.