Quotient Ring of Cauchy Sequences is Normed Division Ring/Lemma 1

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

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

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

For all $\sequence {x_n} \in \mathcal {C}$, let $\eqclass {x_n}{}$ denote the left coset $\sequence {x_n} + \mathcal {N}$

Let $\norm {\, \cdot \,}_1:\mathcal {C} \,\big / \mathcal {N} \to \R_{\ge 0}$ be defined by:


 * $\displaystyle \forall \eqclass {x_n}{} \in \mathcal {C} \,\big / \mathcal {N}: \norm {\eqclass {x_n}{} }_1 = \lim_{n \to \infty} \norm{x_n}$

Then:
 * $\norm {\, \cdot \,}_1$ is well-defined.

That is,
 * $\forall \eqclass {x_n}{}: \lim_{n \to \infty} \norm{x_n}$ exists.
 * $\displaystyle \forall \eqclass {x_n}{}, \eqclass {y_n}{} \in \mathcal {C} \,\big / \mathcal {N}: \eqclass {x_n}{} = \eqclass {y_n}{} \implies \lim_{n \to \infty} \norm{x_n} = \lim_{n \to \infty} \norm{y_n}$

Proof
By Norm Sequence of Cauchy Sequence has Limit then:
 * for each $\eqclass {x_n}{}$ the $\displaystyle \lim_{n \to \infty} \norm{x_n}$ exists.

Suppose $\eqclass {x_n}{} = \eqclass {y_n}{}$.

By Left Cosets are Equal iff Difference in Subgroup then:
 * $\sequence {x_n} - \sequence {y_n} = \sequence {x_n - y_n} \in \mathcal{N}$

By Equivalent Cauchy Sequences have Equal Limits of Norm Sequences then:
 * $\displaystyle \lim_{n \to \infty} \norm{x_n} = \lim_{n \to \infty} \norm{y_n}$

Hence:
 * $\displaystyle \norm { \eqclass {x_n}{} }_1 = \lim_{n \to \infty} \norm{x_n} = \lim_{n \to \infty} \norm{y_n} = \norm { \eqclass {x_n}{} }_1$

The result follows.