Quotient Ring of Cauchy Sequences is Normed Division Ring

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

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

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

Let $\norm {\, \cdot \,}_1:\mathcal {C} \paren {R} \,\big / \mathcal {N} \to \R_{\ge 0}$ be defined by:
 * $\displaystyle \forall \sequence {x_n} + \mathcal {N}: \norm {\sequence {x_n} + \mathcal {N} }_1 = \lim_{n \to \infty} \norm{x_n}$

Then:
 * $\struct {\mathcal {C} \paren {R} \,\big / \mathcal {N}, \norm {\, \cdot \,}_1 }$ is a normed division ring.

Proof
By Quotient Ring of Cauchy Sequences is Division Ring then $\mathcal {C} \paren {R} \,\big / \mathcal {N}$ is a division ring and it only remains to be proved that $\norm{\,\cdot\,}_1$ is a norm.

That is, it needs to be shown $\norm{\,\cdot\,}_1$ satisfies the norm axioms:


 * $\quad \norm{\,\cdot\,}_1$ is Well-defined
 * $\quad \paren {N1}$ : Positive Definiteness
 * $\quad \paren {N2}$ : Multiplicativity
 * $\quad \paren {N3}$ : Triangle Inequality

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

$\norm{\,\cdot\,}_1$ is Well-defined
By Modulus of Cauchy Sequence has Limit then:
 * for each $\left[ {x_n} \right]$ the $\displaystyle \lim_{n \to \infty} \norm{x_n}$ exists.

Suppose $\left[ {x_n} \right] = \left[ {y_n} \right]$.

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 Same Modulus Limit then:
 * $\displaystyle \lim_{n \to \infty} \norm{x_n} = \lim_{n \to \infty} \norm{y_n}$

Hence:
 * $\displaystyle \norm { \left[ {x_n} \right] }_1 = \lim_{n \to \infty} \norm{x_n} = \lim_{n \to \infty} \norm{y_n} = \norm { \left[ {y_n} \right] }_1$

The result follows.

$\paren {N1}$ : Positive Definiteness
By Quotient Ring of Cauchy Sequences is Division Ring the zero of $\mathcal {C} \paren {R} \,\big / \mathcal {N}$ is $\left[ {0} \right]$.

The result follows.

$\paren {N2}$ : Multiplicativity
Let $\left[ {x_n} \right], \left[ {x_n} \right] \in \mathcal {C} \paren {R} \,\big / \mathcal {N}$

$\paren {N3}$ : Triangle Inequality
Let $\left[ {x_n} \right], \left[ {x_n} \right] \in \mathcal {C} \paren {R} \,\big / \mathcal {N}$

By Axiom N3 of norm (Triangle Inequality) then:
 * $\forall n: \norm { x_n + y_n } \le \norm { x_n } + \norm {y_n }$

So: