Quotient Ring of Cauchy Sequences is Normed Division Ring

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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:

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


Then:

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


Corollary

Let $\struct {R, \norm {\, \cdot \,} }$ be a valued field.


Then $\struct {\mathcal {C} \,\big / \mathcal {N}, \norm {\, \cdot \,}_1 }$ is a valued field.


Proof

By Quotient Ring of Cauchy Sequences is Division Ring then $\mathcal C \,\big / \mathcal N$ is a division ring.

It remains to be proved that:

$\norm {\, \cdot \,}_1$ is well-defined
$\norm {\, \cdot \,}_1$ satisfies the norm axioms.


Lemma 1

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

That is,

$(1): \quad \forall \eqclass {x_n}{}: \lim_{n \mathop \to \infty} \norm{x_n}$ exists.
$(2): \quad \displaystyle \forall \eqclass {x_n}{}, \eqclass {y_n}{} \in \mathcal C \,\big / \mathcal N: \eqclass {x_n}{} = \eqclass {y_n}{} \implies \lim_{n \mathop \to \infty} \norm{x_n} = \lim_{n \mathop \to \infty} \norm{y_n}$

$\Box$


Lemma 2

$\norm {\, \cdot \,}_1$ satisfies the norm axiom (N1).

That is:

$\forall \eqclass {x_n}{} \in \mathcal {C} \,\big / \mathcal {N}: \norm {\eqclass {x_n}{} }_1 = 0 \iff \eqclass {x_n}{} = \eqclass {0_R}{} $

$\Box$


Lemma 3

$\norm {\, \cdot \,}_1$ satisfies the norm axiom (N2).

That is:

$\forall \eqclass {x_n}{}, \eqclass {y_n}{} \in \mathcal {C} \,\big / \mathcal {N}: \norm {\eqclass {x_n}{} \eqclass {y_n}{} }_1 = \norm {\eqclass {x_n}{} }_1 \times \norm {\eqclass {y_n}{} }_1$

$\Box$


Lemma 4

$\norm {\, \cdot \,}_1$ satisfies the norm axiom (N3).

That is:

$\forall \eqclass {x_n}{}, \eqclass {y_n}{} \in \mathcal {C} \,\big / \mathcal {N}: \norm {\eqclass {x_n}{} + \eqclass {y_n}{} }_1 \le \norm {\eqclass {x_n}{} }_1 + \norm {\eqclass {y_n}{} }_1$

$\Box$


The result follows.

$\blacksquare$


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