Normed Division Ring Sequence Converges in Completion iff Sequence Represents Limit/Sufficient Condition

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Theorem

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

Let $\CC$ be the ring of Cauchy sequences over $R$

Let $\NN$ be the set of null sequences.

Let $Q = \CC / \NN$ where $\CC / \NN$ denotes a quotient ring.

Let $\norm {\, \cdot \,}_Q: Q \to \R_{\ge 0}$ be the norm on the quotient ring $Q$ defined by:

$\ds \forall \sequence {x_n} + \NN: \norm {\sequence {x_n} + \NN }_Q = \lim_{n \mathop \to \infty} \norm{x_n}_R$

Let $\phi: R \to Q$ be the mapping from $R$ to the quotient ring $Q$ defined by:

$\forall a \in R: \map \phi a = \tuple {a, a, a, \ldots} + \NN$

where $\sequence {a, a, a, \ldots} + \NN$ is the left coset in $Q$ that contains the constant sequence $\sequence {a, a, a, \ldots} $.

Let $\sequence{x_n}$ be a sequence in $R$.

Let $y \in Q$.

If $\sequence{x_n} \in y$ then $\sequence{\map \phi {x_n}}$ converges to $y$

Proof

Lemma 2

For each $n \in \N$:

$\norm{\map \phi {x_n} - y}_Q = \ds \lim_{m \mathop \to \infty} \norm{x_n - x_m}_R$

$\Box$

Let $\sequence{x_n} \in y$.

Then $\sequence{x_n}$ is a Cauchy Sequence by definition of $y$.

Let $\epsilon > 0$ be arbitrary.

By definition of a Cauchy sequence:

$\exists N \in \N: \forall n, m \ge N : \norm{x_n - x_m}_R < \dfrac \epsilon 2$

Let $n \ge N$ be arbitrary.

From Difference Rule for Cauchy Sequences in Normed Division Ring, the sequence $\sequence{x_n - x_m}_{m \in \N}$ is a Cauchy sequence.

Hence:

$\forall m \ge N : \norm{x_n - x_m}_R < \dfrac \epsilon 2$

From Norm Sequence of Cauchy Sequence has Limit:

$\ds \lim_{m \mathop \to \infty} \norm{x_n - x_m}_R$ exists

From Inequality Rule for Real Sequences:

$\ds \lim_{m \mathop \to \infty} \norm{x_n - x_m}_R \le \dfrac \epsilon 2 < \epsilon$

From Lemma 2:

$\norm{\map \phi {x_n} - y}_Q < \epsilon$

By definition of convergent sequence:

$\ds \lim_{n \mathop \to \infty} \norm{\map \phi {x_n} }_Q = y$

$\blacksquare$