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

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

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

Lemma 2
Hence:
 * $\norm{\map \phi {x_n} - y}_Q < \epsilon$

By definition of convergent sequence:
 * $\ds \lim_{n \mathop \to \infty} \norm{\map \phi {x_n} }_Q = y$