Convergent Subsequence of Cauchy Sequence/Normed Division Ring

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

Let $\sequence{x_n}_{n \in \N}$ be a Cauchy sequence in $\struct {R, \norm {\,\cdot\,}}$.

Let $x \in R$.

Then $\sequence{x_n}$ converges to $x$ it has a subsequence that converges to $x$.

Proof
Let $d$ be the metric induced on $R$ be the norm $\norm {\,\cdot\,}$.

By Sequence is Convergent in Norm iff Convergent in Metric then:
 * $\sequence {x_n}$ converges to $x$ in $\struct {R, \norm {\,\cdot\,}}$ $\sequence {x_n}$ converges to $x$ in $\struct {R, d}$.

By Convergent Subsequence of Cauchy Sequence in Metric Space then:
 * $\sequence{x_n}$ converges to $x$ in $\struct {R, d}$ it has a subsequence that converges to $x$ In $\struct {R, d}$.

By Sequence is Convergent in Norm iff Convergent in Metric then:
 * $\sequence {x_n}$ has a subsequence that converges to $x$ In $\struct {R, d}$ $\sequence {x_n}$ has a subsequence that converges to $x$ In $\struct {R, \norm {\,\cdot\,}}$.

The result follows.