# Convergent Subsequence of Cauchy Sequence/Normed Division Ring

< Convergent Subsequence of Cauchy Sequence(Redirected from Convergent Subsequence of Cauchy Sequence in Normed Division Ring)

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## Theorem

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

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

Let $x \in R$.

Then $\sequence {x_n}$ converges to $x$ if and only if $\sequence {x_n}$ has a subsequence that converges to $x$.

## Proof

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

By the definition of a convergent sequence in a normed division ring then:

- $\sequence {x_n}$ converges to $x$ in $\struct {R, \norm {\,\cdot\,} }$ if and only if $\sequence {x_n}$ converges to $x$ in $\struct {R, d}$.

By Convergent Subsequence of Cauchy Sequence in Metric Space:

- $\sequence {x_n}$ converges to $x$ in $\struct {R, d}$ if and only if $\sequence {x_n}$ has a subsequence that converges to $x$ In $\struct {R, d}$.

By the definition of a convergent sequence in a normed division ring:

- $\sequence {x_n}$ has a subsequence that converges to $x$ In $\struct {R, d}$ if and only if $\sequence {x_n}$ has a subsequence that converges to $x$ in $\struct {R, \norm {\,\cdot\,} }$.

The result follows.

$\blacksquare$

## Sources

- 2007: Svetlana Katok:
*p-adic Analysis Compared with Real*... (previous) ... (next): $\S 1.2$: Normed Fields, Exercise $11$ $(2)$