Convergent Sequence is Cauchy Sequence/Normed Division Ring

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Theorem

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


Every convergent sequence in $R$ is a Cauchy sequence.


Proof 1

Let $\sequence {x_n}$ be a sequence in $R$ that converges to the limit $l \in R$.

Let $\epsilon > 0$.

Then also $\dfrac \epsilon 2 > 0$.

Because $\sequence {x_n}$ converges to $l$, we have:

$\exists N: \forall n > N: \norm {x_n - l} < \dfrac \epsilon 2$

So if $m > N$ and $n > N$, then:

\(\displaystyle \norm {x_n - x_m}\) \(=\) \(\displaystyle \norm {x_n - l + l - x_m}\)
\(\displaystyle \) \(\le\) \(\displaystyle \norm {x_n - l} + \norm {l - x_m}\) Triangle Inequality
\(\displaystyle \) \(<\) \(\displaystyle \frac \epsilon 2 + \frac \epsilon 2\) (by choice of $N$)
\(\displaystyle \) \(=\) \(\displaystyle \epsilon\)

Thus $\sequence {x_n}$ is a Cauchy sequence.


Proof 2

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

Let $\sequence {x_n}$ be a sequence in $R$ that converges to the limit $l$ in $\struct {R, \norm {\,\cdot\,}}$.

By the definition of a convergent sequence in a normed division ring then $\sequence {x_n} $ converges to the limit $l$ in $\struct {R, d}$.

By Convergent Sequence is Cauchy Sequence in metric space then $\sequence {x_n} $ is a Cauchy sequence in $\struct {R, d}$.

By the definition of a Cauchy sequence in a normed division ring then $\sequence {x_n} $ is a Cauchy sequence in $\struct {R, \norm {\,\cdot\,}}$.

$\blacksquare$


Also see


Sources