Convergent Sequence is Cauchy Sequence/Normed Division Ring/Proof 1

From ProofWiki
Jump to navigation Jump to search


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

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


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.


Also see