Convergent Sequence is Cauchy Sequence/Normed Division Ring/Proof 1
Jump to navigation
Jump to search
Theorem
Let $\struct {R, \norm {\,\cdot\,}} $ be a normed division ring.
Every convergent sequence in $R$ is a Cauchy sequence.
Proof
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:
\(\ds \norm {x_n - x_m}\) | \(=\) | \(\ds \norm {x_n - l + l - x_m}\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \norm {x_n - l} + \norm {l - x_m}\) | Norm Axiom $\text N 3$: Triangle Inequality | |||||||||||
\(\ds \) | \(<\) | \(\ds \frac \epsilon 2 + \frac \epsilon 2\) | by choice of $N$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \epsilon\) |
Thus $\sequence {x_n}$ is a Cauchy sequence.
$\blacksquare$
Also see
Sources
- 2007: Svetlana Katok: p-adic Analysis Compared with Real: $\S 1.2$: Normed Fields, Exercise $11 \ (1)$