Cauchy Sequence is Bounded/Normed Division Ring/Proof 1
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Theorem
Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring.
Every Cauchy sequence in $R$ is bounded.
Proof
Let $\sequence {x_n} $ be a Cauchy sequence in $R$.
Then by definition:
- $\forall \epsilon \in \R_{\gt 0}: \exists N \in \N : \forall n, m \ge N: \norm {x_n - x_m} < \epsilon$
Let $n_1$ satisfy:
- $\forall n, m \ge n_1: \norm {x_n - x_m} < 1$
Then $\forall n \ge n_1$:
\(\ds \norm {x_n}\) | \(=\) | \(\ds \norm {x_n - x_{n_1} + x_{n_1} }\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \norm {x_n - x_{n_1} } + \norm {x_{n_1} }\) | Norm Axiom $\text N 3$: Triangle Inequality | |||||||||||
\(\ds \) | \(\le\) | \(\ds 1 + \norm {x_{n_1} }\) | as $n, n_1 \ge n_1$ |
Let $K = \max \set {\norm {x_1}, \norm {x_2}, \dots, \norm {x_{n_1 - 1} }, 1 + \norm {x_{n_1} } }$.
Then:
- $\forall n < n_1: \norm {x_n} \le K$
- $\forall n \ge n_1: \norm {x_n} \le 1 + \norm {x_{n_1} } \le K$
It follows by definition that $\sequence {x_n}$ is bounded.
$\blacksquare$
Sources
- 2007: Svetlana Katok: p-adic Analysis Compared with Real: $\S 1.2$: Normed Fields, Exercise $11 \ (2)$