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 $\forall n \ge n_1$:

\(\displaystyle \norm {x_n}\) \(=\) \(\displaystyle \norm {x_n - x_{n_1} + x_{n_1} }\)
\(\displaystyle \) \(\le\) \(\displaystyle \norm {x_n - x_{n_1} } + \norm {x_{n_1} }\) Norm axiom (N3) (Triangle Inequality).
\(\displaystyle \) \(\le\) \(\displaystyle 1 + \norm {x_{n_1} }\) 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.

$\blacksquare$


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