Convergent Sequence in Normed Division Ring is Bounded/Proof 1

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Theorem

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

Let $\sequence {x_n}$ be a sequence in $R$.

Let $\sequence {x_n}$ be convergent in the norm $\norm {\,\cdot\,}$ to the following limit:

$\displaystyle \lim_{n \mathop \to \infty} x_n = l$


Then $\sequence {x_n}$ is bounded.


Proof

Let $\sequence {x_n}$ be convergent in the norm $\norm {\,\cdot\,}$ to the limit $l$, then:

$\forall \epsilon \in \R_{\gt 0}: \exists N \in \N : \forall n \ge N: \norm {x_n - l} < \epsilon$


Let $n_1$ satisfy:

$\forall n \ge n_1: \norm {x_n - l} < 1$


Then $\forall n \ge n_1$:

\(\displaystyle \norm {x_n}\) \(=\) \(\displaystyle \norm {x_n - l + l}\)
\(\displaystyle \) \(\le\) \(\displaystyle \norm {x_n - l} + \norm l\) Norm axiom (N3) (Triangle Inequality)
\(\displaystyle \) \(\le\) \(\displaystyle 1 + \norm l\) as $n \ge n_1$


Let $K = \max \set {\norm {x_1}, \norm {x_2}, \dots, \norm {x_{n_1 - 1} }, 1 + \norm l}$.

Then:

$\forall n < n_1: \norm {x_n} \le K$
$\forall n \ge n_1: \norm {x_n} \le 1 + \norm l \le K$

Thus, by definition, $\sequence {x_n}$ is bounded.

$\blacksquare$


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