Convergent Sequence in Normed Division Ring is Bounded/Proof 1

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.

$\blacksquare$