Cauchy Sequence Is Eventually Bounded Away From Non-Limit

Theorem

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

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

Suppose $\sequence {x_n}$ does not converge to $l \in R$.

Then:

$\exists K \in \N$ and $C \in \R_{>0}: \forall n > K: C < \norm {x_n - l}$

Because $\sequence {x_n}$ does not converge to $l$:

$\exists \epsilon \in \R_{>0}: \forall n \in \N: \exists m \ge n: \norm {x_m - l} \ge \epsilon$

Because $\sequence {x_n}$ is a Cauchy sequence:

$\exists K \in \N: \forall n, m \ge K: \norm {x_n - x_m} < \dfrac \epsilon 2$

Let $M \ge K: \norm {x_M - l} \ge \epsilon$.

Then $\forall n > K$:

$\ds \epsilon$

$\le$

$\ds \norm {x_M - l}$

$\ds $

$=$

$\ds \norm {x_M - x_n + x_n - l}$

$\ds $

$\le$

$\ds \norm {x_M - x_n} + \norm {x_n - l}$

$\ds $

$<$

$\ds \dfrac \epsilon 2 + \norm {x_n - l}$

because $n, M \ge K$

$\ds \leadsto \ \ $

$\ds \dfrac \epsilon 2$

$<$

$\ds \norm {x_n - l}$

subtracting $\dfrac \epsilon 2$ from both sides of the equation

Let $C = \dfrac \epsilon 2$ and the result follows.

$\blacksquare$