# 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 $C \in \R_{\gt 0}: \forall n \gt K: C \lt \norm {x_n - l}$

## Proof

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

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

Since $\sequence {x_n}$ is a Cauchy sequence then:

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

Let $M >= K: \norm {x_M - l} >= \epsilon$

Then $\forall n > K$:

 $\displaystyle \epsilon$ $\le$ $\displaystyle \norm {x_M - l }$ $\displaystyle$ $=$ $\displaystyle \norm {x_M - x_n + x_n - l }$ $\displaystyle$ $\le$ $\displaystyle \norm {x_M - x_n } + \norm {x_n - l }$ Axiom (N3) of norm (Triangle Inequality) $\displaystyle$ $\lt$ $\displaystyle \dfrac \epsilon 2 + \norm {x_n -l }$ Since $n, M \ge K$ $\displaystyle \implies \ \$ $\displaystyle \dfrac \epsilon 2$ $\lt$ $\displaystyle \norm {x_n - l }$ Subtracting $\dfrac \epsilon 2$ from both sides of the equation.

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

$\blacksquare$