# 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.

$\blacksquare$