Cauchy Sequence Is Eventually Bounded Away From Non-Limit

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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 $\forall n > K$:

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

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

$\blacksquare$