Cauchy Sequence Is Eventually Bounded Away From Zero

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Theorem

Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring with zero: $0$.

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

Suppose $\sequence {x_n}$ does not converge to $0$, then:

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


Proof

Since $\sequence {x_n}$ does not converge to $0$ then:

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

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

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

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

Then $\forall n > K$:

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

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

$\blacksquare$