Cauchy Sequence Is Eventually Bounded Away From Non-Limit

From ProofWiki
Jump to navigation Jump to search


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

Then $\forall n > K$:

\(\displaystyle \epsilon\) \(\le\) \(\displaystyle \norm {x_M - l}\)
\(\displaystyle \) \(=\) \(\displaystyle \norm {x_M - x_n + x_n - l}\)
\(\displaystyle \) \(\le\) \(\displaystyle \norm {x_M - x_n} + \norm {x_n - l}\) Norm Axiom $\text N 3$: Triangle Inequality
\(\displaystyle \) \(<\) \(\displaystyle \dfrac \epsilon 2 + \norm {x_n - l}\) because $n, M \ge K$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \dfrac \epsilon 2\) \(<\) \(\displaystyle \norm {x_n - l}\) Subtracting $\dfrac \epsilon 2$ from both sides of the equation.

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