Norm Sequence of Cauchy Sequence has Limit

Theorem
Let $\struct {R, \norm { \, \cdot \, } }$ be a normed division ring.

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

Then $\sequence {\norm {x_n}}$ has a limit in $\R$.

That is,
 * $\exists l \in \R: \displaystyle \lim_{n \to \infty} {\norm {x_n}} = l$

Proof
It is first shown that $\sequence {\norm {x_n}}$ is a real Cauchy sequence in $\R$.

Given $\epsilon \gt 0$.

By the definition of Cauchy sequence then:
 * $\exists N \in \N: \forall n, m > N, \norm {x_n - x_m} \lt \epsilon$

By the reverse triangle inequality, then:
 * $\forall n, m \gt N: \cmod {\norm {x_n} - \norm {x_m}} \le \norm {x_n - x_m} \lt \epsilon$

From the definition of a real Cauchy sequence it follows that $\sequence {\norm {x_n}}$ is a real Cauchy sequence in $\R$.

By Real Sequence is Cauchy iff Convergent the sequence $\sequence {\norm {x_n}}$ has a limit in $\R$.