Norm Sequence of Cauchy Sequence has Limit

From ProofWiki
Jump to navigation Jump to search

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,

$\exists l \in \R: \displaystyle \lim_{n \mathop \to \infty} \norm {x_n} = l$


Proof

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

Let $\epsilon \in \R_{>0}$ be given.

By the definition of Cauchy sequence then:

$\exists N \in \N: \forall n, m > N, \norm {x_n - x_m} < \epsilon$

By the reverse triangle inequality, then:

$\forall n, m > N: \cmod {\norm {x_n} - \norm {x_m} } \le \norm {x_n - x_m} < \epsilon$

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 $\R$.

$\blacksquare$


Sources