Modulus of Cauchy Sequence is Cauchy Sequence

From ProofWiki
Jump to: navigation, search

Theorem

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

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

Then

$\sequence {\norm {x_n}}$ is a real Cauchy sequence in $\R$.

Proof

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 the result follows.

$\blacksquare$


Sources