Cauchy Sequence in Normed Space Let $(E, ||\cdot ||)$ be a normed space and let $(x_n)$ be a sequence in $E$. Show that the following
conditions are equivalent:
(a) $(x_n)$ is a Cauchy sequence.
(b) For every increasing function $p: \mathbb{N} \to \mathbb{N}$ we have $||x_{p(n+1)} - x_{p(n)}||\to 0$ as $n \to \infty$.
I've encountered this problem in my graduate level analysis class.
The definition I have for Cauchy sequence is: $\forall  \varepsilon >0$  $\exists \, n_0$ such that $\forall m,n>n_0$, $||x_n-x_m||< \varepsilon$ (or $||x_n-x_m||\to 0$ as $n,m\to \infty$).
Thank you for any assistance here. 
 A: Remark that if $p$ is an (strictly) increasing function defined on $N$, $p(n)>n$.
Suppose that $x_n$ is a Cauchy sequence Let $c>0$ $\exists N$ such that $n,m>N$ implies that $\|x_n-x_m\|<c$, in particular $\|x_{p(n)}-x_{p(n+1)}\|<c$ if $n>N$ since $p(n),p(n+1)>n>N$. done.
In the other hands, suppose that for every increasing function $p$, $lim\|x_{p(n)}-x_{p(n+1)}\|$ goes to zero as n goes to infinity and the sequence is not Cauchy. This is equivalent to saying that there exists $c>0$ such that for every $M>0$, there exists $n_M<m_M, n_M,m_M>M $ such that $\|x_{n_M}-x_{m_M}\|\geq c$. We construct $f:N\rightarrow N$ such that $f(1)=n_1, f(2)=m_1$ suppose that $f(2m)$ and $f(2m-1)$ are defined such that $\|x_{f(2m)}-x_{f(2m-1)}\|\geq c$ and $f(2m)>f(2m-1)$. There exists $f(2m+2)>f(2m+1)> f(2m)$ such that $\|x_{f(2m+2)}-x_{f(2m+1)}\|\geq c$ since the sequence is not Cauchy. The map $f$ is increasing and $lim_n\|x_{f(n)}-x_{f(n+1)}\|$ is not zero. Contradiction. done.
A: Showing a) implies b) is straightforward. 
Given $\epsilon$ we get a $n_0$ and then select an n with $p(n)>n_0$. For these n we have $||x_{p(n)}-x_{p(n+1)}||<\epsilon$.
For the reverse assume the sequence is not a Cachy sequence and lets construct an increasing sequence p(n) such that $||x_{p(n)}-x_{p(n+1)}||>\epsilon$.
There must exist an $\epsilon>0$ such that for every n there exists an m>n with $||x_n-x_m||>\epsilon$. Now given p(n) use this as n and select p(n+1)>p(n) as the m value in this construction giving the inequality $||x_{p(n)}-x_{p(n+1)}||>\epsilon$. This shows that b) is false if a) is false.
