I am trying to prove the following :
Let $E$ be a topologic vector space. $(x_n)\subset E$ is a Cauchy sequence in $E \ $ iff $\ \lim_{k\to\infty}(x_{m_k}-x_{n_k})=0$ for any $n_k, m_k$ pair of strictly increasing sequences of $\Bbb N$.
I know that $x_n$ is a Cauchy sequence in $E \ $ means
$\forall V\in\mathscr V_0 \ \ \ \exists n_0\in \Bbb N \ \ \textrm{such that} \ \ \forall n,m\geq n_0 \ \ x_n-x_m\in V$ where $\mathscr V_0 $ is collection of nbds of zero.
I am sorry for this easy question but I even don't know how should I start because it seems as if there is nothing to prove. Thanks in advance for any help for any direction of this proposition.
I appreciate any help.
"It seems as if there is nothing to prove" - this is not true, but it is not difficult.
Let $(x_n)$ be a Cauchy sequence in the sense of your definition. Let $V\in\mathscr V_0$ and $(m_k), (n_k)$ be two strictly increasing sequences of integers. We know that $x_m - x_n \in V$ for $n \ge n_0$. Since $(m_k), (n_k)$ are strictly increasing, there exists $k_0$ such that $m_k , n_k \ge n_0$ for $k \ge k_0$. Hence $x_{m_k}-x_{n_k} \in V$ for $k \ge k_0$. This shows that $ \lim_{k\to\infty}(x_{m_k}-x_{n_k})=0$ .
Let the limit condition be satisfied. Let $V\in\mathscr V_0$. Assume that for all $n_0$ there exist $m,n \ge n_0$ such that $x_{m}-x_{n} \notin V$. This can be used to construct inductively strictly increasing sequences of integers such that $x_{m_k}-x_{n_k} \notin V$ for all $k$.Then certainly $(x_{m_k}-x_{n_k})$ cannot have the limit $0$, a contradiction.