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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.

handa
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  • https://math.stackexchange.com/questions/187703/cauchy-nets-in-a-metric-space – Ataulfo Jun 09 '22 at 13:38
  • @Piquito I am in this question does not have answer of mine. I wanted to ask is it possible to get a similar condition contains cofinals like subsequences. But first, I need to understand how should I prove the given equivalence because it seems like definition for me. – handa Jun 09 '22 at 15:52
  • Best regards. Good luck in your purpose. – Ataulfo Jun 09 '22 at 20:35
  • Thank you. Best. – handa Jun 09 '22 at 20:39

1 Answers1

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"It seems as if there is nothing to prove" - this is not true, but it is not difficult.

  1. 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$ .

  2. 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.

Paul Frost
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