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Let $M$ be a metric space and $a \in M$. We say that $V \subseteq M$ is a neighborhood of $a$ when $a \in \operatorname{Int}(V)$. Show that if $(x_n)$ is a sequence in $M$, then the following are equivalent:

  1. $\lim x_n = a$;
  2. For every neighborhood $V$ of $a$ there is $n_{0} \in \mathbb{N}$ such that $x_n \in V$ when $n ≥ n_0$.
Cameron Buie
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Viviane
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  • The second one should probably read "For every neighborhood $V$ of $a$ there is $n_0\in\Bbb N$ such that $x_n\in V$ when $n\ge n_0.$" What have you tried so far? I assume that your definition of $\lim x_n=a$ involves $\epsilon$? – Cameron Buie Oct 13 '13 at 16:36
  • okay. I used the settings as follows: Let M be a metric space and let (xn) be a sequence of points of M. We say that (xn) converges to x belonging to M if for every neighborhood V of x belonging to N exists n0 such that for all n greater than or equal n0 xn belongs to V. As for xn = lim to say that the sequence converges to the twelfth or tends to. A sequence that has limit is convergent then the two statements are equivalent. – Viviane Oct 13 '13 at 16:50
  • I don't understand what you mean by "converges to the twelfth." I'm going to post an answer with some hints, based on what I assume your definition is. Please correct me if I'm wrong. – Cameron Buie Oct 13 '13 at 16:56

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I assume that your definition of convergence is as follows:

If $M$ is a metric space with metric $d,$ and $a\in M$ and $(x_n)$ is a sequence in $M,$ then we say that $\lim x_n=a$ if for all real $\epsilon>0,$ there is some $n_0\in\Bbb N$ such that $d(x_n,a)<\epsilon$ whenever $n\ge n_0$.

Here's what I recommend, then. For each $a\in M$ and each real $\epsilon>0,$ define $$B_d(a;\epsilon):=\{x\in M\mid d(x,a)<\epsilon\}.$$ You should be able to $B_d(a;\epsilon)$ is a neighborhood of $a,$ so one part of the proof is then trivial. For the other direction, use the definition of interior point, together with the definition of convergence.

Cameron Buie
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