neighborhood space metric 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:


*

*$\lim x_n = a$;

*For every neighborhood $V$ of $a$ there is $n_{0} \in \mathbb{N}$ such that $x_n \in V$ when $n ≥ n_0$.


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.
