Two equivalent definitions of convergent sequences? I know that:
Definiton 1. The sequence $(x_n)$ in the metric space $(X,d)$ is said to converge to the point $x_0\in X$ if $$\forall\epsilon>0, \exists n_0\in\mathbb{N} \text{ such that } \forall n\geq n_0, d(x_n,x_0)<\epsilon.$$  
In other words, the sequence $(x_n)$ in the metric space $(X,d)$  converges to the point $x_0\in X$ if $d(x_n,x_0)\rightarrow 0$ with $n\rightarrow\infty.$
Definiton 2. The sequence $(x_n)$ in the metric space $(X,d)$ converges to the point $x_0\in X$ if in every neighborhood $U_{x_0}$ of $x_0$ there exists a natural number $n_0$ such that it is satisfied $\forall n\geq n_0\Rightarrow x_n\in U_{x_0}.$     
The point $x_0$ is said to be the limit of the sequence $(x_n)$. Write $x_n\rightarrow x_0, (n\rightarrow\infty)$ or $\lim_{n\to\infty}x_n=x_0.$
Can these two definitions be proved to be equivalent?
 A: Where are you stuck? As some starting food for thought, notice that
1) For every open neighborhood $U_{x_0}$ of $x_0$, there exists an $\epsilon>0$ with $(x_0-\epsilon,x_0+\epsilon)\subset U_{x_0}$.
2) For every $\epsilon>0$, the set $(x_0-\epsilon,x_0+\epsilon)$ is an open neighborhood of $x_0$.
These two facts should suggest that both definitions have the same content.
A: Let $x_{n}$ converge to $x_{0}$ according to definition 1) and let
$N$ be a neighbourhood of $x_{0}$. 
Since we are working in a metric
space this means exactly that some $\epsilon>0$ exists such that
$d\left(x_{0},y\right)<\epsilon\Rightarrow y\in N$. 
Then $n_{0}\in\mathbb{N}$
exists such that $n\geq n_{0}$ implies that $d(x_{0},x_{n})<\epsilon$
and consequently $x_{n}\in N$. 
Conversely let $x_{n}$ converge to
$x_{0}$ according to definition 2) . 
Note that for every $\epsilon>0$
the set $N=\left\{ y\in X\mid d\left(x_{0},y\right)<\epsilon\right\} $
is a neighbourhood of $x_{0}$, hence some $n_{0}\in\mathbb{N}$ exists
such that $n\geq n_{0}$ implies that $x_{n}\in N$ or equivalently
$d(x_{0},x_{n})<\epsilon$.
