I cannot prove this statement, I tried to prove by using the definition of open sets however i feel that it is necessary prove it in two directions since it's an iff statement.
The question is,
Let $X$ be a metric space with metric $d$, and let $x_n$ be a sequence of points in $X$. Prove that $x_n\rightarrow a$ if and only if for every open set $U$ with $a\in U$, there is a number N such that whenever n > N we have $x_n\in U$
What I have done is, by using the definition of open sets, $B(a,r)\cap U$ is not an empty set and $r > 0$. So by choosing $r=1$ we obtain $x_1\in B(a,1)$ which intersects with $U$. Secondly I chose $r=1/2$ so $x_2\in B(a,1/2)$ which again intersects with $U$. By continuing like this eventually I chose $r=1/n$ so $x_n\in B(a,1/n)$ which intersects with $U$. Therefore $d(x_n,a) <\frac{1}{n}$. Therefore I conclude that when $n$ goes to infinity that $x_n\rightarrow a$.
Am I right up to this point? If so how should I proceed and what should I do to prove that for opposite inclusion?
I'm so new, so let me know if I do or did something inconvenient. Thanks!