Equivalent definitions of limiting point of function. 
Let $(X,d_X)$, $(Y,d_Y)$ be metric spaces, let $E$ be a subset of $X$ and let $f:X\to Y$ be a function. Let $x_0\in X$ be an adherent point of $E$ and $L\in Y$. Prove the following are equivalent:


a. For every sequence $(x^{(n)})_{n=1}^\infty$ in $E$, converging to $x_0$, the sequence $f((x^{(n)}))_{n=1}^\infty$ converges to $L$.


b. For every open set $V\subseteq Y$ which contains $L$, there exists an open set $U\subseteq X$ containing $x_0$ such that $f(U\cap E)\subseteq V$

Assuming a. I want to prove by contradiction,
Soo suppose $L\in V$ which is open in $Y$ and assume there is no open set $U\subseteq X$ containing $x_0$ such that $f(U\cap E)\subseteq V$
Define the sequence $(x_n)_{n=1}^\infty$ by $x_n\in B_{\frac{1}{n}}(x_0)$
By archimedean property for any $\epsilon >0$ there will exist an $n$ large enough that $d_X(x^{(n)},x+0)<\epsilon$, so this sequence converges to $x_0$.
Since $V$ is open and $L\in V$, there exists an open ball $B_r(L)\subseteq V$
By assuming $a.$ however we know that $f((x^{(n)}))_{n=1}^\infty$ converges thus there exists an $m\in \mathbb{N}$ such that if $n\geq m$, $d_Y(f(x^{(n)}),L)<r$, so it is in $B_r(L)$, or in other words, I know there is a ball small enough $B_{\frac{1}{m}}(x_0)$ such that all point of my sequence within this ball have image in $B_r(L)$.
How do I go from saying I know all the points which belong to my sequence which are in $B_{\frac{1}{n}}(x_0)$ have image in $B_r(L)$ to know the entire open ball is, regardless of whether the point is in my sequence or not?
 A: You need to choose the points $x_n$ a bit more carefully. You’ve assumed that there is no open nbhd $U$ of $x_0$ such that $f[U\cap E]\subseteq V$, and each $B_{1/n}(x_0)$ is an open nbhd of $x_0$, so for each $n\in\Bbb Z^+$ there must be at least one point $x_n\in B_{1/n}(x_0)\cap E$ such that $f(x_n)\notin V$. Now $\langle x_n:n\in\Bbb Z^+\rangle$ is a sequence converging to $x_0$ with the added property that $f(x_n)\notin V$ for each $n\in\Bbb Z^+$. This clearly implies that the sequence $\langle f(x_n):n\in\Bbb Z^+\rangle$ does not converge to $L$, contradicting the assumption (a).
A: Don't just pick any points $x_n$ in $B_{\frac 1 n} (x_0)$. That will yield  a proof.
We know that $f(U \cap E)$ is not contained in $V$ where $U=B_{\frac 1 n} (x_0)$. So there exists $x_n \in U \cap E$ such that $f(x_n) \notin V$. Now $x_n \to x_0$ so $f(x_n) \to L$. Also $V$ is open and $L \in V$. By defintion of limit this implies that $f(x_n)$ must be in $V$ for $n$ sufficiently large  but $f(x_n) \notin V$ for any $n$. This contradiction proves that a) implies b).
