How do I prove the direct implication i.e if $ x \in \bar{E}$ then there is a filterbase in E converging to $x$? I can only prove the converse implication, as follows: (any feedback on this would be great)
Conversely, suppose there is a filterbase $\mathcal{B}$ in $E$ converging to $x$. We have to show that $x$ is a point of closure of $E$. Let $U(x)$ be an open neighbourhood of $x$. Since $\mathcal{B}$ converges to $x$, there exists $B' \in \mathcal{B}$ such that $b \in B' \subset U(x)$. It follows that $b \in U(x)\cap E$. Thus, $U(x) \cap E \neq \emptyset$ for every open neighbourhood $U(x)$ of $x$. Thus, $x$ is a point of the closure of $E$, and hence $x \in \bar{E}$.
