# If $E \subset X$ then $x \in \bar{E}$ iff there is a filterbase in E converging to $x$.

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 $x \in \bar{E}$.

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Hint: If $x \in \overline{E}$, then $\mathcal{B} = \{ U \cap E : U \text{ is an open neighbourhood of } x \}$ is a filterbase.
... Since $\mathcal B$ converges to $x$, there exists a $B^\prime \in \mathcal B$ such that $B^\prime \subseteq U(x)$, and so in particular $\emptyset \neq B^\prime \subseteq U(x) \cap E$. ....