If $b \in \bar{A}$ then there exists a filter $G$ such that $A \in G$ and $G \longrightarrow b$ Let $E=(E,\tau)$ be a topological space and $A \subset E$. I want to prove that if $b \in \bar{A}$ then there exists a filter $G$ in $E$ such that $A \in G$ and $G \longrightarrow b$. For this, note that by definition for the closure (by filters) we have
$$A \cap U \neq \emptyset,\; \forall \; U \in \mathcal{F}(b), $$
where $\mathcal{F}(b)$ denote the filter of neighborhoods of the point b. Hence, I could prove that the set
$$ \mathcal{B}:=\{ A \cap U \subset E \; ; \;  U \in \mathcal{F}(b)\}$$
is a basis of filter on A.
Considering $G$ the filter generated by $\mathcal{B}$, I am not able to show that $A \in G$ and $G \longrightarrow b$. How to proceed?

A better notation is $$\mathcal{B} = \{A \cap U\mid U \in \mathcal{N}(b) \}$$
where $\mathcal{N}(b)$ is the neighbourhood filter at $b$. Then it's easy to see that $\mathcal{B}$ is a filter base: all sets are non-empty by the assumption $b \in \overline{A}$ and the collection is closed under (finite) intersections, as $\mathcal{N}(b)$ is. Denote by $\mathcal{F}$ the filter it generates (i.e. the supersets of the members of $\mathcal{B}$).
Taking $E \in \mathcal{N}(b)$, we already have that $A \cap E = A \in \mathcal{B} \subseteq \mathcal{F}$.
And to see $\mathcal{F} \to b$, let $U$ be any neighbourhood of $b$. Then by definition $A \cap U \in \mathcal{B}$ and so $U (\supseteq A \cap U) \in \mathcal{F}$. So $\mathcal{N}(b) \subseteq \mathcal{F}$ or equivalently $\mathcal{F} \to b$.
Just apply the definitions.
