Equivalence of Definitions of Limit Point of Filter Basis

Theorem
Let $T = \struct {S, \tau}$ be a topological space.

Let $\FF$ be a filter on the underlying set $S$ of $T$.

Let $\BB$ be a filter basis of $\FF$.

Proof
Let $\FF$ be a filter on $S$.

Let $\BB$ be a filter basis of $\FF$.

$(1)$ implies $(2)$
Let $x \in S$ be a limit Point of $\BB$ by definition $1$.

Then by definition $\FF$ converges on $x$.

By definition of convergent filter:
 * $\forall N_x \subseteq S: N_x \in \FF$

where $N_x$ is a neighborhood of $x$.

That is, every neighborhood of $x$ is an element of $\FF$.

By definition of filter basis:
 * $\forall U \in \FF: \exists V \in \BB: V \subseteq U$

But we have shown that $N_x \in \FF$.

Therefore:
 * $\exists V \in \BB: V \subseteq N_x$

That is, every neighborhood of $x$ contains a set of $\BB$.

Thus $x \in S$ is a limit Point of $\BB$ by definition $2$.

$(2)$ implies $(1)$
Let $x \in S$ be a limit Point of $\BB$ by definition $2$.

Then by definition every neighborhood of $x$ contains a set of $\BB$.

By definition of filter basis, $\FF := \set {V \subseteq X: \exists U \in \BB: U \subseteq V}$ is a filter on $S$ :
 * $\forall V_1, V_2 \in \BB: \exists U \in \BB: U \subseteq V_1 \cap V_2$

But then letting $V_1 = V_2 = V$, we have:
 * $\forall V \in \BB: \exists U \in \BB: U \subseteq V$

and so $V \in \FF$.

Thus every neighborhood of $x$ is an element of $\FF$.

Thus, by definition, $\FF$ converges on $x$.

Thus $x \in S$ is a limit Point of $\BB$ by definition $1$.