Coarser Between Generator Set and Filter is Generator Set of Filter

Theorem
Let $L = \left({S, \wedge, \preceq}\right)$ be a meet semilattice.

Let $F$ be a filter on $L$.

Let $G$ be a generator set of $F$.

Let $A$ be a subset of $S$ such that
 * $G$ is coarser than $A$ and $A$ is coarser than $F$.

Then $A$ is generator set of $F$.

Proof
By definition of generator set of filter:
 * $F = \left({\operatorname{fininfs}\left({G}\right)}\right)^\succeq$

where
 * $\operatorname{fininfs}\left({G}\right)$ denotes the finite infima set of $G$,
 * $A^\succeq$ denotes the upper closure of $A$.

By Finite Infima Set of Coarser Subset is Coarser than Finite Infima Set:
 * $\operatorname{fininfs} \left({G}\right)$ is coarser than $\operatorname{fininfs} \left({A}\right)$

Thus by Upper Closure of Coarser Subset is Subset of Upper Closure:
 * $F \subseteq \left({\operatorname{fininfs} \left({G}\right)}\right)^\succeq$

By filter in ordered set:
 * $F$ is an upper set.

By Set Coarser than Upper Set is Subset:
 * $A \subseteq F$

Thus by Finite Infima Set and Upper Closure is Smallest Filter:
 * $\left({\operatorname{fininfs}\left({A}\right)}]right)^\succeq \subseteq F$

Thus by definition of set equality:
 * $F = \left({\operatorname{fininfs}\left({A}\right)}]right)^\succeq$

Hence $A$ is generator set of $F$.