Finite Infima Set and Upper Closure is Smallest Filter

Theorem
Let $L = \struct {S, \wedge, \preceq}$ be a meet semilattice.

Let $X$ be a non-empty subset of $S$.

Then
 * $X \subseteq \map {\operatorname {fininfs} } X^\succeq$ and
 * for every a filter $F$ in $L$: $\paren {X \subseteq F \implies \map {\operatorname {fininfs} } X^\succeq \subseteq F}$

where
 * $\map {\operatorname {fininfs} } X$ denotes the finite infima set of $X$,
 * $X^\succeq$ denotes the upper closure of $X$.

Proof
By Set is Subset of Finite Infima Set:
 * $X \subseteq \map {\operatorname {fininfs} } X$

By Upper Closure of Subset is Subset of Upper Closure:
 * $X^\succeq \subseteq \map {\operatorname {fininfs} } X^\succeq$

By Set is Subset of Upper Closure:
 * $X \subseteq X^\succeq$

Thus by Subset Relation is Transitive:
 * $X \subseteq \map {\operatorname {fininfs} } X^\succeq$

Let $F$ be a filter in $L$ such that
 * $X \subseteq F$

Let $x \in \map {\operatorname {fininfs} } X^\succeq$

By definition of upper closure of subset:
 * $\exists y \in \map {\operatorname{fininfs} } X: y \preceq x$

By definition of finite infima set:
 * $\exists A \in \map {\operatorname {Fin} } X: y = \inf A \land A$ admits an infimum

where $\map {\operatorname {Fin} } X$ denotes the set of all finite subsets of $S$.

By Subset Relation is Transitive:
 * $A \subseteq F$

By Filtered in Meet Semilattice with Finite Infima:
 * $A \ne \O \implies \inf A \in F$

By Infimum of Empty Set is Greatest Element and Top in Filter:
 * $A = \O \implies \inf A = \top \in F$

where $\top$ denotes the greatest element of $S$.

So
 * $y \in F$

Thus by definition of upper set:
 * $x \in F$