Finite Infima Set and Upper Closure is Filter

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

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

Then
 * $\map {\operatorname {fininfs} } X^\succeq$ is filter in $P$.

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

Proof
By Finite Infima Set and Upper Closure is Smallest Filter:
 * $X \subseteq \map {\operatorname {fininfs} } X^\succeq$

By definition of non-empty set:
 * $\map {\operatorname {fininfs} } X^\succeq$ is a non-empty set.

We will prove that
 * $\map {\operatorname {fininfs} } X$ is filtered.

Let $x, y \in \map {\operatorname {fininfs} } X$

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

and
 * $\exists B \in \map {\mathit {Fin} } X: y = \inf B \land B$ admits an infimum

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

Define $C = A \cup B$.

By Union of Subsets is Subset:
 * $C \subseteq X$

By Union of Finite Sets is Finite:
 * $C$ is finite.

Then
 * $C \in \map {\mathit {Fin} } X$

By Existence of Non-Empty Finite Infima in Meet Semilattice:
 * $C \ne \O \implies C$ admits an infimum.

By Union is Empty iff Sets are Empty:
 * $C = \O \implies A = \O$

So
 * $C$ admits an infimum.

By definition of finite infima set:
 * $\inf C \in \map {\operatorname {fininfs} } X$

By Set is Subset of Union:
 * $A \subseteq C$ and $B \subseteq C$

Thus by Infimum of Subset:
 * $\inf C \preceq x$ and $\inf C \preceq y$

Hence $\map {\operatorname {fininfs} } X$ is filtered.

By Filtered iff Upper Closure Filtered:
 * $\map {\operatorname {fininfs} } X^\succeq$ is filtered.

By Upper Closure is Upper Section:
 * $\map {\operatorname {fininfs} } X^\succeq$ is upper.

Hence $\map {\operatorname {fininfs} } X^\succeq$ is filter in $P$.