Finite Infima Set of Coarser Subset is Coarser than Finite Infima Set

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

Let $A, B$ be subsets of $S$ such that
 * $A$ is coarser than $B$.

Then $\map {\operatorname{fininfs} } A$ is coarser than $\map {\operatorname{fininfs} } B$

where $\map {\operatorname{fininfs} } B$ denotes the finite infima set of $B$.

Proof
Let $x \in \map {\operatorname{fininfs} } A$

By definition of finite infima set:
 * $\exists Y \in \map {\operatorname {Fin} } A: x = \inf Y$ and $Y$ admits an infimum,

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

By definition of coarser subset:
 * $\forall y \in Y: \exists z \in B: z \preceq y$

By Axiom of Choice:
 * $\exists f: Y \to B: \forall y \in Y: \map f y \preceq y$

We will prove that
 * $f \sqbrk Y$ admits an infimum.

In the case when $Y = \O$:

By Image of Empty Set is Empty Set: Corollary 1:
 * $f \sqbrk Y = \O$

Thus
 * $f \sqbrk Y$ admits an infimum.

In the case when $Y \ne \O$

By definition of non-empty set:
 * $\exists y: y \in Y$

By definition of image of set:
 * $\map f y \in f \sqbrk Y$

By Image of Mapping from Finite Set is Finite:
 * $f \sqbrk Y$ is a finite set.

Thus by Existence of Non-Empty Finite Infima in Meet Semilattice:
 * $f \sqbrk Y$ admits an infimum.

We will prove that:
 * $\inf f \sqbrk Y$ is lower bound for $Y$.

Let $y \in Y$.

By definition of $f$:
 * $\map f y \preceq y$

By definition of image of set:
 * $\map f y \in f \sqbrk Y$

By definitions of infimum and lower bound:
 * $\inf f \sqbrk Y \preceq \map f y$

Thus by definition of transitivity:
 * $\inf f \sqbrk Y \preceq y$

By definition of infimum:
 * $\inf f \sqbrk Y \preceq x$

By definition of finite infima set:
 * $\inf f \sqbrk Y \in \map {\operatorname{fininfs} } B$

Thus
 * $\exists y \in \map {\operatorname{fininfs} } B: y \preceq x$