Theorem: Mapping Preserves Finite and Filtered Infima

Let $\struct {S_1, \preceq_1}$, $\struct {S_2, \preceq_2}$ be meet semilattices.
Let $f: S_1 \to S_2$ be a mapping.
Let $f$ preserve finite infima and preserve filtered infima.
Then $f$ also preserves all infima

Proof:
Assume that
:$(1): \quad f$ preserves finite infima
and
:$(2): \quad f$ preserves filtered infima.
Let $X$ be a subset of $S_1$.
Let $X$ admits an infimum in $\struct {S_1, \preceq_1}$
Define $Z := \set {\inf A: A \in \map {\operatorname {Fin} } X \land A \ne \O}$
where
:$\inf A$ denotes the infimum of $A$ in $\struct {S_1, \preceq_1}$
:$\map {\operatorname {Fin} } X$ denotes the set of all finite subsets of $X$
By Existence of Non-Empty Finite Infima in Meet Semilattice:
:for every non-empty finite subset $A$ of $X$, $A$ admits an infimum in $\struct {S_1, \preceq_1}$
By Infimum of Infima:
:$\inf Z = \inf \bigcup \paren {\map {\operatorname {Fin} } X \setminus \set \O} = \inf X$
:$Z$ admits an infimum in $\struct {S_1, \preceq_1}$
We will prove that
:$Z$ is filtered
Let $x, y \in Z$.
By definition of $Z$:
:$\exists A \in \map {\operatorname {Fin} } X \setminus \set \O: x = \inf A$
and
:$\exists B \in \map {\operatorname {Fin} } X \setminus \set \O: y = \inf B$
By Finite Union of Finite Sets is Finite:
:$A \cup B$ is finite
:$A \cup B \ne \O$
By definition of $Z$:
:$\map \inf {A \cup B} \in Z$
By Infimum of Infima:
:$\inf A \wedge \inf B = \map \inf {A \cup B}$
Thus by definition of infimum:
:$\map \inf {A \cup B} \preceq_1 x \land \map \inf {A \cup B} \preceq_1 y$
Thus by definition
:$Z$ is filtered
By $(2)$ and definitions of mapping preserves filtered infima and mapping preserves infimum on subset:
:$f^\to \sqbrk Z$ admits an infimum in $\struct {S_2, \preceq_2}$ and $\map \inf {f^\to \sqbrk Z} = \map f {\inf Z}$
We will prove that
:$X \subseteq Z$
Let $x \in X$.
By definition of $\map {\operatorname {Fin} } X$:
:$\set x \in \map {\operatorname {Fin} } X$ and $\set x \ne \O$
By definition of $Z$:
:$\inf \set x \in Z$
Thus by Infimum of Singleton:
:$x \in Z$
By Image of Subset under Mapping is Subset of Image:
:$f^\to \sqbrk X \subseteq f^\to \sqbrk Z$
By definition of infimum:
:$\map \inf {f^\to \sqbrk Z}$ is a lower bound for $f^\to \sqbrk Z$
By Lower Bound is Lower Bound for Subset
:$\map \inf {f^\to \sqbrk Z}$ is a lower bound for $f^\to \sqbrk X$
We will prove that
:for every lower bound $a$ for $f^\to \sqbrk X$, $a$ is lower bound for $f^\to \sqbrk Z$
Assume that
:$a$ is lower bound for $f^\to \sqbrk X$
Let $x \in f^\to \sqbrk Z$.
By definition of image of set:
:$\exists y \in S_1: y \in Z \land \map f y = x$
By definition of $Z$:
:$\exists A \in \map {\operatorname {Fin} } X: y = \inf A \land A \ne \O$
By $(1)$ and definition of mapping preserves finite infima:
:$f$ preserves the infimum of $A$
By Existence of Non-Empty Finite Infima in Meet Semilattice
:$A$ admits an infimum in $\struct {S_1, \preceq_1}$ and $\map \inf {f^\to \sqbrk A} = \map f {\inf A} = x$
By Image of Subset under Mapping is Subset of Image:
:$f^\to \sqbrk A \subseteq f^\to \sqbrk X$
By Lower Bound is Lower Bound for Subset
:$a$ is lower bound for $f^\to \sqbrk A$
Thus by definition of infimum:
:$a \preceq_2 x$
Thus by definition
:$a$ is lower bound for $f^\to \sqbrk Z$
By definition of infimum:
:for every lower bound $a$ for $f^\to \sqbrk X$, $a \preceq_2 \map f {\inf X}$
Again by definition of infimum:
:$f^\to \sqbrk X$ admits an infimum in $\struct {S_2, \preceq_2}$
:$\map \inf {f^\to \sqbrk X} = \map f {\inf X}$
Thus the result follows by definition of all infima preserving mapping.
{{qed}}