Mapping Preserves Finite and Filtered Infima

Theorem
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.