Infima Preserving Mapping on Filters Preserves Filtered Infima

Theorem
Let $\left({S, \preceq}\right)$, $\left({T, \precsim}\right)$ be ordered sets.

Let $f: S \to T$ be a mapping.

Let every filter $F$ in $\left({S, \preceq}\right)$ be such that $f$ preserves the infimum on $F$.

Then $f$ preserves filtered infima.

Proof
Let $F$ be a filtered subset of $S$ such that
 * $F$ admits an infimum in $\left({S, \preceq}\right)$.

By Filtered iff Upper Closure Filtered:
 * $F^\succeq$ is filtered

where $F^\succeq$ denotes the upper closure of $F$.

By Upper Closure is Upper Set:
 * $F^\succeq$ is upper.

Because filtered is bob-empty therefore by definition
 * $F^\succeq$ is filter in $\left({S, \preceq}\right)$.

By Infimum of Upper Closure of Set:
 * $F^\succeq$ admits an infimum in $\left({S, \preceq}\right)$

and
 * $\inf\left({F^\succeq}\right) = \inf F$

By assumption and mapping preserves the infimum on subset:
 * $f^\to\left({F^\succeq}\right)$ admits an infimum in $\left({T, \precsim}\right)$.

and
 * $\inf\left({f^\to\left({F^\succeq}\right)}\right) = f\left({\inf\left({F^\succeq}\right)}\right)$

By Upper Closure is Closure Operator:
 * $F \subseteq F^\succeq$

By Image of Subset under Relation is Subset of Image/Corollary 2:
 * $f^\to\left({F}\right) \subseteq f^\to\left({F^\succeq}\right)$

By definition of infimum:
 * $f\left({\inf F}\right)$ is lower bound of $f^\to\left({F^\succeq}\right)$

By Lower Bound is Lower Bound for Subset:
 * $f\left({\inf F}\right)$ is lower bound of $f^\to\left({F}\right)$

We will prove that
 * for every element $x$ of $T$ if $x$ is lower bound of $f^\to\left({F}\right)$, then $x \precsim f\left({\inf F}\right)$