Infima Preserving Mapping on Filters is Increasing

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 $\left({S, \preceq}\right)$, $f$ preserve the infimum on $F$.

Then $f$ is increasing.

Proof
Let $x, y \in S$ such that
 * $x \preceq y$

By Infimum of Singleton:
 * $\left\{ {x}\right\}$ and $\left\{ {y}\right\}$ admit infima in $\left({S, \preceq}\right)$

By Infimum of Upper Closure of Set:
 * $\left\{ {x}\right\}^\preceq$ and $\left\{ {y}\right\}^\preceq$ admit infima in $\left({S, \preceq}\right)$

where $\left\{ {x}\right\}^\preceq$ denotes the upper closure of $\left\{ {x}\right\}$

By Upper Closure of Singleton
 * $x^\preceq$ and $y^\preceq$ admit infima in $\left({S, \preceq}\right)$

By Upper Closure of Element is Filter:
 * $x^\preceq$ and $y^\preceq$ are filter in $\left({S, \preceq}\right)$

By assumption and definition of mapping preserves the infimum on subset:
 * $f^\to\left({x^\preceq}\right)$ and $f^\to\left({y^\preceq}\right)$ admit infima in $\left({T, \precsim}\right)$

and
 * $\inf\left({f^\to\left({x^\preceq}\right)}\right) = f\left({\inf\left({x^\preceq}\right)}\right)$ and $\inf\left({f^\to\left({y^\preceq}\right)}\right) = f\left({\inf\left({y^\preceq}\right)}\right)$

By Infimum of Upper Closure of Element:
 * $\inf\left({x^\preceq}\right) = x$ and $\inf\left({y^\preceq}\right) = y$

By Upper Closure is Decreasing;
 * $y^\preceq \subseteq x^\preceq$

By Image of Subset under Relation is Subset of Image/Corollary 2
 * $f^\to\left({y^\preceq}\right) \subseteq f^\to\left({x^\preceq}\right)$

Thus by Infimum of Subset:
 * $f\left({x}\right) \precsim f\left({y}\right)$

Thus by definition
 * $f$ is increasing.