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.

For every filter $F$ in $\left({S, \preceq}\right)$, let $f$ 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:
 * $\left\{ {x}\right\}^\succeq$ and $\left\{ {y}\right\}^\succeq$ admit infima in $\left({S, \preceq}\right)$

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

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

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

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

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

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

By Upper Closure is Decreasing:
 * $y^\succeq \subseteq x^\succeq$

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

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

Thus by definition:
 * $f$ is increasing.