Mapping at Element is Supremum implies Mapping is Increasing

Theorem
Let $\left({S, \vee, \wedge, \preceq}\right)$ be a lattice.

Let $\left({T, \vee_2, \wedge_2, \precsim}\right)$ be a complete lattice.

Let $f: S \to T$ be a mapping such that
 * $\forall x \in S: f\left({x}\right) = \sup \left\{ {f\left({w}\right): w \in S \land w \ll x}\right\}$

Then $f$ is an increasing mapping.

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

By Preceding implies Way Below Closure is Subset of Way Below Closure:
 * $x^\ll \subseteq y^\ll$

By definitions of image of set and way below closure:
 * $f\left[{x^\ll}\right] = \left\{ {f\left({w}\right): w \in S \land w \ll x}\right\}$

and
 * $f\left[{y^\ll}\right] = \left\{ {f\left({w}\right): w \in S \land w \ll y}\right\}$

where $f\left[{x^\ll}\right]$ denotes the image of $x^\ll$ under $f$.

By Image of Subset under Relation is Subset of Image/Corollary 2:
 * $f\left[{x^\ll}\right] \subseteq f\left[{y^\ll}\right]$

Thus