Subset and Image Admit Suprema and Mapping is Increasing implies Supremum of Image Precedes Mapping at Supremum

Theorem
Let $\struct {S, \preceq}$, $\struct {T, \precsim}$ be ordered sets.

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

Let $D \subseteq S$ such that
 * $D$ admits a supremum in $S$ and $f \sqbrk D$ admits a supremum in $T$.

Then:
 * $\map \sup {f \sqbrk D} \precsim \map f {\sup D}$

Proof
By definition of supremum:
 * $\sup D$ is upper bound for $D$.

By Increasing Mapping Preserves Upper Bounds:
 * $\map f {\sup D}$ is upper bound for $f \sqbrk D$.

Thus by definition of supremum:
 * $\map \sup {f \sqbrk D} \precsim \map f {\sup D}$