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

## Theorem

Let $\left({S, \preceq}\right)$, $\left({T, \precsim}\right)$ 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 $f\left[{D}\right]$ admits a supremum in $T$.

Then $\sup \left({f\left[{D}\right]}\right) \precsim f\left({\sup D}\right)$

## Proof

By definition of supremum:

$\sup D$ is upper bound for $D$.
$f\left({\sup D}\right)$ is upper bound for $f\left[{D}\right]$.

Thus by definition of supremum:

$\sup \left({f\left[{D}\right]}\right) \precsim f\left({\sup D}\right)$

$\blacksquare$