# Extension of Directed Suprema Preserving Mapping to Complete Lattice Preserves Directed Suprema

## Theorem

Let $L_1 = \left({S_1, \preceq_1}\right)$, $L_2 = \left({S_2, \preceq_2}\right)$ be ordered sets.

Let $L_3 = \left({S_3, \preceq_3}\right)$ be a complete lattice.

Suppose that

$L_2$ is directed suprema inheriting ordered subset of $L_3$.

Let $f:S_1 \to S_2$ be a mapping such that

$f$ preserves directed suprema.

Then $f:S_1 \to S_3$ preserves directed suprema.

## Proof

By definition of ordered subset:

$S_2 \subseteq S_3$

Then define $g = f:S_1 \to S_3$

Let $X$ be a directed subset of $S_1$ such that

$X$ admits a supremum in $L_1$.

Thus by definition of complete lattice:

$g\left[{X}\right]$ admits a supremum in $L_3$.

By definition of mapping preserves directed suprema:

$f\left[{X}\right]$ admits a supremum in $L_2$ and $\sup_{L_2} \left({f\left[{X}\right]}\right) = f\left({\sup_{L_1} X}\right)$
$f$ is an increasing mapping.
$f\left[{X}\right]$ is directed.

Thus by definition of directed suprema inheriting:

$\inf_{L_3} \left({g\left[{X}\right]}\right) = g\left({\inf_{L_1} X}\right)$

$\blacksquare$