Increasing Mapping Preserves Lower Bounds

Theorem
Let $L = \left({S, \preceq}\right)$, $L' = \left({S', \preceq'}\right)$ be ordered sets.

Let $f:S \to S'$ be an increasing mapping.

Let $x \in S$, $X \subseteq S$ such that
 * $x$ is lower bound for $X$.

Then $f \left({x}\right)$ is lower bound for $f \left[{X}\right]$.

Proof
Let $y \in f\left[{X}\right]$.

By definition of image of set:
 * $\exists z \in X: y = f \left({z}\right)$

By definition of lower bound:
 * $x \preceq z$

Thus by definition of increasing mapping:
 * $f \left({x}\right) \preceq' y$