Order Isomorphism Preserves Infima and Suprema

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

Let $f:S \to S'$ be an order isomorphism between $L$ and $L'$.

Then $f$ preserves infima and suprema.

$f$ preserves infima
Let $X$ be a subset of $S$ such that
 * $X$ admits an infimum in $L$.

By definition of infimum:
 * $\inf X$ is lower bound for $X$.

Thus by Order Isomorphism Preserves Lower Bounds:
 * $f\left({\inf X}\right)$ is lower bound for $f\left[{X}\right]$.

We will prove that
 * $\forall x \in S': x$ is lower bound for $f\left[{X}\right] \implies x \preceq' f\left({\inf X}\right)$

Let $x \in S'$ such that
 * $x$ is lower bound for $f\left[{X}\right]$.

By definitions of order isomorphism and bijection:
 * $f$ is a surjection.

By definition of surjection:
 * $S' = f\left[{S}\right]$

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

By Order Isomorphism Preserves Lower Bounds:
 * $y$ is lower bound for $X$.

By definition of infimum:
 * $y \preceq \inf X$

Thus by definition of order isomorphism:
 * $x \preceq' f\left({\inf X}\right)$

Thus by definition of infimum:
 * $f\left[{X}\right]$ admits an infimum in $L'$ and $\inf\left({f\left[{X}\right]}\right) = f\left( {\inf X}\right)$

$f$ preserves suprema
This fallows by mutatis mutandis.