Order Isomorphism 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 order isomorphism between $L$ and $L'$.

Let $x \in S$, $X \subseteq S$.

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

Proof
By definition of order isomorphism:
 * $f$ is an order embedding.

Sufficient Condition
Assume that
 * $x$ is lower bound for $X$.

By Order Embedding is Increasing Mapping:
 * $f$ is an increasing mapping.

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

Necessary Condition
Assume that
 * $f\left({x}\right)$ is lower bound for $f\left[{X}\right]$.

Let $y \in X$.

By definition of image of set:
 * $f\left({y}\right) \in f\left[{X}\right]$

By definition of lower bound:
 * $f\left({y}\right) \preceq' f\left({x}\right)$

Thus by definition of order embedding:
 * $y \preceq x$