Order Isomorphism Preserves Lower Bounds

Theorem
Let $L = \struct {S, \preceq}$, $L' = \struct {S', \preceq'}$ 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 $X$ $\map f x$ is lower bound for $f \sqbrk X$.

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:
 * $\map f x$ is lower bound for $f \sqbrk X$.

Necessary Condition
Assume that
 * $\map f x$ is lower bound for $f \sqbrk X$.

Let $y \in X$.

By definition of image of set:
 * $\map f y \in f \sqbrk X$

By definition of lower bound:
 * $\map f y \preceq' \map f x$

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