Bottom is Way Below Any Element

Theorem
Let $\left({S, \preceq}\right)$ be a bounded below ordered set.

Let $x \in S$.

Then
 * $\bot \ll x$

where $\bot$ denotes the bottom in $S$.

Proof
Let $D$ be directed subset of $S$ such that
 * $D$ admits a supremum and $x \preceq \sup D$

Because $D$ is non-empty, therefore
 * $\exists a: a \in D$

By definition of bottom:
 * $\bot \preceq a$

Thus by definition of way below relation;
 * $\bot \ll x$