Bottom in Compact Closure

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

Let $x \in S$.

Then $\bot \in x^{\mathrm{compact} }$

where $\bot$ denotes the smallest element in $L$,
 * $ x^{\mathrm{compact} }$ denotes the compact closure of $x$.

Proof
By Bottom is Compact:
 * $\bot$ is a compact element.

By definition of the smallest element:
 * $\bot \preceq x$

Thus by definition of compact closure:
 * $\bot \in x^{\mathrm{compact} }$