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:

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

$\blacksquare$