Bottom in Ideal

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

Let $I$ be a ideal in $S$.

Then $\bot \in I$

where $\bot$ denotes the smallest element of $S$.

Proof
By definition of ideal in ordered set:
 * $I$ is non-empty and lower.

By definition of non-empty set:
 * $\exists x: x \in I$

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

Thus by definition of lower set:
 * $\bot \in I$