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 smallest element 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 smallest element:

$\bot \preceq a$

Thus by definition of way below relation;

$\bot \ll x$

$\blacksquare$

Sources

• 1980: G. Gierz, K.H. Hofmann, K. Keimel, J.D. Lawson, M.W. Mislove and D.S. Scott: A Compendium of Continuous Lattices

• Mizar article WAYBEL_3:4