Bounded Lattice has Both Greatest Element and Smallest Element

Theorem

Let $\struct {S, \vee, \wedge, \preceq}$ be a bounded lattice.

Then:

$(1): \quad \struct {S, \preceq}$ has a smallest element, namely:

$\bot := \sup \O$

$(2): \quad \struct {S, \preceq}$ has a greatest element, namely:

$\top := \inf \O$

Proof

By definition of bounded lattice:

$\bot = \sup \O$ and $\top = \inf \O$ exist

From Supremum of Empty Set is Smallest Element:

$\bot = \sup \O$ if and only if $\bot$ is the smallest element of $\struct {S, \preceq}$

From Infimum of Empty Set is Greatest Element:

$\top = \inf \O$ if and only if $\top$ is the greatest element of $\struct {S, \preceq}$

$\blacksquare$