Supremum does not Precede Infimum

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

Let $T \subseteq S$ admit both a supremum $M$ and an infimum $m$.

Then $m \preceq M$.

Proof
By definition of supremum:


 * $\forall a \in T: a \preceq M$

By definition of infimum:


 * $\forall a \in T: m \preceq a$

The result follows from transitivity of ordering.