Supremum Not Less than Infimum

Theorem

Let $\left({S, \preceq}\right)$ be a partially 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.

$\blacksquare$

Sources

• James M. Hyslop: Infinite Series (1942): $\S 3$