Upper Bound is Lower Bound for Inverse Ordering

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

Let $T \subseteq S$.

Let $M$ be an upper bound for $\left({T, \preceq}\right)$.

Let $\succeq$ be the dual ordering of $\preceq$.

Then $M$ is a lower bound for $\left({T, \succeq}\right)$.

Corollary
Let $m$ be a lower bound for $\left({T, \preceq}\right)$.

Then $m$ is an upper bound for $\left({T, \succeq}\right)$.

Proof
Let $M$ be an upper bound for $\left({T, \preceq}\right)$.

That is:


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

By definition of dual ordering, it follows that:


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

That is, $M$ is a lower bound for $\left({T, \succeq}\right)$.

Proof of Corollary
We have that $\succeq$ is an ordering whose inverse is $\preceq$.

The result follows by reversing the argument of the main proof.