Supremum of Subset Product in Ordered Group

Theorem

Let $\struct {G, \circ, \preceq}$ be an ordered group.

Suppose that subsets $A$ and $B$ of $G$ admit suprema in $G$.

Then:

$\sup \paren {A \circ_\PP B} = \sup A \circ \sup B$

where $\circ_\PP$ denotes subset product.

Proof

Let $a \in A$, $b \in B$.

Then:

 $\ds a \circ b$ $\preceq$ $\ds \sup A \circ b$ Definition of Supremum of Set $\ds$ $\preceq$ $\ds \sup A \circ \sup B$ Definition of Supremum of Set

Hence $\sup A \circ \sup B$ is an upper bound for $A \circ_\PP B$.

Suppose that $u$ is an upper bound for $A \circ_\PP B$.

Then:

 $\ds \forall b \in B: \forall a \in A: \,$ $\ds a \circ b$ $\preceq$ $\ds u$ $\ds \leadsto \ \$ $\ds \forall b \in B: \forall a \in A: \,$ $\ds a$ $\preceq$ $\ds u \circ b^{-1}$ $\ds \leadsto \ \$ $\ds \forall b \in B: \,$ $\ds \sup A$ $\preceq$ $\ds u \circ b^{-1}$ Definition of Supremum of Set $\ds \leadsto \ \$ $\ds \forall b \in B: \,$ $\ds b$ $\preceq$ $\ds \paren {\sup A}^{-1} \circ u$ $\ds \leadsto \ \$ $\ds \sup B$ $\preceq$ $\ds \paren {\sup A}^{-1} \circ u$ Definition of Supremum of Set $\ds \leadsto \ \$ $\ds \sup A \circ \sup B$ $\preceq$ $\ds u$

Therefore:

$\sup \paren {A \circ_\PP B} = \sup A \circ \sup B$

$\blacksquare$