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:

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:

Therefore:


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

Also see

 * Infimum of Product