Supremum of Subset Product in Ordered Group

Theorem
Let $\left({G, \circ, \preceq}\right)$ be a totally ordered group.

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

Then:
 * $\sup \left({A \circ_{\mathcal P} B}\right) = \sup A \circ \sup B$

where $\circ_{\mathcal P}$ denotes subset product.

Proof
It is trivial that $\sup A \circ \sup B$ is an upper bound for $A \circ_{\mathcal P} B$.

Suppose that $u$ is an upper bound for $A \circ_{\mathcal P} B$.

Then:

Therefore:


 * $\sup \left({A \circ_{\mathcal P} B}\right) = \sup A \circ \sup B$

Also see

 * Infimum of Product