# Supremum of Product

## Theorem

Let $\left({G, \circ, \preceq}\right)$ be an 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:

 $\displaystyle \forall b \in B: \forall a \in A: \ \$ $\displaystyle a \circ b$ $\preceq$ $\displaystyle u$ $\displaystyle \implies \ \$ $\displaystyle \forall b \in B: \forall a \in A: \ \$ $\displaystyle a$ $\preceq$ $\displaystyle u \circ b^{-1}$ $\displaystyle \implies \ \$ $\displaystyle \forall b \in B: \ \$ $\displaystyle \sup A$ $\preceq$ $\displaystyle u \circ b^{-1}$ Definition of Supremum $\displaystyle \implies \ \$ $\displaystyle \forall b \in B: \ \$ $\displaystyle b$ $\preceq$ $\displaystyle \left({\sup A}\right)^{-1} \circ u$ $\displaystyle \implies \ \$ $\displaystyle \sup B$ $\preceq$ $\displaystyle \left({\sup A}\right)^{-1} \circ u$ Definition of Supremum $\displaystyle \implies \ \$ $\displaystyle \sup A \circ \sup B$ $\preceq$ $\displaystyle u$

Therefore:

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

$\blacksquare$