# Supremum of Product

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## 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:

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

$\blacksquare$