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:

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

$\blacksquare$


Also see