Infimum 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 infima in $G$.

Then the infimum of the subset product $A \circ_{\mathcal P} B$ exists and is equal to $\inf A \circ \inf B$.

Proof
Follows by the same argument as in Supremum of Product, with the symbol $\preceq$ replaced with $\succeq$, and the supremum replaced with the infimum.

Also see

 * Supremum of Product