Supremum of Subset Product in Ordered Group
Jump to navigation
Jump to search
Theorem
Let $\struct {G, \circ, \preceq}$ be an ordered group.
Suppose that subsets $A$ and $B$ of $G$ admit suprema in $G$.
Then:
- $\sup \paren {A \circ_\PP B} = \sup A \circ \sup B$
where $\circ_\PP$ denotes subset product.
Proof
Let $a \in A$, $b \in B$.
Then:
| \(\ds a \circ b\) | \(\preceq\) | \(\ds \sup A \circ b\) | Definition of Supremum of Set | |||||||||||
| \(\ds \) | \(\preceq\) | \(\ds \sup A \circ \sup B\) | Definition of Supremum of Set |
Hence $\sup A \circ \sup B$ is an upper bound for $A \circ_\PP B$.
Suppose that $u$ is an upper bound for $A \circ_\PP B$.
Then:
| \(\ds \forall b \in B: \forall a \in A: \, \) | \(\ds a \circ b\) | \(\preceq\) | \(\ds u\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \forall b \in B: \forall a \in A: \, \) | \(\ds a\) | \(\preceq\) | \(\ds u \circ b^{-1}\) | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \forall b \in B: \, \) | \(\ds \sup A\) | \(\preceq\) | \(\ds u \circ b^{-1}\) | Definition of Supremum of Set | |||||||||
| \(\ds \leadsto \ \ \) | \(\ds \forall b \in B: \, \) | \(\ds b\) | \(\preceq\) | \(\ds \paren {\sup A}^{-1} \circ u\) | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \sup B\) | \(\preceq\) | \(\ds \paren {\sup A}^{-1} \circ u\) | Definition of Supremum of Set | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \sup A \circ \sup B\) | \(\preceq\) | \(\ds u\) |
Therefore:
- $\sup \paren {A \circ_\PP B} = \sup A \circ \sup B$
$\blacksquare$