Upper Bound is Lower Bound for Inverse Ordering
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Definition
Let $\left({S, \preceq}\right)$ be an ordered set.
Let $T \subseteq S$.
Let $M$ be an upper bound for $\left({T, \preceq}\right)$.
Let $\succeq$ be the dual ordering of $\preceq$.
Then $M$ is a lower bound for $\left({T, \succeq}\right)$.
Proof
Let $M$ be an upper bound for $\left({T, \preceq}\right)$.
That is:
- $\forall a \in T: a \preceq M$
By definition of dual ordering, it follows that:
- $\forall a \in T: M \succeq a$
That is, $M$ is a lower bound for $\left({T, \succeq}\right)$.
$\blacksquare$
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