Duality Principle (Order Theory)/Local Duality

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This proof is about Local Duality in Order Theory. For other uses, see Duality Principle.

Theorem

Let $\Sigma$ be a statement about ordered sets (whether in natural or a formal language).

Let $\Sigma^*$ be the dual statement of $\Sigma$.

Let $\left({S, \preceq}\right)$ be an ordered set, and let $\left({S, \succeq}\right)$ be its dual.


Then the following are equivalent:

$(1): \quad \Sigma$ is true for $\left({S, \preceq}\right)$
$(2): \quad \Sigma^*$ is true for $\left({S, \succeq}\right)$


Proof

$(1)$ implies $(2)$

By assumption, $\Sigma$ is true for $\left({S, \preceq}\right)$.

By Dual of Dual Ordering, the dual statement $\Sigma^*$ applied to $\left({S, \succeq}\right)$ is the same as $\Sigma$ applied to $\left({S, \preceq}\right)$.


Hence $\Sigma^*$ is true for $\left({S, \succeq}\right)$.

$\Box$


$(2)$ implies $(1)$

From Dual of Dual Statement (Order Theory), $\left({\Sigma^*}\right)^* = \Sigma$.

From Dual of Dual Ordering, $\left({S, \preceq}\right)$ is the dual of $\left({S, \succeq}\right)$.


The result thus follows from applying the other implication to $\Sigma^*$ and $\left({S, \succeq}\right)$.

$\blacksquare$


Also see