Lower Bound is Upper 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 a lower bound for $\left({T, \preceq}\right)$.

Let $\succeq$ be the dual ordering of $\preceq$.


Then $m$ is an upper bound for $\left({T, \succeq}\right)$.


Proof

Let $m$ be a lower bound for $\left({T, \preceq}\right)$.

That is:

$\forall a \in T: m \preceq a$

By definition of dual ordering, it follows that:

$\forall a \in T: a \succeq m$

That is, $M$ is an upper bound for $\left({T, \succeq}\right)$.

$\blacksquare$


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