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How can I prove that for any given Poset $(A,\preceq)$, $\preceq$ is a total order implies that $\forall a\in\preceq$, if a is a maximal, then a is maximum? Same goes for minimal/minimum.

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up vote 1 down vote accepted

I assume that you are asking the following:

In a totally ordered set, why is every maximal element a greatest element?

The answer is simple: Assume $a$ is maximal. Let $b$ be an arbitrary element of our set. Since the set is totally ordered set we have either $a\leq b$ or $b\leq a$. Since $a$ is maximal, we must have $b\leq a$. Since $b$ was arbitrary, it follows that $a$ is a greatest element.

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We say that a set $A$ admits a maximal element $a$ with respect to a binary relation $\succeq$ on $A$ if there exists no element $b\in A$ such that $b\succ a$. We want to show that $a\succeq b$ for all $b$ in $A$. Since $\succeq$ total either $a\succeq b$ or $b\succeq a$. Let $b\in A$ be such that $b\succeq a$ since $a$ is maximal not $b\succ a$ therefore whenever $b\succeq a$ a it implies that $b=a$. Hence $a$ is maximum.

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