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Show there exist partially ordered sers with more than one maximal element and /or more than one minimal element. This is not a Homework question so completeness is appreciated.

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Let $X$ be any set with more than one element, consider the relation $\preceq$ to be simply $\{\langle x,x\rangle \mid x\in X\}$, that is the identity relation.

Then $\preceq$ is certainly reflexive and it is also transitive and symmetric by vacuous arguments. If $x\preceq y$ then $x=y$ and therefore $y\preceq x$ as well; similarly for transitivity.

In this relation every element is both minimal and maximal. Since we assumed $X$ has more than one element, there is more than one minimal and more than one maximal.

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Take any set $X$ with more than one element and order $\mathcal {P}(X)\setminus \{\varnothing,X\}$ by inclusion.


The non trivial divisors of any natural composite number which isn't a power of a prime number with the order $a|b\iff (\exists k\in\Bbb N)(b=ak)$, for any $a,b\in \Bbb N$.

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  • $\begingroup$ $+1^{{{+^+}^+}^+}\quad\ddot\smile\quad$ $\endgroup$ – Namaste Apr 18 '13 at 17:11

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