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Am I correct in saying that, given a and b are elements of S, the join(a, b) and meet(a, b) won't be either a or b, if there is no order between a and b?

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In fact, $a\land b = a$ if and only if $a\leq b$; $a\lor b = a$ if and only if $a\geq b$. So if $a$ and $b$ are not comparable in the lattice, then you are correct, $a\land b\notin\{a,b\}$ and $a\lor b\notin\{a,b\}$. –  Arturo Magidin Nov 20 '11 at 6:33

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up vote 4 down vote accepted

(So the question doesn't remain without answers).

Yes, you are correct. If $a$ and $b$ are comparable, say $a\leq b$, then $a\land b = a$ and $a\lor b= b$; conversely, since $a\land b\leq a$ and $a\land b\leq b$ for all $a$ and $b$, if $a\land b\in\{a,b\}$, then $a$ and $b$ are comparable (similar if $a\lor b\in{a,b}$).

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thanks! I was just waiting for you to make an answer so I could checkmark it hehe. –  cesar Nov 21 '11 at 7:49

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