in Set Theory & Algebra
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Is (S, R) a poset if S is the set of all people in the world and (a, b) R, where a and b are people,
if a is not taller than b?
 

in Set Theory & Algebra
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No. It is not a POSET. The "is not taller than" relation looks like "less than or equal to" operator. And it is already known fact (from wikipedia) that the real numbers ordered by the standard less-than-or-equal relation ≤ is a POSET. However POSET has a requirement that the relation must be ANTISYMMETRIC. That is for (S, R) to be POSET       a R b   &&   b R a . =>  a=b.

If we look at the real numbers ordered by the ≤ relation, then obviously consider two numbers a and b. If a  ≤ b and b ≤ a => a=b.

But this does not hold if a and b are "persons". In this question the relation R is "not taller than". It is not required that for two persons a and b of the same height to be exactly the same person. Consider a=You and b=Me. Suppose both of us have the same height of 4 feet :-) Now obviously a not equals b. But R(a, b) and R(b, a) both holds (Neither of us is taller than the other). Thus the antisymmetric nature of R is violated. Consequently it is not a POSET
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good way of explaining :)
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