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The question is, "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) a is taller than b?

b)a is not taller than b?

c) $a=b$ or a is an ancestor of b?

d) a and b have a common friend?"

The only ones that I am having trouble with are c) and d).

For c, do both conditions have to be meet, or only one?

For d, I can see how it is reflexive; and I can see that isn't antisymmetric. But is it transitive? To me, it would seem like it wasn't, because both a and b could have a different group of friends, the intersection of the groups being the null set.

I would appreciate the help. Thank you!

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For (c) only one condition has to be met: the $a=b$ part is just to allow the ancestor relation to be reflexive. In effect, it’s redefining ancestor to include you as well as your actual ancestors. I agree with you that (d) is not transitive. It’s perfectly possible for $b$ to be a friend of both $a$ and $c$ without $a$ and $c$ being friends. – Brian M. Scott Nov 13 '12 at 20:36
Thank you so much! – Mack Nov 13 '12 at 20:50
You’re welcome! – Brian M. Scott Nov 13 '12 at 20:56
up vote 2 down vote accepted

In $(c)$ You have a disjunctive statement (OR), which holds if either, or both, parts hold.

For $(d)$ Let's say $a \sim b$. That is, Ann ($a$) is friends with Bob ($b$). Can there be another person $c =$ Curt who is friends with Bob, but not with Ann? That is can it be the case that $a\sim b$ AND $b\sim c$ but $a \not\sim c$?

What does that tell you about whether relation $(d)$ is transitive?

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