Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
3  
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

1 Answer 1

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?

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.