Question

Determine Poset properties of the given relation.

Short answer

\(R\) is not a poset, as it is not reflexive, not antisymmetric, not transitive.

Step-by-step solution

Given data

\({\rm{R}}\) is the relation, a related to \(b\) if \(a\) and \(b\) have \(a\) common friend.

Concept used of partially ordered set

A relation\(R\)is a poset if and only if,\((x,x)\)is in\({\rm{R}}\)for all x (reflexivity)

\((x,y)\)and \((y,x)\) in R implies \(x = y\) (anti-symmetry),\((x,y)\) and \((y,z)\) in R implies \((x,z)\) is in \({\rm{R}}\) (transitivity).

Find  if the \(R\) is poset

\(R\)is not reflexive as \(x\) and \(x\) always have \(x\) as a common friend.

\(R\)is not antisymmetric as a and \(b\) may not have \(a\) common friend without a and \(b\) being the same person.

\(R\)is not transitive: a and \(b\) may not have common friend, \(b\) and \(c\) may not have a common friend, but a and \(c\) may have a common friend.

\(R\) is not a poset, as it is not reflexive, not antisymmetric, not transitive.