Question

Determine Poset properties of the given relation.

Short answer

\(R\) is not a poset, as it is 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 reflexive as \((x,x)\) is in \(R\) for all \(x\).

\(R\)is not antisymmetric as a and b can have a common friend without a being equal to b.

\(R\)is not transitive: a and \(b\) may have \(a\) common friend, \(b\) and \(c\) may have a common friend, without a and \(c\) having a common friend.

\(R\) is not a poset, as it is not anti symmetric, not transitive.