# Partitioning a set with a relation on it

Let R be a relation on a set A that is reflexive and transitive but not symmetric. Let R(x) = {y: xRy}. Does the set a = {R(x): x ∈ A} always form a partition of A?

I really don't know where to start with this one. I know that R(x) is the same as x/R except R is not an equivalence relation.

HINT: What happens if $R$ is the relation $\le$ on $\Bbb Z$, say? Start by finding $R(0)=\{n\in\Bbb Z:0\le n\}$ and $R(1)$.
• @Robert: In fact you get a partition if and only if $R$ is symmetric. Whenever $R$ is not symmetric, you have some $x,y\in A$ such that $y\in R(x)$ but $x\notin R(y)$, so that $R(x)$ and $R(y)$ overlap without being equal; that’s impossible for a partition. – Brian M. Scott Nov 4 '13 at 6:26