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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.

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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 $R(1)$.

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  • $\begingroup$ I think i see where to go. I went through it and got that it did not always form a partition. $\endgroup$ – Robert Eccain Nov 4 '13 at 6:19
  • $\begingroup$ @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. $\endgroup$ – Brian M. Scott Nov 4 '13 at 6:26
  • $\begingroup$ @Robert: You’re very welcome. $\endgroup$ – Brian M. Scott Nov 4 '13 at 6:35

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