TITLE: Definability using rudimentary function
QUESTION [6 upvotes]: Denote by RUD the set of all rudimentary functions, together with the function that takes any set to its transitive closure. 
Assume that I know that a binary relation $R$ is definable by some function in RUD. Is there a way to define $R^+$, the transitive closure of the relation $R$, using functions in RUD? Maybe there isn't one, but I strongly feel there should be a way and I cannot seem to find it.
I can be even more specific, I want to be able to define the function that given a set $x$ returns the set $\{ y | R^+(x,y) \}$, given that I know there is a function in RUD which defines the function that takes $x$ to $\{ y | R(x,y) \}$.
Thanks for your help!

REPLY [8 votes]: If I understand you correctly I believe the answer is no.  Let $S(x)$ be the rud. function that returns $x \cup \{x\}$.  Let $R(x,y) \leftrightarrow y = S(x)$.  Then $F(x) = \{y\mid R(x,y)\} = \{S(x)\}$ is rud.  But $G(x)=\{y\mid R^+(x,y)\}\supseteq \{y \mid \exists p<\omega\,\, y = SS...SS(x)\, (p \mbox{ times }) \}$.  However then there is no $q$ so that $\forall x : rk(G(x))\leq rk(x)+q$.  Hence $G$ cannot be rud.