TITLE: Is Robinson Arithmetic biinterpretable with some theory in LST?
QUESTION [16 upvotes]: Let ZFC$^{\text{fin}}$ be ZFC minus the axiom of infinity plus the negation of the axiom of infinity.  It is well-known that ZFC$^{\text{fin}}$ is biinterpretable with Peano Arithmetic.  In this sense one could argue that ZFC$^{\text{fin}}$ is PA couched in the language of set theory (ie one nonlogical binary relation, $\in$) rather than the language of arithmetic ($+$, $\cdot$, $0$, $S$).  This gives us some confidence that "there exists an infinite set" -- and the hierarchy of large cardinal axioms beyond -- is an at least somewhat-natural extension of arithmetic.
In precise terms, every theory in this hierarchy proves the consistency of all those before it.  In vague terms, each theory in this hierarchy adds "more infiniteness" than those before it. 
Does the hierarchy start at PA, or is there a step below it?  Robinson Arithmetic is a theory in the language of arithmetic; among its properties are:

Robinson Arithmetic is essentially undecidable (as PA and all stronger theories are)
PA proves the consistency of Robinson Arithmetic
Robinson Arithmetic is finitely axiomatizable

The first point might be considered an argument for why Robinson Arithmetic is part of the same hierarchy as PA/ZFC$^{\text{fin}}$ -- it has enough coding power to express primitive recursion.  The second point shows why Robinson Arithmetic is strictly below PA/ZFC$^{\text{fin}}$ on this hierarchy.  The third point explains -- in vague terms -- what sort of "infiniteness" PA/ZFC$^{\text{fin}}$ add to Robinson Arithmetic: it adds infinite collections of axioms.
From PA on up, all theories on the hierarchy are biinterpretable with some theory in the language of set theory.

Question: is Robinson Arithmetic biinterpretable with some theory in the language of set theory?

REPLY [5 votes]: Since the other (excelent) answers didn’t touch the biinterpretability aspect of the question, let me comment on that.
First, a caveat: arbitrary structures (say, in a finite language) can be encoded by directed graphs. Thus, it might very well be the case that $Q$ is biinterpretable with a theory in a language with one binary predicate (i.e., in the language of set theory) which looks nothing like a “set theory” at all, and is not included in or consistent with any recognizable theory of sets like ZFC.
So, a more natural question is if $Q$ is biinterpretable with any theory that contains at least a modicum of set theory. While I didn’t exactly specified what this means, I propose that the answer is negative because of the following result of Albert Visser: $Q$ is not biinterpretable with any theory which includes the axiom of pairing $\forall x,y\,\exists z\,\forall t\,(t\in z\leftrightarrow t=x\lor t=y)$. See On $Q$.