A choice rule $C$ satisfies path independence if for all $A, B \in 2^X \setminus \emptyset$, $C(A \cup B) = C(C(A) \cup C(B))$. Prove that if $C$ is nonempty and rationalizable, then $C$ satisfies path independence.

My attempt:

Since $C$ is rationalizable, there exists a preference relation $\succeq$ such that $C = C_{\succeq}$. Now let $A, B \in 2^{X} \setminus \emptyset$ and $x \in C(A \cup B) = C_{\succeq}(A \cup B) \subseteq A \cup B$. So $x \succeq y$ for all $y \in A \cup B$. Since $C(A) \cup C(B) = C_{\succeq}(A) \cup C_{\succeq}(B) \subseteq A \cup B$, then $x \succeq y$ for all $y \in C(A) \cup C(B)$, hence $x \in C(C(A) \cup C(B))$, which shows that $C(A \cup B) \subseteq C(C(A) \cup C(B))$. Now let $x \in C(C(A) \cup C(B)) = C_{\succeq}(C_{\succeq}(A) \cup C_{\succeq}(B))$, which implies that $x \succeq y$ for all $y \in C(A) \cup C(B)$. But if $y \in C(A) \cup C(B)$ then this means that $y \succeq z$ for all $z \in A \cup B$. By transitivity, we have $x \succeq z$ for all $z \in A \cup B$, so $x \in C(A \cup B)$. So, $C(C(A) \cup C(B)) \subseteq C(A \cup B)$. Therefore, together we have $C(A \cup B) = C(C(A) \cup C(B))$.

Is my proof correct?

Furthermore, can anyone provide an example of a nonempty choice rule that satisfies path independence but is not rationalizable?

  • $\begingroup$ You may have a look at this paper plott 73. Specifically, Remark 1 establishes that there exists a choice rule which is not rationalizable but which satisfies path independence. $\endgroup$ – Fato Aug 30 '17 at 10:09

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