Reformulation of the Weak Axiom of Revealed Preference

Question

This question about foundations of mathematical economics.

Let $X$ be some set, $\mathcal{B}\subset 2^{X}$ and $C:\mathcal{B}\rightarrow 2^{X}$ such that for all $B\in\mathcal{B}$ we have

1) $C(B)\neq\emptyset$;

2) $C(B)\subset B$.

The Weak Axiom of Revealed Preference (see http://en.wikipedia.org/wiki/Revealed_preference) states that for all $B_{1},B_{2}\in\mathcal{B}$ with $x,y\in B_{1}$, $x,y\in B_{2}$ and $x\in C(B_{1})$, $y\in C(B_{2})$ it holds that $x\in C(B_{2})$.

My question is following

Does there exists purely set-theoretic reformulation of the Weak Axiom of Revealed Preference?

I'm looking for something like this

$$B_{1},B_{2}\in\mathcal{B},\ x\in C(B_{1})\cap B_{2},\ y\in C(B_{2})\cap B_{1}\Rightarrow \{x,y\}\in C(B_{1})\cap C(B_{2}).$$

but in more "closed-formula" way (i.e. without $x$ and $y$).

Answer 1

The axiom says in effect that for any $B_1,B_2\in\mathscr{B}$, if $$C(B_1)\cap(B_1\cap B_2)\ne\varnothing\ne C(B_2)\cap(B_1\cap B_2)\;,$$ then $C(B_1)\subseteq C(B_2)$. Since it’s symmetric in $B_1$ and $B_2$, it’s equivalent to the statement that for any $B_1,B_2\in\mathscr{B}$, if $$C(B_1)\cap(B_1\cap B_2)\ne\varnothing\ne C(B_2)\cap(B_1\cap B_2)\;,$$ then $C(B_1)=C(B_2)$. Since $C(B)\subseteq B$ for each $B\in\mathscr{B}$, this can be reduced to the statement for any $B_1,B_2\in\mathscr{B}$, if

$$C(B_1)\cap B_2\ne\varnothing\ne C(B_2)\cap B_1\;,$$

then $C(B_1)=C(B_2)$:

$$\forall B_1,B_2\in\mathscr{B}\Big(C(B_1)\cap B_2\ne\varnothing\ne C(B_2)\cap B_1\to C(B_1)=C(B_2)\Big)\;.$$