quasi rationality, interesting axiom of revealed preferences So imagine there is a notion of rationality that captures the idea of "thresholds in preference." For example, let $\mathbb{Z}$ be the integers: $\mathbb{Z} = \{\dots, -10, -9, \dots, 0, 1, 2, \dots\}$ and write $xBy$ if $x-1 > y$. Thus $9B7$, $9B6$, and $9B5$ but not $9B8$. $B$ is transitive and irreflexive ($\forall x$ not $xBx$). Since it is irreflexive it is not complete.
Define the choice function generated by $B$ as follows. For all $A \in \mathcal{P}(\mathbb{Z})$ define $$c(A, B) = \{x \in A \text{ }|\text{ there is no }y \in A \text{ s.t. }yBx\}.$$Can we redo the standard theory of rational preferences by inventing a condition a condition on choice functions $c: \mathcal{P}(\mathbb{Z}) \to \mathcal{P}(\mathbb{Z}$ such that if (i) if $B$ is a transitive and irreflexive relation, then $c(\cdot, B)$ satisfies the condition and (ii) if $c$ satisfies the condition, then $c$ is transitive and irreflexive generated (i.e., there exists a transitive and irreflexive relation $B$ such that for all $A \subset X, A \neq X$, it is the case that $c(A, B) = c(A)$?
 A: Addressing ii.) first, any $c(A)$ can be generated by a transitive and irreflexive relation $B$ so long as it is nonempty for finite $A$. Given nonempty $c(A)$, let $xBy$ for any $x\in c(A)$ and $y\in A-c(A)$. Then leave all other comparisons incomplete. This is transitive and then we can add not $xBx$ for irreflexivity.
If $c(A)$ is empty, then for any $a\in A$, there exists $x\in A$ such that $xBa$. Continuing with the assumption that $c(A)$ is empty, there must be a $z\in A$ such that $zBx$. We cannot have $z=a$ or else $xBx$ by transitivity. So, $z$ is unique. Proceeding along, there must always exist a new point in $A$, so $A$ must not be finite.
i.) I'm not immediately sure if there's anything interesting or surprising here. WARP won't hold of course. If $x,y\in U,S$ and $x\in c(U,B), y\in c(S,B)$ then neither $x\in c(S,B)$ nor $y \in c(U,B)$ need hold. 
Consider $S=\{-4,-2,0,2\}$ and $\{-2,0,2,4\}$. Let $B$ relate only integers of the same sign (call zero positive for the sake of it). Let $xBy$ iff $|x|>|y|$. Then $c(S,B)=\{-4,2\}$ and $c(U,B)=\{-2,4\}$. 
If you want to address revealed preferences, you'll have to weaken the weak axiom. But to me, it's not clear what would be left with any revealed preference flavor without additional completeness assumptions. 
At least there is the property where if $x,y\in c(A,B)$, then we cannot have $x\in c(A\cup \{z\}, B)$, not $z\in c(A \cup \{z\}, B)$, and not $y \in c(A \cup \{z\}, B)$.
Then $zBy$, $aBz$ for some $a\in A$. By transitivity, $aBy$, which rules out $y\in c(A,B)$. So this is something like the the independence of irrelevant alternatives.
