trying to work out this problem, please find the theorem, the problem and my attempt below.
Theorem: Let Z be a finite set and let X be the set of all non-empty subsets of Z. Let $ \ge $ be a preference relation on X (not on Z). Let A, B, C $ \in$ X and consider the following two properties of the preference relation on X:
i) If A $ \ge $ B and C is disjoint to both A and B then $ A \cup C \ge B \cup C$. Also, if A $\gt$ B and C is a disjoint set then $A \cup C \gt B \cup C$.
ii) If $x \in Z$ and {x} > {y} $ \forall y \in A$ then $ A \cup ${x} > A
Problem: Show that if there are an x, y, z $\in$ Z such that {x} > {y} > {z} then there is no preference relation satisfying both properties.
My attempt: If we let A = x, B = y and C = Z then all the conditions specified are satisfied, as for the properties, I get that the first one holds, and not sure about the second. Any help would be great.