In decision theory, condition $ \beta $ is defined as follows: If $a,b \in A \subset B, a, b \in C(A)$, and $b \in C(B)$, then $a \in C(B) $. $C(.)$ here is the choice correspondence of a decision maker from a set.
The decision maker has in mind a complete directed graph $\rightarrow$. If the set A contains an element $a$ such that $a \rightarrow x$ for all $ x \in A − \{a\} $, then $C(A) = \{a\}$. Otherwise, $C(A) = A$. Unless $\rightarrow$ is transitive, the correspondence satisfies condition $\beta$.
I don't understand why this is true. For example, I have a counterexample: for $A \subset B$, $x$ is connected to all elements in B and it is the only element to be so, which makes $C(B) = \{x\}$. Consider another element $y$ that is connected to all elements in $A$. However, since $x$ is connected to all elements in B, it is also connected to all elements in A. Since there are two elements (not one) in A that satisfy the results , $C(A) = A$, which also includes $x$ and $y$. However, $y \notin C(B)$. Where did I go wrong?
I understand "complete directed graph" here to be meant as follows: take a complete graph, and orient each edge as you like. (In other words, a tournament.) In such a case, it is impossible to have $v \to w$ and $w \to v$ at the same time.
The proposed counterexample does not work, because we cannot have two elements $x \in A$ and $y \in A$ such that $x \to a$ for all $a \in A - \{x\}$ and $y \to a$ for all $a \in A - \{y\}$. If this were true, then we would have to have $x \to y$ and $y \to x$, which is impossible. In general, for any tournament and any set of vertices $A$, exactly one of the two cases holds:
Here is a proof that the choice correspondence $C$ defined from a tournament always satisfies condition $\beta$.
Suppose we have two sets $A, B$ with $A \subset B$. If $C(A) = A$ and $C(B)=B$, then condition $\beta$ definitely holds for all $a,b \in C(A)$. If $|C(A)|=1$, then condition $\beta$ holds trivially: there are no pairs $a,b \in C(A)$ for it to apply to. Therefore the only case we can be worried about is the case that $C(A)=A$, but $C(B) = \{x\}$ for some $x \in B$.
What does such a situation mean? It means that $x \to b$ for all $b \in B - \{x\}$. In particular, if $x$ were an element of $A$, because $x \to a$ for all $a \in A-\{x\}$, we'd conclude that $C(A) = \{x\}$ as well. But we're assuming $C(A) = A$, so we conclude that $x$ cannot be an element of $A$. In particular, $x \notin C(A)$, because $x \notin A$.
Therefore condition $\beta$ still holds for the sets $A$ and $B$. It holds because there cannot be $a,b \in C(A)$ with $b \in C(B)$ as well: $b \in C(B)$ can only happen if $b=x$, but $b \in C(A)$ is incompatible with $b=x$.
As a side note, the claim is "unless $\to$ is transitive, $C(\cdot)$ satisfies condition $\beta$", but it satisfies condition $\beta$ if $\to$ is transitive as well. For a transitive tournament, we can number the vertices $x_1, x_2, \dots, x_n$ such that $x_i \to x_j$ whenever $i < j$. In this case, for every set $A$, there will be a vertex $x \in A$ such that $x \to a$ for all $a \in A-\{x\}$: it will be the vertex with the smallest label. Therefore $|C(A)|=1$ for all sets $A$, and condition $\beta$ holds trivially: there are no pairs $a,b$ for it to apply to.