If $\succsim$ is rational, then $B \mapsto C^*(B, \succsim)$ satisfies the weak axiom, and $\succsim=\succsim^*$

Question

If $\succsim$ is rational, then $B \mapsto C^*(B, \succsim)$ satisfies the weak axiom, and $\succsim=\succsim^*$

Previously in the same theorem actually, they proved the following:

If $C$ is a choice rule satisfying the weak axiom, then $\succsim^*$ is rational, and for all $B \in \mathcal{P}(X), C(B) = C^*(B, \succsim^*).$

Note: $C^*(B) = C^*(B, \succsim) = \{ x \in B: \forall y \in B, x \succsim y \}$, and the weak axiom is referring to the weak axiom of revealed preference which a choice rule satisfies if $[x \succsim^* y ] \Rightarrow \neg (\exists B)[y \succ_B x]$

My try of the proof:

First suppose $B \mapsto C^*(B, \succsim)$ does not satisfy the weak axiom, then $\succsim^*$ is not rational by the earlier part of the theorem, so if we can prove $\succsim = \succsim^*$ we have a contradiction which will prove the statement.

As $\succsim$ is rational, then there is a utility function $u: X \to \mathbb{R}$ that represents $\succsim$, then $\{x \in B: \forall y \in B, u(x) \geq u(y) \} = C^*(B, \succsim)$ but then $\exists B$ s.t. $x,y \in B$ and since $B \mapsto C^*(B, \succsim)$ then $u(x) \in C^*(B, \succsim)$ so $x \in C(B)$ hence $\succsim = \succsim^*$

Answer 1

Here is an attempt at a direct proof:

Let all the assumptions hold, and further assume that $x \succcurlyeq^* y$. Then for some $B \supseteq \{x,y\}$, $C^*(B) \supseteq \{x\}$ (by definition of $\succcurlyeq^*$), which further implies $x \succcurlyeq z$ for all $z \in B$, and therefore $x \succcurlyeq y$ (by definition of $C^*$). (This shows that $\succcurlyeq^* \subseteq \succcurlyeq$; the other direction follows from looking at doubleton sets $B = \{x,y\}$).

What is left to show is that $C^*$ satisfies the weak axiom. So take any $B' \supseteq \{x,y\}$, such that $y \in C^*(B')$. If no such set exists, then $y$ is not revealed preferred to $x$ and we are done. If such a set exists then by definition $y \succcurlyeq z$ for all $z \in B'$. We already know $x \succcurlyeq y$ so by transativity $x \succcurlyeq z$ for all $z \in B'$ and hence $x \in C^*(B')$. Again, $y$ is not revealed preferred to $x$ and we are done.

Notice, while we used transitivity, we did not use completeness. This hints at the fact that the statement can actually be generalized to pre-orders (reflexive and transitive), by allowing $C^*$ to be empty. This point is somewhat lost by appealing to the second result.