Preference Maximizing Choice Rule Definition: A Choice Rule is a function $ C: \mathcal{P}(X) \to \mathcal{P}(X) $ such that $ C(B) \subset B, $ $\forall B \in \mathcal{P} (X) $ and $ C(B) \neq \emptyset $ if $ B \neq \emptyset $
The interpretation is that $C(B)$ is the set of options that may be chosen from the menu of $B$ options.
Let $C^*(B)$ be the class of choice rules such that $C^*(B)=C^*(B,\succsim)=\{x\in B : \forall y \in B, x \succsim y \}$
This is called the Preference Maximizing Choice Rule
And define $ \displaystyle S^* = \cap_{x \in B} \{y \in B: y \succsim x \} $
Show $S^* = C^*(B)$
Tips and hints how to show this please, I am confused how the x's and y's may change position going from one side to the other.
 A: It seems that the chosen variables are causing some confusion, so allow me to rewrite the problem slightly to dissipate this problem.

Define $C^*(B)=\{y\in B\colon \forall x \in B(y\succsim x)\}$ (note that this is exactly the same set as in your question, despite $x$ and $y$ having swapped their roles) and $S^*=\bigcap\limits_{x\in B}\left(\{y\in B\colon y\succsim x\}\right)$.
Prove that $C^*(B)=S^*$

Start by noting that $\forall s\left(s\in S^*\iff \forall x\in B\left(s\in \{y\in B\colon y\succsim x\}\right)\right)$
$\bbox[5px,border:2px solid #000000]{C^*(B)\subseteq S^*}$
Let $y\in C^*(B)$. One gets $y\in B$ and $\color{blue}{\forall x\in B(y\succsim x)}$.
(Remember that the goal is to prove that $\forall x\in B\left(y\in \{y\in B\colon y\succsim x\}\right)$).
Take $x\in B$. Due to the blue hypothesis above, $y\succsim x$ follows. But this means that $y\in \{y\in B\colon y\succsim x\}$. Since $x$ was arbitrary, this means that $y\in S^*$.
I'll leave the other inclusion for you.
A: Thank you, the second answer helps clear somethings for me, I'll try and answer the other direction with my comments to see how well I understand this.
$\boldsymbol{\Leftarrow C^*(B) \subseteq S^*}$
Let $s \in C^*(B)$ Then $s \in \{z \in B: \forall y \in B, z \succsim y \}$ This says that z are all of the elements of B that are preferred to all of the y elements in B, and s is one such of these elements equivalent to z.
We want to prove $s \in S^*$ which can be done by showing $s \in \cap_{y \in B} \{ z \in B: z \succsim y \}$ i.e. $\forall y \in B, s \in B$ and $s \succsim y.$  Here we replaced y with z, and x with y from the definition of $S^*$.
Take $y \in B$ by the hypothesis this gives us $s \in B$ and $\forall y \in B, s \succsim y$, as desired.  Please explain what "take" means, is it like consider the arbitrary element y in an arbitrary set B?
