Properties satisfied by an order relation Let $X$ be a finite set and $2^{X}- \{\emptyset\}$  denote its power set excluding the empty set.

Definition 1: A choice correspondence over $X$ is a map $C:2^{X}- \{\emptyset\}\mapsto 2^{X}- \{\emptyset \} $ such that 
  $C(A)\subseteq A$ for all non-empty $A\subseteq X$.

Consider the following order relation ($\succeq$) defined in $2^{X}- \{\emptyset\}$ as follows:

Let $x^*\notin X$ and $C$ a choice correspondence over $X^*=X\cup\lbrace x^*\rbrace$. Let $A, B$ be non-empty subsets of $X$
$$A\succeq B\; \Longleftrightarrow\;  C(A\cup x^*) - C(A)
\subseteq C(B\cup x^*)- C(B)$$


My question is: What properties does the relation $\succeq$ satisfies?


*

*It is easy to see it is reflexive and transitive.  So it is a pre-order.

*EDIT: The answer by Jing Zhang proves it is not antisymmetric.

*Does it satisfies the following 'upper bound' relation?


$$A\succeq B; A\succeq C\;\implies A\succeq B\cup C$$


*What other 'functional properties' does it satisfy? (Bounty will be offered to the most interesting property and/or the most complete 'characterization')



If we add some structure to $C$ how does the answer change?
For example what if $C$ satisfies any (or all, or some) of the following?


*

*$C(A)\cap C(B)\subseteq C(A\cup B)$ $\quad$  $\;\forall\;A, B\in 2^{X}- \{\emptyset\}$

*$A\subseteq B  \implies C(B)\cap A\subseteq C(A) $ $\quad$  $\;\forall\;A, B\in 2^{X}- \{\emptyset\}$

*$C(B)\subseteq A\subseteq B \implies C(A)\subseteq C(B)$                $\quad$  $\;\forall\;A, B\in 2^{X}- \{\emptyset\}$


Any help or hints would be highly appreciated.

First of all it is not antisymmetric.
Think about X as a finite set, along with $x^*$, let $<$ be a well-order put on $X\cup \{x^*\}$ such that $x^*$ is the third element under $<$ (say there are 5 elements to be concrete, so $x_0<x_1<x^*<x_2<x_3<x_4$.) Then $C$ picks the $<$-least element each time (to be precise, it picks the singleton that contains the $<$-least element). Then $\{x_0\}\preceq \{x_1\}, \{x_1\}\preceq \{x_0\}$ but they are not equal.
Now slightly change $C$ such that it still picks as before except that if the input contains $x_0,x_1$ simultaneously, then it returns the largest element. Now $A=\{x_0\}$ and $B=\{x_0\}, C=\{x_1\}$, we know $B,C\prec A$, but $B\cup C\not \prec A$ as $C(B\cup C \cup \{x^*\})-C(B\cup C)=\{x^*\}, C(A\cup \{x^*\})-C(A)=\emptyset$.
