A model of extensionality Suppose I have the set
$X = \{\varnothing ,\{ \varnothing\} ,\{\{\varnothing\}\} , \{\{\{\varnothing\}\}\} , ..., x_n , \{x_n\}, ... \}$
My book asks which finite and which infinite subsets are models of extensionality? If $Y \subseteq X$ then $Y \vDash$ extensionality iff $Y \cap ((x\setminus y) \cup (y\setminus x)) \ne \varnothing$ for all distinct $x, y \in Y$. It seems to me that all non-empty subsets are models of extensionality, whether they are finite or infinite (except that finite sets must have at least two elements). If $x=x_m$ and $y=x_n$ (the $m$ and $n$ indicate the number of braces around the empty set), then $x_m \setminus x_n = x_{m-n-1} \ne \varnothing$ for $m > n$; and $x_m \setminus x_n = x_m \ne \varnothing$ for $m < n$. 
 A: Note that extensionality holds when $(x\setminus y) \cup (y\setminus x)$ is not empty for $Y$ for $x\ne y$, that means that if $y\ne x$, then there exists some $z\in Y(!)$ such that $z\in x\setminus y$ or $z\in y\setminus x$.
Take $Y=\{\emptyset,\{\{\emptyset\}\}\}$ for example, in this case both $\emptyset$ and $\{\{\emptyset\}\}$ contains no elements in $Y$, so both are empty sets for $Y$, and so if $x=\emptyset,y=\{\{\emptyset\}\}$, we have $\left((x\setminus y) \cup (y\setminus x)\right)^Y$ is empty so extensionality does not hold.
A: After thinking about @Holo's reply, I have the full answer to my question (I think):


*

*The only finite subsets of $X$ that model extensionality are the sequential subsets $Y=\{x_0, x_1, ..., x_n \}$ where $x_0=\varnothing$ and $x_{m+1}=\{x_m\}$. If an element $x_m$ for $m < n$ is missing from $Y$, then $x_m \notin Y$ but $x_{m+1}\in Y$, in which case $Y \cap (x_{m+1} \setminus x_0) = Y \cap x_{m+1} = \varnothing$ (and $x_0 \setminus
x_{m+1}$ is itself empty). 

*The only infinite subset of $X$ that models extensionality is $X$ itself. Again, this is because the elements must be sequential as in the finite argument.
