If $Y$ not transitive then $\in$ not extensional? Let $W$ be a binary relation on a set $Y$. The relation $W$ is called extensional if $$ \forall x,y \in Y (x \neq y \rightarrow \exists z \in Y (( \langle z,x \rangle \in W \land \langle z,y \rangle \notin W) \lor ( \langle z,x \rangle \notin W \land \langle z,y \rangle \in W )))$$ Consider the $\in$ relation. Let $Y$ be a set that is not transitive. This means that there is $y$ in $Y$ such that $x \in y$ but $x \notin Y$. (Right?) How does this make $\in$ non-extensional? (As I understand extionsionality means that two sets are equal if and only if they contain the same elements. How is this violated if $\in$ is not transitive?) Thanks. Here is a copy of the exercise, page 64, Just/Weese: !enter image description here

Consider $Y=\\{\varnothing,\\{1\\}\\}$.

From the point of view of $Y$ neither contain any elements, because $1\
otin Y$. But these are different sets.

To say that $\langle Y,\in\rangle$ is extensional is to say that the following is true: $$\forall x\in Y\forall y\in Y(x=y\leftrightarrow\forall z\in Y(z\in x\leftrightarrow z\in y))$$

This clearly fails in our case.