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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:

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    Note that the Exercise does not say that _all_ non-transitive sets have non-extensional $\in$-relations. But only that _some_ do (this is the phrase "$\in$ _may not_ be an extensional relation on $Y$" [emphasis mine]). Read Asaf's answer below.2012-11-12

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

From the point of view of $Y$ neither contain any elements, because $1\notin Y$. 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.

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    @AsafKaragila : )2012-11-12