Extensionality of a relation and Axiom of Extensionality

Question

What's the difference between the Axiom of Extensionality $(A1)$ and an extensional relation?

The definitions are

$(A1) \forall x,y ( x = y \leftrightarrow \forall z ( z \in x \leftrightarrow z \in y ))$

and

Let $W$ be a binary relation on a set $Y$. The relation $W$ is called extensional if $\forall x,y \in Y ( x \neq y \rightarrow \exists z \in Y (( \langle z,x \rangle \in W \land \langle z,y \rangle \notin W) \lor (\langle z,x \rangle \notin W \land \langle z,y \rangle \in W))$.

Writing the negations inside the definition above as not-negations one can restate it as

Let $W$ be a binary relation on a set $Y$. The relation $W$ is called extensional if $\forall x,y \in Y ( x = y \leftarrow \forall z \in Y ( \langle z,x \rangle \in W \leftrightarrow \langle z,y \rangle \in W) $.

Well. Of course $\rightarrow$ is always true so that extensionality of a relation seems to be the same as the axiom of exteniosnality:

Let $W$ be a binary relation on a set $Y$. The relation $W$ is called extensional if $\forall x,y \in Y ( x = y \leftrightarrow \forall z \in Y ( \langle z,x \rangle \in W \leftrightarrow \langle z,y \rangle \in W) $.

But I must be missing something since I just proved the following claim (which was an exercise in Just/Weese):

Claim: Let $N$ be a set. Then $\langle N , \overline{\in} \rangle \models (A1)$ iff $\overline{\in}$ is an extensional relation on $N$.

Here "$\overline{\in}$" is used to indicated that $N$ is a standard model and $\overline{\in}$ is the actual $\in$ on $N$. But if my musings above are right then of course this claim holds. Furthermore, a much more general claim would hold:

If $N$ is a class then $\langle N , E \rangle \models (A1)$ iff $E$ is an extensional relation on $N$.

That is, even if $N$ is a non-standard class model of set theory, the claim still holds.

Where did I go wrong? Would you point out the difference between $(A1)$ and an extensional relation to me? Many thanks for your help.

Answer 1

The axiom of extensionality concerns the universe of all sets. An extensional relation is defined on a set. Formally, the relation is a subset of $W\times W$, but $\epsilon\subseteq V\times V$ is formally meaningless.

That the concepts are very similar is of course on purpose.

Answer 2

"Extensionality of a relation seems to be the same as the axiom of exteniosnality". Not so.

An observation that might help. Consider the promiscuous relation $U$ on $Y$ which relates any member of $Y$ to any other member of $Y$. So $U = \{\langle x, y\rangle\,|\,x, y \in Y\}$. Then $U$ isn't an 'extensional relation', by the definition. But of course the axiom of extensionality applies to $U$ as to any set.

(The idea is that an extensional relation $W$ treats distinct $x$ and $y$ differently -- there is some difference between the objects that $W$ relates to $x$ and the objects $W$ relates to $y$: and not every relation on $Y$ is like that.)

Answer 3

I think you're overthinking things here.

Yes, it is completely right that

If $\mathbf N$ and $\mathbf E$ are classes, then $\langle\mathbf N,\mathbf E\rangle\vDash (A1)$ iff $\mathbf E$ is an extensional relation on $\mathbf N$.

But that doesn't seem to contradict anything else you've said or been asked to do here.

Why do you think you've "gone wrong" somewhere? It looks like you've convinced yourself that you must be misunderstanding something simply because you find the exercise easy and can see, correctly, that some of its assumptions can be relaxed. This is a fallacy.