Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

The definitions are

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


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.

share|improve this question

3 Answers 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.

share|improve this answer

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.

share|improve this answer

"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.)

share|improve this answer
Huh? If $Y$ has more than one member, then $\langle Y,U\rangle$ is not a model of the Axiom of Extensionality. –  Henning Makholm Dec 20 '12 at 12:54
@HenningMakholm And I didn't say it was! :-) Look again at that sentence from the OP which I quoted. It suggests (I thought/still think, maybe wrongly) that saying a relation $W$ is extensional says nothing over and above what is already given by the axiom of extensionality applied to the set $W$. The simple example $U$ shows not so. So my answer was addressing that first part of the OPs remarks: of course you are quite right in your response to the second part. –  Peter Smith Dec 20 '12 at 13:23
I don't see what your point is, unless you're deliberately making "the axiom of extensionality applied to the set $W$" mean something different from its natural meaning, i.e. "the axiom of extensionality evaluated in an interpretation of the language of set theory that has $W$ as its membership relation". –  Henning Makholm Dec 20 '12 at 14:03
@HenningMakholm I think you may be over interpreting what is going on. Forget everything after the OP's "But I must be missing something." What is going on before? The OP seems puzzled about the defn of an 'Extensional relation' (capitalise for this notion). He seems to think that because of extensionality (in the sense of A1) it is trivially true that all relations (being just more sets) are Extensional. Not so. U is a binary (extensional!) relation on Y but not Extensional. No? Perhaps the OP can help us out at this point!! –  Peter Smith Dec 20 '12 at 15:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.