# Is set equality a primitive notion or a relation whose definition is given by the axiom of extensionality?

I saw some sources claiming that the equality of sets is a primitive notion (i.e. according to Wikipedia, a concept which is not defined in terms of previously defined concepts). However, the axiom of extensionality gives that two sets are said to be equal iff $$\forall x(x \in A \iff x \in B)$$ . Isn't set equality then a concept which can be defined in terms of the the primitive notion of set equality together with the concepts from first order logic? (i.e. isn't set equality not a primitive notion?)

• Im not sure; I havent studied ZFC well enough. Frankly Im not sure what the distinction is between a primitive notion and an axiomatic one. In my mind they are one and the same. I do know that the concept of element containment within a set is pretty fundamental. From that you can define set containment (i.e. subsets). And set equality is thus definable as mutual containment. Jan 22 at 19:31
• Whether equality is a primitive notion or defined from the membership relation depends on the formulation… it’s not a question that has a right or wrong answer. It is possible to formulate set theory in first order logic without equality and then define equality. But most authors choose to work in first order logic with equality, in which case equality is automatically primitive. This is partly due to familiarity and partly for technical reasons. Jan 22 at 19:53
• Equality is neither a primitive notion nor a definition. It is a symbol of the logical calculus. Equality means that two things are the same thing. Equal is equal in all respects and that is built into first order logic and has nothing to do with set theory. Extensionality says that if two sets have the same elements (which is a primitive notion of set theory) then they are equal. Jan 22 at 21:28

This is a good question. Although there is something missing. When you state the Axiom of Extensionality, you are stating it as an “if and only if” condition. This axioms actually states that $$\forall x \forall y ((\forall z (z \in x \leftrightarrow z \in y)) \rightarrow x = y)\,.$$ The converse of this axiom, which is also true, is a particular instance of a standard theorem in any first order theory with equality (in this case, we consider the $$\in$$ as our relational symbol).
Putting both together we in fact obtain that $$\forall x \forall y ((\forall z (z \in x \leftrightarrow z \in y)) \leftrightarrow x = y)\,.$$