# How does the Axiom of Extensionality prove the uniqueness of Specified sets?

Note that there is one axiom for every such predicate φ; thus, this is an axiom schema.

To understand this axiom schema, note that the set B must be a subset of A. Thus, what the axiom schema is really saying is that, given a set A and a predicate P, we can find a subset B of A whose members are precisely the members of A that satisfy P. By the axiom of extensionality this set is unique

My question is why, by the axiom of extensionality, is this set unique? The Axiom of Extensionality is about set equality; what is the relation between the Axiom of Extensionality with uniqueness of this set?

• How would you prove uniqueness without proving equality? And how would you prove equality without the axiom of extensionality? – Asaf Karagila Sep 22 '12 at 17:33
• What "unique" means is that there are not two non-equal things that both have the property in question. So "unqueness" is inherently about equality. – Henning Makholm Sep 23 '12 at 13:39

Suppose we have some predicate $\varphi$ and let $X, Y \subseteq Z$ be such that $\forall{x} (x \in X \leftrightarrow \varphi(x))$ and $\forall{x} (x \in Y \leftrightarrow \varphi(x))$ (by specification). By a simple substitution of equivalents,
$$x \in X \leftrightarrow \varphi(x) \leftrightarrow x \in Y.$$
By extensionality, $\forall{x} (x \in X \leftrightarrow x \in Y) \rightarrow X = Y$. But $\forall{x} (x \in X \leftrightarrow x \in Y)$. So $X = Y$.
Intuitively, extensionality asserts that if two sets contain the same elements then they are the same set. Note that the $\in$ relation is just a binary relation. You can find binary relations R and element x and y such that $(\forall c)(c R x \Leftrightarrow c R y)$, but $x \neq y$. So extensionality is necessary if your set theory is to behave as usual.
Now suppose that Z is a set and $\varphi$ is a formula in the language of set theory. Suppose A and B are two subsets of Z satisfying the axiom of specification for the formula $\varphi$. A and B consist of exactly those elements of Z that satisfies $\varphi$. In particular, A and B consist of exact the same elements. By extensionaly, it can be concluded that $A = B$.
Suppose $B_1$ is a set such that every member is a member of $A$ satisfying $P$ and $B_2$ is also a set such that every member is a member of $A$ satisfying $P$. Then $x \in B_1$ iff $x \in A \land Px$ iff $x \in B_2$. So by extensionality $B_1 = B_2$.