Axiomatic Set Theory (ZFC): Intersection I'm currently reading "Axiomatic Set Theory" by Suppes, and the book gives a proof of the existence of the intersection (which relies on the Axiom of Separation).  While I understand the idea of the proof, I don't understand why it is the Theorem is stated as follows:
$$\forall x \exists y(x\in y \Leftrightarrow x\in A \wedge x\in B)$$
By the definition of the Axiom of Separation don't we need a "$z$" such that $x\in z$ (?)
The formula would then be:
$$\forall x \exists y (x \in y \Leftrightarrow x\in z\wedge(x\in A\wedge x\in B))$$
Why is the proof not given this way?  Also, in the same way that a separate "Axiom of Union" is introduced for the union, is the above method of "defining" the intersection really the best way to do it?
 A: Now, I have been reading the book. I summarize

Axiom 1 (of Separation). Let $A$ a set, and for each $x \in A$, let $\varphi(x)$ a property pertaining to $x$. Then there exists a set $C := \{x \in A : \varphi(x) \text{ is true} \}$ (or $\{x \in A : \varphi(x)\}$ for short),  whose elements are precisely the elements $x$ in $A$ for which $\varphi(x)$ is true.
  $$\exists C \, \forall x \; (x \in C \iff x \in A \;\land\; \varphi(x) ).$$
Axiom 2 (of Extensionality). Two sets $A$ and $B$ are equal, $A = B$, if every element $x$ of $A$ belongs also to $B$, and every element $y$ of $B$ belongs also to $A$.
  $$\forall x \; (x \in A \iff x \in B) \implies A = B.$$

Your theorem

Theorem 3. Let $A$ and $B$ be sets. Then exists a unique set $C$ whose elements belongs to both $A$ and $B$.
  $$\exists ! C \, \forall x \; (x \in C \iff x \in A \;\land\; x \in B).$$

Now, to prove the theorem we need show the existence and uniqueness. First the existence. Let $A$ and $B$ be sets, and let $\varphi(x) := x \in B$. Then, by Axiom 1, we have
$$\exists C \, \forall x \; (x \in C \iff x \in A \;\land\; \varphi(x) ),$$
i.e.,
$$\exists C \, \forall x \; (x \in C \iff x \in A \;\land\; x \in B ).$$
This prove the existence of the intersection set $C$ for any sets $A, B$.
We now show the uniqueness. Let $A$ and $B$ sets. Suppose there exists two sets $C, C'$ such that
$$\exists C \, \forall x \; (x \in C \iff x \in A \;\land\; x \in B )$$
and
$$\exists C' \, \forall x \; (x \in C' \iff x \in A \;\land\; x \in B ).$$
Using the notation, we have
$$x \in A \;\land\; x \in B \iff x \in C',$$
i.e.,
$$x\in C \iff x \in A \;\land\; x \in B \iff x \in C'.$$
This means, the statements $x\in C$, $x \in A \;\land\; x \in B$, and $x \in C'$ are equivalents. Thus, we have
$$x \in C \iff x \in C'.$$
By Axiom 2, we have $C = C'$ as desired. $\;\Box$
Now, note that to use the Axiom 1, you need two sets: a reference set from which construct the intersection, and another to define the property. Also, another valid definition of intersection is
$$A \cap B = \{x \in A : x \in B\}.$$
So the set $z$ that you mention is $A$, and your property $x \in A \; \land \; x \in B$ should be just $x \in B$.
Finally, in Set theory, the Axiom of union exists because this does not follow from the other axioms (empty set, extensionaity, separation, etc). These axioms allow us to build smaller sets from other reference sets. But the union set is a larger set from anothers, because of that this axiom is necessary.
