$z \cap x$ denotes a set whose existence is proved using the axiom of replacement?

Question

I am reading the book Mathematical Logic and Model Theory by Prestel and Delzell and after they talked about first order semantics there was a section on the axiomatization of set theory. They present the foundation axiom in the following way:

$$\forall x (x\neq\emptyset \longrightarrow \exists z (z\in x \wedge z \cap x=\emptyset))$$

Then they go on to say that $z\cap x$ (for intersection) denotes a set whose existence is proved using the axiom of replacement and whose uniqueness is guaranteed by the axiom of extensionality.

I don't get this at all. The more I think about it the more confused I am.They don't define $z\cap x$. Is the axiom supposed to be the definition of it or are we supposed to take the usual definition?

Could someone explain how to prove existence and uniqueness of $z\cap x$ ?

Answer 1

In fact you (and they) don't need to use $\cap$ in order to state Regularity. As they haven't defined the symbol, they shouldn't use it. You can eliminate intersection simply by expanding its definition: $$ \text{AxRegularity}\iff \forall x (x\neq\emptyset \longrightarrow \exists z (z\in x \wedge \forall u \neg (u\in z \wedge u\in x)) $$

Their statement about why $z\cap x$ exists is, frankly, a little murky. Given two sets $X,Y$, Separation guarantees existence of a set $Z$ such that $$ u\in Z \longleftrightarrow (u\in X \wedge u\in Y) $$ i.e. $Z = \{u\in X\mid u\in Y\}$. Uniqueness of $Z$ follows by Extensionality. Of course, $Z$ is just $X\cap Y$.

You need Replacement to show that $\{z\cap x\mid z\in x\}$ exists, but it's not necessary to show that: all you need is existence of the intersection of any two sets.

Answer 2

I just wanted to add one thing to BrianO's nice answer. It's certainly strange to say that the existence of $z\cap x$ should be proved by Replacement, when it most naturally follows from Separation, as in BrianO's answer. It's strange, that is, unless Separation doesn't appear on your list of axioms.

I don't have a copy of the book by Prestel and Delzell, so I can't check whether they use this approach, but some people prefer to prove Separation from Replacement and Empty Set as follows:

Suppose $X$ is a set and $\varphi(x)$ is a formula, possibly with parameters. We want to form the set $Y = \{a\in X\mid \varphi(a)\}$. If no element of $X$ satisfies $\varphi$, then $Y = \emptyset$, which exists by the Empty Set axiom. Otherwise, there is some $b\in X$ such that $\varphi(b)$ holds. Let $\psi(x,y)$ be the formula $$(\varphi(x) \land (y = x))\lor (\lnot \varphi(x) \land (y = b)).$$ Then $\psi(x,y)$ defines a function $F(x) = y$ (for all $x$, there is exactly one $y$ such that $\psi(x,y)$ holds), and the image of $X$ under $F$ is $Y$.