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

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 $z\cap x$ ?

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.

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 $Y$.

• Thanks, Alex. This is a darn nice addendum, as well. I hadn't thought that the authors might take such a slick approach. But if so, we can infer that they haven't even derived Separation, and are leaving that as an exercise for the reader, as merely part of the proof that intersections exist. Weird, maaan. Thankfully, I don't know their book either ;/ – BrianO Nov 12 '15 at 18:06
• Yes you are both correct. They don't include the axiom of separation and they also don't show it follows from other axioms. Thank you for clarifying. – Student Nov 12 '15 at 18:10
• @Student cc Alex: I'm slackjawed. – BrianO Nov 12 '15 at 18:11