# Question about ZF set theory and the way of defining a set

I have some questions about ZF set theory. I was reading some courses about the construction of $$\mathbb{N}$$, $$\mathbb{Z}$$, $$\mathbb{Q}$$ and $$\mathbb{R}$$ and something disturbes me a bit.

I saw things like : $$\mathbb{N} = \{n \mid n \in I \text{ for every inductive set } I\},$$ or $$\mathbb{Q} = \left\{\frac{p}{q} \mathrel{}\middle|\mathrel{} p \in \mathbb{Z}, \ q \in \mathbb{N}^{\ast}\right\},$$ and I was wondering if it is really correct to write these sets in such a way.

If I'm not wrong, in ZF, if we consider a set $$A$$, we can always construct a set $$B$$ such that : $$B = \{x \in A \mid \phi(x)\} ,$$ where $$\phi$$ is a formula in ZF language (I'm not really a specialist in logic, but I get the idea of "formula") and it is called the "axiom schema of specification". But we cannot construct set of the form : $$\{x \mid \phi(x)\}$$ with this axiom schema right (otherwise, we can find things like Russell's paradox) ?

Now, if we take for example the way I wrote $$\mathbb{N}$$ above, there are two possibilities for me :

1. This way of writing is not correct. In that case, how should we write in particular the set $$\mathbb{N}$$ ? I read some things about von Neumann's construction of $$\mathbb{N}$$ and about Peano's axioms, but I didn't see exactly a way to "write succinctly" $$\mathbb{N}$$...

2. This way of writing is correct. In that case, which axiom of ZF theory is used to write for example $$\mathbb{N}$$ or $$\mathbb{Q}$$ in that way ?

The axioms of ZF only talk about abstraction of the form $$\{x\in X\:|\:\varphi\}$$, true; but that doesn't mean we can't write the same set another way. In particular, $$\{x\:|\:\varphi\}$$ is still a perfectly acceptable way to write something when you know it's a set. It's worth noting, since writers and instructors don't always make this clear, that $$\{x\:|\:x\in X\wedge\varphi\}$$ is exactly the same thing as $$\{x\in X\:|\:\varphi\}$$, so nothing specifically rides on whether the membership claim is in that first space. The important thing is that $$\{x\:|\:\varphi\}$$ is a set exactly when there's an $$X$$ such that $$x\in X\wedge\varphi\iff\varphi$$, which is exactly what's happening in the two examples you give.

In the case of $$\mathbb{N}$$ we are given by the ZF axioms that there is an inductive set $$I$$; the elements that are in every inductive set will also all be in $$I$$, so there's no difference in the between the extension of "$$x$$ is a member of every inductive set" and "$$x\in I$$ and $$x$$ is a member of every inductive set."

In the case of $$\mathbb{Q}$$, as it's written above it's a little bit question-begging. We might write it that way if we're working in $$\mathbb{R}$$ and defining the rational numbers in $$\mathbb{R}$$ by the naturals and integers in $$\mathbb{R}$$; in which case because we've taken two sets of reals and applied an operation defined only on the reals, under which the reals are closed, "$$x$$ is a quotient of an integer by a non-zero natural" and "$$x\in \mathbb{R}$$ and $$x$$ is a quotient of an integer by a non-zero natural" once again refer to exactly the same collection.

You are right to observe that in a fully formal proof, you would probably prove existence via first using that $$\{x\:|\:x\in X\wedge\varphi\}$$ is an instance of separation, and then proving that $$x\in X\wedge\varphi\iff\varphi$$ to justify later use of $$\{x\:|\:\varphi\}$$; but in practice it's usually obvious (or presumed obvious) how to supply the appropriate $$X$$ if we really need to, so we skip it and just write the more natural form.

• Okay, I understand now. Thank you ! – deeppinkwater Nov 21 '18 at 7:40
• Just a little thing, is there a link between “axiom schema of replacement” in ZF ? – deeppinkwater Nov 21 '18 at 7:48
• @deeppinkwater - A link between the axiom schema of replacement and what? I'm not sure I understand the question. – Malice Vidrine Nov 21 '18 at 9:58
• Sorry, between the axiom schema of replacement and the fact that you said : $x \in X \wedge \varphi \Leftrightarrow \varphi$ ? When I ask if "there is a link", it is in the sense that I'm not sure to have really understand this axiom, but I have the feeling that there is something similar. – deeppinkwater Nov 21 '18 at 10:45
• No, this doesn't necessarily involve replacement. It's because "$\{x\:|\:\varphi\}$ exists" is short for (the universal closure of) "$\exists y\forall x(x\in y\Leftrightarrow \varphi)$." If a sentence of this form is a theorem, and $\forall x(\varphi\Leftrightarrow\psi)$ is a theorem, then one can infer "$\{x\:|\:\psi\}$ exists," too. It's just a matter of the logic. – Malice Vidrine Nov 21 '18 at 10:51