What is the definition of a set in ZFC?

Question

I ask this as I am confused about empty sets. I assume there is some formalisation past 'a collection of objects'. As then the empty set doesn't really make sense. Dealing with the empty set all the time in mathematics, I have become very comfortable with its manipulation. But upon trying to explain the axiom of choice to an undergrad philosophy student, he had a lot of issues with the concept of an empty set as a whole, making me question it myself.

Even if included as an axiom, what justifies its existence. The axiom of choice does seem very obvious when you are a person very used to manipulating the empty set, but if not it has little to no intuitive understanding.

Don't you like it when the intersection of two sets is a set? — Angina Seng, Apr 27 '20 at 17:04
Obviously I do, sets manipulate very nicely. But when creating axioms what justifies this, why should sets be closed under intersection — Joshua Farrell, Apr 27 '20 at 17:09
A set is something which satisfies the axioms. There does not need to be an axiom of the empty set if, for example, you can deduce the existence of an empty set by constructing it as a subset of another set and then using an axiom of extensionality to show there is only one empty set. — Henry, Apr 27 '20 at 17:11
A collection of objects that satisfy a certain property; if the property is contradictory, there is no object satisfying it. This is the "intuition" with the empty set. — Mauro ALLEGRANZA, Apr 27 '20 at 17:11
In "ancient time", when the unrestricted Comprehension axiom was available, every contradictory property (like e.g. $x \ne x$) can be used to prove the existence of the empty set. Now, with theories like $\mathsf {ZFC}$, where the Comprehension axiom is no longer valid, we need a specific axiom for the empty set. — Mauro ALLEGRANZA, Apr 27 '20 at 17:14
Does the Separation schema seem intuitive to you? If there are any sets at all, and the separation schema holds, you get the empty set as a consequence. — Malice Vidrine, Apr 27 '20 at 19:01

score 1 · Answer 1 · answered Apr 27 '20 at 17:09

The Axiom of Choice doesn't really have anything to do with the existence of the empty set.

What does it mean to define an object in a theory $T$? It means write a formula $\phi(x)$ in one free variable such that $$ T\vdash \exists ! x \phi(x) $$ In the language of set theory $\{\in\}$; let $\phi(x)$ be the formula $$ \neg \exists y (y\in x) $$ Then $$ ZF \vdash \exists ! x \phi(x) $$ The fact that $ZF\vdash\exists \phi(x)$ follows from the Axiom of Infinity, although some authors include an Axiom of the Empty Set asserting exactly that $\exists x\phi(x)$. The fact that there is only one "empty set", i.e. that $$ ZF \vdash \forall x\forall z [(\phi(x)\wedge \phi(z))\leftrightarrow x=z] $$ is a consequence of the Axiom of Extensionality.

Once all this is done, we can state the Axiom of Choice as "Every family of non-empty sets has a transversal" (or a choice function, they're equivalent), which we can write as $$ \forall F \left[\emptyset \not\in F \to \exists T \forall x\in F (\text{sing}(x\cap F))\right] $$ where $\text{sing}(v)$ is a formula which asserts that $v$ is a singleton. In the displayed sentence above, $\emptyset\not\in F$ is a convenient abbreviation for $$ \neg \exists y\in F \phi(y), $$ which "says" the same thing.

I think my confusion arises from mixing basic logic knowledge and layman intuition — Joshua Farrell, Apr 27 '20 at 17:15

score 1 · Answer 2 · answered Apr 27 '20 at 17:27

What is an "element of a group"?

We have a theory (Called GRP), in the sense of first order logic, whose models are exactly groups. The Language of this theory is

$$(1, \cdot, {}^{-1})$$

where $1$ is a constant symbol, $\cdot$ is an (infix) binary function, and ${}^{-1}$ is a (postfix) unary function.

The Axioms are well known:

$\forall x . 1 \cdot x = x \cdot 1 = x$
$\forall x . x \cdot x^{-1} = x^{-1} \cdot x = 1$
$\forall x. \forall y . \forall z . (x \cdot y) \cdot z = x \cdot (y \cdot z)$

I ask again, what is an "element of a group"? A model theorist would say it is a member of a Model of the theory of groups.

Now - what is a "set"?

We have a theory (called ZFC), in the sense of first order logic, whose models are exactly "set theories". The Language of this theory is

$$(\in)$$

where $\in$ is a (infix) binary relation.

The Axioms are well known (and too long for me to bother writing down).

I ask again, what is a "set"? A model theorist would say it is a member of a Model of the theory ZFC.

The only hiccup is that to formalize the notion of a "model", we use sets. So really we're working on two different levels of sets. We have the "external" sets, which we use to talk about groups (and internal sets), and the "internal" sets, which are analogous to group elements, and are the things we talk about.

The great tragedy of mathematical philosophy is that there is no end to this rabbit hole. It's turtles all the way down, and you have to accept that, at some point, we must be informal with our notion of "set". There has been much ink spilled over this, so I won't waste any more, but it's a very interesting issue, and has led to lots of interesting math.

I hope this helps ^_^

What is the definition of a set in ZFC?

2 Answers2