# Axiom of infinity and empty set

The axiom of infinity is formulated as

$$\exists S ( \varnothing \in S \wedge (\forall x \in S) x \cup \{x\} \in S)$$

Can someone explain why the use of $\varnothing$ in the axiom of infinity makes sense, when the very existence of $\varnothing$ is predicated on it?

• Existence of emptyset, in axiomatizations I have used, does not use the Axiom of Infinity. Nov 19, 2014 at 4:42
• @AndréNicolas That's interesting. Are we talking about the usual axioms of ZFC? How would one prove the existence of the empty set without the axiom of infinity? Nov 19, 2014 at 4:48
• There are several axiomatiatins, all relatively mild variants of each other. One can prove (in any standard theory, set-theoretic or not) that there is an object. I think of it model-theoretically, the underlying set of any $L$-structure is non-empty. Then one can pick out the empty set using Separation by saying it consists of all objects not equal to themselves. Nov 19, 2014 at 4:56
• When you say there are several axiomatizations, what do you mean? My book Set Theory by Jech and the Wikipedia page on ZFC says there are 9 axiom/axiom schemas which are extensionality, regularity, schema of specification, pairing, union, replacement, infinity, power set, and AC. Without the axiom of infinity, it does not follow that a single set exists. Nov 19, 2014 at 5:01
• Nvm. I found the answer. $\exists x (x = x)$ is apparently a theorem of first order logic, so it's not even needed as an axiom. Nov 19, 2014 at 5:13

You don't need the axiom of the empty set for the axiom of infinity.

$$\exists S(\exists x(x\in S\land\forall y(y\notin x)\land\forall z(z\in S\rightarrow\exists u(u\in S\land\forall w(w\in u\leftrightarrow w\in z\lor w=z))))$$

The axiom states that there exists $$S$$ such that there is an element of $$S$$ which has no members, and $$S$$ is closed under successorship.

• Why is it not $\forall w(w\in u\leftrightarrow (w\in z)\vee (w=z))$ ? I'm just trying to parse the last bit and I can't really see how $u\in z$ or $u=z$ implies that $w\in u$ for any set $w$. Jun 4, 2021 at 3:51
• Yeah, that's definitely a typo. Thanks. Jun 4, 2021 at 9:16
• You have made the same typo here as well. Was trying to derive some stuff about ordinals from ZFC axioms and came across these answers, which have helped a lot. Jun 4, 2021 at 9:22
• I'm glad they helped, and thanks for the corrections. Jun 4, 2021 at 13:28
• Just to confirm, no other axioms (for example replacement or separation) are required other than this version of axiom infinity to prove "there exists an empty set"? Then infinity + extensionality is enough to prove the existence of a unique empty set? Dec 19, 2022 at 3:55

The Axiom of Existence states that the empty set exists. If you don't accept the Axiom of Existence as axiomatic, the Axiom of Infinity implies the existence of $\varnothing$, though you need another axiom to "extract" it from $S$.

• And which axiom is that? Nov 19, 2014 at 19:32
• Axiom of Subsets works. Nov 19, 2014 at 20:00
• Right. But since the axiom of infinity already tells us that there is a set such that one of its elements is empty, why do we need an additional axiom? If something is an element of another set, then it exists. Nov 19, 2014 at 20:04
• "If something is an element of another set, then it exists." - I don't think you can state this without an axiom asserting it directly or indirectly. Nov 19, 2014 at 20:11
• Let's go at it using quantifier instantiations. You agree that the axiom of infinity, as written in my answer on this page, asserts the existence of $S$ which is inductive and has an empty set as a member? Nov 19, 2014 at 20:14

The existence of the empty set can be proved from predicate calculus, the Rule of Generalization, and the Axiom of Separation; you don't need the Axiom of Infinity. You can find a formalized proof here:

http://us.metamath.org/mpegif/axnul.html

Of course in many treatments it is simply taken as an axiom.