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?

  • $\begingroup$ Existence of emptyset, in axiomatizations I have used, does not use the Axiom of Infinity. $\endgroup$ – André Nicolas Nov 19 '14 at 4:42
  • $\begingroup$ @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? $\endgroup$ – user193756 Nov 19 '14 at 4:48
  • $\begingroup$ 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. $\endgroup$ – André Nicolas Nov 19 '14 at 4:56
  • $\begingroup$ 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. $\endgroup$ – user193756 Nov 19 '14 at 5:01
  • $\begingroup$ 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. $\endgroup$ – user193756 Nov 19 '14 at 5:13

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

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

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.


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 $S$.

  • $\begingroup$ And which axiom is that? $\endgroup$ – Asaf Karagila Nov 19 '14 at 19:32
  • $\begingroup$ Axiom of Subsets works. $\endgroup$ – GFauxPas Nov 19 '14 at 20:00
  • $\begingroup$ 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. $\endgroup$ – Asaf Karagila Nov 19 '14 at 20:04
  • $\begingroup$ "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. $\endgroup$ – GFauxPas Nov 19 '14 at 20:11
  • $\begingroup$ 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? $\endgroup$ – Asaf Karagila Nov 19 '14 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:


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


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