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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?

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  • $\begingroup$ 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
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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.

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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 $S$.

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  • $\begingroup$ 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
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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.

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