My understanding is that ZF posits the existence of just one set, the empty set $\emptyset$, and the von Neumann hierarchy is constructed starting from there using the axioms of power set and infinity.

The axiom of regularity $$ \forall x (x \neq \emptyset \implies \exists y (y \in x \land \forall z (z \in x \implies z \notin y)))$$ asserts the existence of an $\epsilon$-minimal member $y$ of every non-empty set $x$.

Can one prove that every set in the von Neumann hierarchy has an $\epsilon$-minimal member? If so then why do we need the axiom of regularity? Or is the von Neumann hierarchy just some sort of "minimal model" which satisfies the ZF axioms and the whole point is to prove theorems about other models which you don't know a priori satisfy regularity?

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    $\begingroup$ Regularity is equivalent to the fact that every set eventually appears in von Neumann hierarchy. $\endgroup$ – Wojowu Dec 16 '15 at 18:40

The axiom of regularity is the assertion that every set is in the Von Neumann hierarchy, sometimes denoted $V = WF$ where $WF$ is the class of well-founded sets.

The proof that every member of the Von Neumann hierarchy has an $\in$-minimal member basically follows from the fact that each stage $V_\alpha$ of the hierarchy is well-founded. Observe that each stage is constructed from either the powerset or the union operation of the previous stage(s), and note the basic facts $x \in WF \rightarrow P(x) \in WF$ and $x \in WF \rightarrow \bigcup x \in WF$. Thus if $x \in V$, then there is a minimal $\alpha$ such that $x \in V_\alpha$, and since $V_\alpha$ is well-founded, so is $x$.

The reason we "need" Regularity is because it is independant of the other axioms of ZF (sometimes denoted $ZF^-$), which means without taking it as an axiom, the existence of non-well-founded sets is consistent with $ZF^-$. I put "need" in quotes because it is perfectly fine to work in set theory with non-well-founded sets, but it doesn't buy us much (basically all of regular mathematics exists within $V$, and mostly within $V_{\omega + \omega}$) and makes everything needlessly complicated.


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