Besides the axioms of extensionality and regularity, all of the axioms of ZFC either postulate the existence of a set or give a method for generating new sets from existing sets.

Extensionality then gives an equivalence relation on sets which allow us to confirm our definitions we create are well defined.

Regularity then gives a rule that all sets must satisfy.

From all the other axioms (besides regularity) we can start generating sets and decide which generated sets are equivalent. For all generated sets they either will satisfy the proposition in the axiom of regularity or they will not.

If all sets generated do satisfy the axiom of regularity then what is the point of the axiom of regularity? It adds no additional structure to the sets of ZFC.

If there exists a generated set that does not satisfy the axiom of regularity then aren't the axioms inconsistant?

I believe my misunderstanding stems from a misunderstanding with how logic/formal systems work. Any clarification would be greatly appreciated.

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    $\begingroup$ If you start with regular sets, then the sets constructed from the other axioms will be regular. But the other axioms do not exclude the existence of irregular sets $\endgroup$ – Hagen von Eitzen Jan 5 '18 at 20:29
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    $\begingroup$ Your process of "generating sets" is not part of ZFC. You could add a process like that to the axioms, for example the constructibility axiom, and then indeed all sets would be well founded. But apart from that, you don't have any control over what sets are in a given universe and this process of "generating sets" will generally miss lots of interesting stuff in the universe. $\endgroup$ – Lee Mosher Jan 5 '18 at 20:30
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    $\begingroup$ @JacobSchneider Axioms don't generate structures at all - a set of axioms describes a class of structures. The fact that regularity is independent of the rest of ZFC just means that there are some models of ZFC-without-Regularity in which Regularity holds, and some in which it fails; similarly, the sentence $"\forall x\forall y(x*y=y*x)"$ is independent of the group axioms since there are some groups which are abelian and some which aren't. $\endgroup$ – Noah Schweber Jan 5 '18 at 20:41
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    $\begingroup$ @JacobSchneider Models of ZFC are extremely complicated. It's better to start with a simpler example: a model of the theory of groups (say) is just a group. There is a single theory of groups, but of course there are tons of different groups. $\endgroup$ – Noah Schweber Jan 5 '18 at 21:57
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    $\begingroup$ Now, models of ZFC are incredibly complicated to describe. However, there are some understood ways to get a new model of ZFC from an old one (or old ones): the most important being initial segments of the cumulative hierarchy, ultraproducts, inner models (especially Godel's $L$), and forcing. $\endgroup$ – Noah Schweber Jan 5 '18 at 21:59

True it adds nothing and can mostly be dispensed with. However it is equivalent to $$\forall x(x\in V)$$ and this is sometimes useful in set theory. In other words it says that every set has a rank.


By transfinite recursion over $\Omega$ (the class of all ordinals) we can build the following sets in $ZFC$

  1. $V_0:=\emptyset$
  2. $V_{\alpha+1}=\mathcal{P}(V_{\alpha})$
  3. $V_{\gamma}=\bigcup_{\alpha<\gamma}V_\alpha$ if $\gamma$ is a limit ordinal.

With this, you can build the class $WF:=\bigcup\{V_\alpha:\alpha\in\Omega\}$. This class is known as the class of well founded sets and this satisfies tha $x\in WF\Leftrightarrow (x$ is a well founded set).

You can prove that, under regularity axiom, $V=WF$ (even more, regularity is equivalen to $V=WF$). Maybe, in this context, is a bit clearest that says regularity axiom. Because, if regularity is not true, then $V\neq WF$, so in the universe of sets, there is a set that don't have a minumun element respect to $\in$.


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