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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.

Jacob Schneider
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    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 – Hagen von Eitzen Jan 05 '18 at 20:29
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    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. – Lee Mosher Jan 05 '18 at 20:30
  • So you are allowed to start with any arbitrary set? As long as it does not break the axiom of regularity? For example, I can't start with the set of all sets but I could start with {a, b}? Therefore anything that passes the axiom of regularity is a set? – Jacob Schneider Jan 05 '18 at 20:37
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    @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 $"\forall x\forall y(xy=yx)"$ is independent of the group axioms since there are some groups which are abelian and some which aren't. – Noah Schweber Jan 05 '18 at 20:41
  • @HagenvonEitzen I know what the axiom of regularity says, but I'm not sure what a "regular set" or an "irregular set" is. How are they defined? – bof Jan 05 '18 at 21:08
  • @HagenvonEitzenSo than what is the point of the axiom of infinity if you can just select some sets to start with? – Jacob Schneider Jan 05 '18 at 21:29
  • @NoahSchweber So is it meaningful to ask does ZFC allow for the set of {a, b, c} to exist? How would one prove that? I guess I am a bit confused about the difference b/w a model and the axiomatic system itself. What would be 2 different models of ZFC? – Jacob Schneider Jan 05 '18 at 21:34
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    @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. – Noah Schweber Jan 05 '18 at 21:57
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    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. – Noah Schweber Jan 05 '18 at 21:59
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    The latter three are technical (especially forcing). However, an example of the first is easy to produce: suppose I have a model $M$ of ZFC with an inaccessible cardinal $\kappa$ - that is, there is some $\kappa\in M$ such that $M\models$"$\kappa$ is an inaccessible cardinal." Then $V_\kappa^M$ is also a model of ZFC, even though it's a proper substructure of $M$. – Noah Schweber Jan 05 '18 at 22:01
  • Ok I think I am starting to understand models better thank you. So would you say the purpose of the axiom of infinity is to ensure that all models have at least one set? Without the axiom of infinity you could have an empty class satisfy all the axioms? I always thought of different axiomatic systems (group axioms, peano axioms, etc) as having models but I guess I just ignored models for ZFC b/c it felt more fundamental in that there was only one model. Understanding what a model truly is helped answer my question. – Jacob Schneider Jan 05 '18 at 22:27
  • @NoahSchweber One final question would be are there any theorems that require the axiom of regularity? Can you base all of math on ZFC - {Regularity}? Also would you say that a model has to define how to generate the sets or not necessarily (its just a generic interpretation)? – Jacob Schneider Jan 05 '18 at 22:32
  • @JacobSchneider It is part of the definition of a model that it in nonempty, so you need no axiom to ensure that. Besides, $\exists x (x=x)$ is a valid sentence in Logic so it holds in all theories. – Rene Schipperus Jan 05 '18 at 22:59

2 Answers2

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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.

Rene Schipperus
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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 $\in$.

YCB
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