What is the purpose of the Axiom of regularity/foundation?

Question

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.

Answer 1

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.

Answer 2

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 $\in$.