If we abandon the axiom of regularity, can the cumulative hierarchy just become a definition? In ZFC, the axiom of regularity is used to prove that every set is an element of some stage of the cumulative hierarchy. The index of the least such stage is, by definition, the rank of that set.
Now suppose we're working in ZFC minus Regularity. Can this idea just become a definition? Like, what happens if we define that a set is ranked iff its an element of some stage of the cumulative hierarchy? I imagine that we'd be able to prove the axiom of regularity relativized to ranked sets.
Suppose this idea works. Then, we'd like a nice characterization of the property of being ranked that can be used before the ordinals are even in place, because, for example, if we define $\mathrm{S}(x) := x \cup \{x\},$ well it would be nice to be able to prove that if $x$ is ranked, then $\mathrm{S}(x)$ is distinct from $x$, and that $\mathrm{S}(x)$ is also ranked. So I'm curious as to whether there's a nice characterization of the property of being ranked that does not rely on the ordinals. e.g. does 'no infinite descending membership chains' do the trick?
 A: The usual formulation of the axiom of regularity actually contains the essence of a definition of "ranked" that works in the absence of that axiom.  A set $x$ is ranked iff every set $u$ that has $x$ as a member also has a minimal member (i.e., a member $y\in u$ such that $y\cap u=\varnothing$).  It is not hard to show that this notion of "ranked" is equivalent to being in some stage of the cumulative hierarchy.  
Note, though, that when you define the cumulative hierarchy, you must first define the notion of "ordinal" in a way that ensures that the ordinals are well-ordered.  "Transitive set of transitive sets" won't do, though it becmes OK if you add that ordinals should be ranked in the sense that I defined above.
With the definition of "ranked" that I suggested above, it is fairly easy to prove, in ZFC without regularity, that all the ZFC axioms (including regularity) are true when restricted to ranked sets.  I believe this is the simplest way to prove the relative consistency of the axiom of regularity.
A: Yes. The sets in the cumulative hierarchy are called "well-founded", or said to "have a rank", and the class of all such may be called $\mathrm{WF}$.
A: The equivalence of $\sf Reg$ and "No decreasing $\in$-chains" requires the axiom of choice, and can be shown to be false without it.
If you want to give a definition which captures being ranked, note that it suffices to require that $\operatorname{TC}(x)$ is well-founded, in order that $x$ has a rank. Where $\operatorname{TC}(x)$ is the transitive closure of $x$. That you can define without using ordinals.
