Why is the Axiom of Regularity preferred over the Axiom (schema) of Induction, or the claim that no downward infinite membership chain exists? Most of the axioms in $\text{ZFC}$ seem intuitive and sensible to me, including the historically contrioverial Axiom of Choice, yet I struggle to find the Axiom of Regularity as intuitive as the rest. It is worth mentioning that the equivalent statement of the nonexistence of downward membership chains is intuitive to me, but that poses the question as to why the Axiom of Regularity is preferred over the previous statement itself.

To be somewhat clearer: let $(\text{ZFC}-\text{R})$ be the axiomatic system of $\text{ZFC}$ without the Axiom of Regularity. In $(\text{ZFC}-\text{R})$ the following are equivalent:

*

*Axiom of Regularity.

*Axiom (schema) of Induction.

*No downward infinite membership chain exists.

It seems to me that $(2)$ and $(3)$ are more intuitive than $(1)$, so I wonder:

*

*Why is $(1)$ preferred (as an axiom) over  $(2)$ and $(3)$?


*If the answer to the above is that $(1)$ is -to others- more or just as intuitive as the other equivalent statements, then could someone provide some intuitive argument for $(1)$ being intuitively true that goes beyond it being equivalent to $(2)$ and $(3)$?
 A: First of all, note that $(1) ⇒ (3)$ can be done in ZF while $(3) ⇒ (1)$ requires some choice, so in the context of ZF those two are not equivalent. We would want our axioms to be the same between ZF and ZFC, so we really need to ask, which formulation is better in ZF-R.
I actually believe that $(3)$ is simpler to understand conceptually. A sequence is something studied relatively early in mathematician's life. To understand $(1)$ you need to think about $∈$ as an order on a set, which is not something people are automatically comfortable with.
That being said, $(1)$ formulates the idea that "every set is $∈$-well founded". The ability to take a minimal element gives us immense power, which results in $(2)$, $\in$-induction. The idea of induction over the universe of sets is something very intuitive: If $P$ holds for $\emptyset$, and if $P$ holds for any set $X$ made out of $Y$'s for which $P$ holds, then $P$ holds for all sets. This idea is much more useful than $(3)$ in ZF, and even if it is a bit more complicated to understand, it is worth it.
Here is another equivalent to the axiom of regularity. Let $V$ be the class of all sets. Let $WF$ be the von Neumann universe, aka cumulative hierarchy, defined by:

*

*$V_0=∅$

*$V_{α+1}=\mathcal P(V_α)$

*If $β$ is limit ordinal then $V_β=\bigcup_{α<β} V_\alpha$

*$WF=\bigcup_{α∈ Ord}V_α$
Then $(1)$ is equivalent over ZF-R to: $(4)$ $V=WF$. ($WF$ stands for "well-founded". In fact, in ZF-R we have that $WF$ satisfies all of ZF, even if $V$ doesn't satisfy regularity.)
This construction is very intuitive and helps define the rank function: If $X$ is a set, then the $∈$-rank, or just rank, of $X$ is the minimal ordinal $α$ such that $X\in V_{α+1}$. Scott's trick also depends on the  $V_\alpha$ hierarchy. (Note that these are weaker than the axiom of regularity, but using the axiom of regularity it is very simple to get them. I am not sure if we can get them from $(3)$, but at the very least doing so is not simple or intuitive.)
I hope I convinced you that $(3)$ is not what we want. So what about $(1),(2),(4)$?
Well, $(4)$ is very nice to have and intuitive, but it is much more complicated to actually define. You need to define ordinals first, then show that recursion over the ordinals is well-defined, then formulate $V=WF$ in the language of sets, and yada yada yada.
$(2)$ is arguably more intuitive than $(1)$ but not only is it a schema, on face value it doesn't tell us anything about individual sets. You need to work to get to that.
$(1)$ is very simple once you get used to looking at $∈$ as an order, and not just a relation. It is very simple to formulate, and it conveys a very simple idea about how sets should look, so $(1)$ seems to be the best choice.

@DanielWainfleet gave another very good point against using $(3)$: it requires the axiom of infinity. Being well founded shouldn't be coupled to an existential axiom like the axiom of infinity--they are very different by nature and by purpose.
A: The other answer is much more informative, but in case you were like me and just wanted some intuition for the first formulation, consider the simple proof of $(1)⇒(3)$:

If we collect the members of an infinite descending membership chain into a set, we get a nonempty set $C$ with the property that for each $x\in C$, there is a $y\in C$ with $y\in x$. The axiom of regularity states that no such set exists.

In other words, the axiom of regularity directly expresses "there are no infinite downward membership structures". To me, this makes it as intuitive as $(3)$.
