Help understanding the axiom of regularity If 0={}, 1={0}, 2={0,1} and so forth, the axiom of regularity holds for every natural number, since for each natural number n, there is an element x in n which is disjoint from n.
However, if we define 0={1}, 1={2}, 2={3} and so on (infinite descending sequence), regularity does no hold, given that there is not a single element of the set S = { all n : n is a natural number }  which is disjoint of S. Is that correct?
 A: Your example is correct (strictly speaking you are not defining correctly, but your general idea is right). There are models in which Regularity does not hold. Regularity is an Axiom, meaning we assume it is true, however there is no proof of that. So there is plenty of examples where regularity does not hold (e.g. just consider $x=\{x\}$ and regularity does not hold).
Mathematicians assume Regularity because such examples are 'not nice'. That is to say they are unexpected and unwanted. Regularity is not needed for some important basic theorems (induction, calculus, arithmetic) but it imposes 'nicer' sets, those which follow common mathematical intuition. Per @Hanul s comment, it is necessary for some other important theorems though.
A: As people have noted in the comments your definition doesn’t make a lot of sense (even if sets don’t need to be well founded, recursive definitions need to be). And more generally, you will never succeed at defining an infinite descending membership sequence in ZF minus regularity, because it is consistent that no such sequence exists. 
Now, it is also consistent with ZF minus regularity that there is a countable sequence of distinct sets $x_n$ such that $x_n =\{x_{n+1}\}$ for all $n$. But even still, you can’t just define the natural numbers as $n=x_n$ unless you actually add antifoundation axioms to the theory that will allow you not only to prove the existence of such a sequence but to define a canonical one. 
But this whole thing about the natural numbers feels like a non sequitur. It seems like the content of your question is really whether the existence of an infinite sequence such that $x_n=\{x_{n+1}\}$  implies that regularity fails. The answer to that is yes: in ZF minus regularity we can define the set $\{x_n:n\in\omega\}$ from 
the sequence, and this set violates regularity. This follows more generally from any infinite descending sequence where $x_{n+1}\in x_n.$ If we also assume choice in the base theory (or even just dependent choice) then the implication goes the other way and the failure of the axiom of regularity implies the existence of an infinite descending sequence. (Thanks to Hanul Jeon in the comments for pointing out that the converse requires choice.)
