Can a self-containing set be used to define natural numbers? I was wondering now (trying to escape from Santa and Jesus at this Christmas Eve) if the existence, in $\mathsf{ZF}-$(Infinity$+$Regularity), of a set $x$ such that $x=\{x\}$, can be used in some way to prove the existence of a minimal $\in$-inductive set.
My motivation? well... for a such set $x$ we have $x=\{x\}=\bigl\{\{x\}\bigr\}=\Bigl\{\bigl\{\{x\}\bigr\}\Bigr\}=\cdots$, and I would like to exploit this to decodify and collect, in some set, the number of pairs of braces. 
 A: It is consistent with ZF-Inf-Reg that there are two sets $a$ and $b$ with $a\not=b$ but $a=\{a\}$ and $b=\{b\}$. Each is an inductive set, and neither contains the other; so in such a model there is no minimal inductive set.
In particular, this means that the existence of a Quine atom in the first place cannot imply the existence of a minimal inductive set.
Meanwhile, I have no idea how to use a Quine atom in the way you outline; maybe you can be a bit more specific? How exactly do you intend to use a Quine atom for a "decoding" process?
A: This seems to be akin to Zermelo's definition of the integers: $0=\varnothing; n+1=\{n\}$. Of course, we can still define the set $\Bbb N$ as all the sets defined for the natural numbers. However this set will not be the limit, in any reasonable set-theoretic sense, of the natural numbers themselves.
This is why von Neumann's construction of the ordinals is superior.
But let me add two things to Noah's answer here:

*

*An inductive set also has $\varnothing$ as an element. And the reason is exactly because we want to ensure that there is a set which is closed under the definable function $x\mapsto x\cup\{x\}$, and that this function is injective but not surjective. This is not the case with Quine atoms. So you might want to talk about a different type of "inductive" here.


*It is impossible to use a Quine atom to define the natural numbers. Easily we can prove by induction that since $x=\{x\}$, it follows that $\{\ldots\{x\}\ldots\}=x$ again for any finite number of applications of the singleton operation. So using $x$ instead of $0$ to model the natural numbers would result in $0=1=2=3=\ldots$ and the core idea that $0\neq 1$ will be false now.
Note that this would also be the case if we take the von Neumann reinterpretation: $0=x, n+1=n\cup\{n\}$; as in this case we can also prove by induction that $n=n+1=x$, since $x=\{x\}$. So trying with a Quine atom but a different direction might not be useful either.
So even if Quine atoms do exist, they do not interfere with the standard definition of inductive sets, and they are not useful for defining the natural numbers in a simple set theoretic way.
