Is $x=\{x,\{x\},\{x,\{x\}\},...\}$ a thing? $\lor$ Do fractal sets exist? I am currently learning about hypersets. So far I have learned that:

*

*Non-well-founded set theories are variants of axiomatic set theory that allow sets to contain themselves and otherwise violate the rule of well-foundedness.

*Axioms contradicting the axiom of regularity are known as anti-foundation axioms, and a set that is not necessarily well-founded is called a hyperset

*It can be shown that the so-called Quine atom, formally defined by Q={Q}, exists and is unique.

[The points above come directly from Wikipedia]
With the previous in mind, I am wondering if something like $x=\{x,\{x\},\{x,\{x\}\},...\}$ is a thing that already exists / has been studied.
Additionally, I am curious about the idea of fractal sets & am curious if $x=\{x,\{x\},\{x,\{x\}\},...\}$ is one.
 A: Under the Aczel antifoundation axiom (AFA), there is a unique Quine atom. Moreover, there is a result called the solution lemma that says that all systems of equations of a certain type have a unique solution. For example, there is a unique set such that $x=\{x,\emptyset\}$ and given any infinite sequence of sets $a_1,a_2,\ldots,$ there is a unique solution $x_1,x_2,\ldots$ to the system of equations $x_i = \{a_i,x_{i+1}\}.$
As others have noted, your equation (which is not quite in a form we can apply the above result to) has a boring solution that is just a Quine atom, and this is the unique solution under AFA. The "problem" is that AFA is really powerful in enforcing sameness of things with similar structure. One can view this as a strong version of extensionality.
But there are antifoundation axioms that have weaker forms of extensionality. The Boffa antifoundation axiom (BAFA) has nothing but plain old extensionality as a constraint. And in fact any extensional graph is isomorphic to a transtive set.
For instance, there are a proper class of Quine atoms, since any graph consisting of a bunch of self-loops must be isomorphic to a set, which will be a set of many distinct Quine atoms. So we're far away from AFA where $x=\{x\}$ has a unique solution. And not only does your equation have any of these Quine atoms as a solution, but it has "nontrivial" solutions as well.
For instance, consider a solution to the system $$ x = \{x,y\}\\ y = \{x\}$$ with distinct $x,y.$ There is no such solution in AFA since the solution is unique and $x=y=Q$ is a solution. But there is in Boffa: just draw a graph with two dots with $x$ pointing to itself and $y$ and $y$ just pointing to $x.$
You can expand out to see that this $x$ is a solution to your equation, and it's certainly not a Quine atom since it has two elements. But it's still a bit trivial because most of the elements on the right hand side of the equation are redundant, though now at least we have two sets.
But then you can keep going, and draw a slightly more complicated graph with three elements satisfying $$ x = \{x,y,z\}\\ y = \{x\}\\ z = \{x,y\}.$$ And then hopefully from there it's clear that there is a countably infinite graph corresponding to a countably infinite set (actually many such sets) that obeys your equation "faithfully". (Whether some other well-studied antifoundation axiom will give a unique “faithful” solution, I’m not sure offhand. These are the only two I’ve really made an effort to understand.)
A: The answer is in your question! Just take $x=Q$:
We have $Q = \{Q\}$, but then
$\{Q,\{Q\}\} = \{Q, Q\} = \{Q\} = Q$
So $\{Q, \{Q\}, \{Q,\{Q\}\}, \ldots \} = \{Q, Q, Q, \ldots\} = \{Q\} = Q$.
