Does there exist a set $A$ such that $A=\{A\}$ ? 


Edit(Peter LL): Such sets are called Quine atoms.


[Naive set theory By Paul Richard Halmos](http://dc108.4shared.com/download/YNLXDssM/Halmos_-_Naive_set_theory.djvu?tsid=20100725-105252-871bd162) On page three, the same question is asked.


Using the usual set notation, I tried to construct such a set: First 
with finite number of brackets and it turns out that after deleting 
those finite number of pairs of brackets, we circle back to the original
 question.


For instance, assuming $A=\{B\}$, we proceed as follows: 
$\{B\}=\{A\}\Leftrightarrow
B=A \Leftrightarrow B=\{B\}$ which is equivalent to the original 
equation.


So the only remaining possibility is to have infinitely many pairs of 
brackets, but I can't make sense of such set. (Literally, such a set is 
both a subset and an element of itself. Further more, It can be shown 
that it is singleton.)


For some time, I thought this set is unique and corresponded to $\infty$
 in some set-theoretic construction of naturals. 


To recap, my question is whether this set exists and if so what 
"concrete" examples there are. (Maybe this set is axiomatically 
prevented from existing.)



Can we have A={A} ? 
In standard set theory (ZF) this kind of set is forbidden because of the [axiom of foundation](http://en.wikipedia.org/wiki/Axiom_of_foundation).


There are alternative axiomatisations of set theory, some of which do
not have an equivalent of the axiom of foundation. This is called
non-well-founded set theory. See e.g. [Aczel's anti-foundation\_axiom](http://en.wikipedia.org/wiki/Aczel%27s_anti-foundation_axiom), where there is a unique set such that $x = \{x\}$.


