Given the definition of subsets and equality of sets:


* A $\subset$ B, if x $\epsilon$ A $\rightarrow$ x $\epsilon$ B for every set x.
* A = B, if A $\subset$ B and B $\subset$ A


Why is it impossible to decide whether two circular sets I = {I} and J = {J} are equal.


I mean, the way is see it is that I is not an element of J, since only J is an element of J, so the two circular sets are not equal.


What's wrong in my reasoning?



Equality of two circular sets
In ZF minus the axiom of foundation there is no way of proving
that all "circular" sets are equal. You could take a model of set theory
allowing ur-elements and replace some or all of these by circular sets.


If you accept the popular alternative to Foundation, Peter Aczel's
[Anti-Foundation axiom](http://en.wikipedia.org/wiki/Anti-Foundation_Axiom), then there is just one circular set.
Effectively the AFA decrees that two sets with the same membership digraph
on its transitive closure are equal.


