Solving systems of set equations 1) Is there any set A such that A={A}?
2) Is there any set A such that {A} is a subset of A?
3) Is there any set A such that each of its elements is a subset of A?
4) Is there any set A such that each of its elements is a bijection from A to A?
 A: In the modern context of set theory, namely $\mathsf{ZFC}$, the answer to the first two is no. The reason is that we require $\in$ to be well-founded which forbids $A\in A$, and therefore forbids $\{A\}\subseteq A$ (which is really the same thing).
To the third question consider $\varnothing$, or $\{\varnothing\}$ or $\{\varnothing,\{\varnothing\}\}$. A set is called transitive if every element is also a subset. Transitive sets are used often in set theory, for example the von Neumann ordinals are transitive sets which are well-ordered by $\in$.
To the final question, the answer is again no if we require the set to be non-empty, and again this is a consequence of the axiom of regularity (one of the axioms of $\mathsf{ZFC}$). If $f\in A$ is a bijection from $A$ to $A$ then $f\in\operatorname{dom}(f)$, which means (in the standardn Kuratowski definition of an ordered pair) that $\{\{f\},\{f,x\}\}\in f$, which means that $$\dots\in f\in\{f\}\in\{\{f\},\{f,x\}\}\in f\in\dots$$ is a counterexample to the well-foundedness of $\in$.

In contexts outside of $\mathsf{ZFC}$ it is possible to have sets of the form $A=\{A\}$ or $A\in A$. For example in the set theory $\mathsf{NF}$ there is a universal set, i.e. the set of all sets. Call this universal set $\cal U$ then $\cal U\in U$, because $\cal U$ is a set, and every set is an element of $\cal U$.
Furthermore if we take a theory in which there is some $x=\{x\}$ then we have that $\langle x,x\rangle=\{\{x\},\{x,x\}\}=\{x,x\}=\{x\}$ and therefore $x=\{\langle x,x\rangle\}$. It follows that $x$ is a bijection of itself with itself, and it is also an element of itself.

Also related:


*

*When is $x=\{ x\}$?

*How to construct $\{\{\{...\}\}\}$ in ZF without axiom of foundation
