# Proving that a set A cannot contain another set B which contains A?

From the ZFC axiom of regularity, which states that every non-empty set contains an element disjoint from it, we can deduce that there is no set $A$ such that $A \in A$.

A proof is outlined here:

Let $A$ be any set. Then $\{A\}$ is a set, and by regularity $\{A\}$ must contain an element disjoint from $\{A\}$. The only element of $\{A\}$ is $A$, so $A\cap\{A\}=\varnothing$, and it follows immediately that $A\notin A$.

I am trying to convince myself that there cannot exist distinct sets $A$ and $B$ such that $B \in A$ and $A \in B$ using the same axiom, but I am no expert in logic and wonder whether this is true, and how this can be deduced.

The main issue I see is we end with infinite recursion. Suppose $A = \{B,C\}$ and $B=\{A,D\} = \{ \{B,C\}, D \}$.

Do we arrive at a contradiction somehow, and is $B$ considered a member of $B$?

Yes, it's true. Apply the axiom of regularity to the set $\{A, B\}$.
You ask whether you can prove $B \in B$. Of course, this is not a meaningful question with the axiom of regularity, since your premise is false. However, without the axiom, I believe the relations $B \in A \in B$ do not imply that $B \in B$.
Certainly there could be alternative axioms that incorporate your point of view but part of learning to use $ZFC$ and axioms is realizing that you are ultimately restricted to the basic assumptions: axioms.