Cycle of Subsets implies Set Equality

Theorem
Let $A_1, A_2, \dotsc, A_n$ be sets.

Let:
 * $\forall k \in \set {2, 3, \dotsc, n}: A_{k - 1} \subseteq A_k$

and:
 * $A_n \subseteq A_1$

Then:
 * $\forall j, k \in \set {1, 2, \dotsc, n}: A_j = A_k$

Proof
Consider the set of sets $\mathbb A = \set {A_1, A_2, \dotsc, A_n}$

Consider the relational structure $S = \struct {\mathbb A, \subseteq}$.

We have from Subset Relation is Ordering that $S$ is an ordered structure.

The result follows from Ordering Cycle implies Equality.