TITLE: Transitive set ordered by epsilon
QUESTION [7 upvotes]: This problem is from here

Definition) A set t is called transitive if every element of every element of t is itself an  element of t, or equivalently, if every
  element of t is a subset of t. A set t is said to be  ordered by
  epsilon if for any two elements x and y of t, either x ∈ y or x = y or
  y ∈ x
5.  Let x and y be transitive sets, each ordered by epsilon. 
(a) Show using foundation that if y - x ≠ Ø, then x ∩ y ∈ y. [Hint:
  Let z be an epsilon-minimal element of y - x, and show z = x ∩ y.]
(b) Show that either x ∈ y or x = y or y ∈ x. [Hint: A proof like that
  of (a) shows also that if x - y ≠ Ø, then x ∩ y ∈ x] 
Can there be two distinct sets each of exactly four elements, each
  both transitive and  ordered by epsilon?

I'm solving (a), and I found out that $z\subseteq x \cap y$. (Let $w$ be an element of $z$, then since $z \in y$ implies $z \subseteq y$, $w$ is an element of $y$. If $w$ is not in $x$, then $w \in z \cap (y-x)$, a contradiction. Hence $w \in x$ and $z\subseteq x$, consequently $z \subseteq x \cap y$.)
But I don't know how to show that $z=x \cap y$. Is the problem okay? If I take $x$ to be empty set, then the conclusion is $\emptyset \in y$, which seems to be false. 
Edit Thanks to hints, I solved (a),(b) and also I think that a set of exactly four elements, both transitive and ordered by epsilon is unique: Obviously $4=\{ 0,1,2,3 \}$ is one. Let $y$ be another one. If $4\in y$, then $0,1,2,3,4 \in y$, a contradiction. If $y \in 4$, then $y$ is one of $0,1,2,3$ which do not have 4 elements, a contradiction.
Also I found that every (nonempty) transitive set has the empty set: If $x$ is a nonempty transitive set, then $y \cap x = \emptyset$ for some $y\in x$. If $z\in y$, then $z \in x$, a contradiction. Hence the $\in $-minimal element of $x$ should be the empty set.
Then my question: Every transitive set ordered by epsilon is an ordinal?

REPLY [5 votes]: How do you define an ordinal? Often the ordinals are defined as transitive sets which are well-ordered by $\in$, these are known as the von Neumann ordinals.
However $\in$ is well-founded, so well-ordered by $\in$ is equivalent to being [linearly] ordered by $\in$.
So yes, every transitive set which is ordered by $\in$ is a von Neumann ordinal, and vice versa.