Transitive sets: problem in proof of Lemma I.8.6 of Kunen's 'Foundations of Mathematics' I've been studying Kunen's notes titled 'The Foundations of Mathematics'.
Definition I.8.1 in Kunen says

$z$ is a transitive set iff $\forall y \in z\, [y \subseteq z]$

In the proof of Lemma I.8.6, $\alpha$ is a transitive set and $x,y,z \in \alpha$. Kunen claims that 

… we have $x \in y \in z \rightarrow x \in z$ because the $\in$
  relation is transitive on $\alpha$ …

But this does not seem to follow from Kunen's definiton of transitivity. Consider $\alpha=\{\emptyset, \{\emptyset\},\{\{\emptyset\}\}\}$. Then it is transitive by Kunen's definition but if we take $x$, $y$ and $z$ to be the three elements in the order given then $x \in y \in z$ but $x \notin z$.
Am I reasoning correctly?
I also have the print edition of the notes and the definition and proof are identical there.

I agree with you that the end of the proof of Lemma I.8.6 in Kunen is misleading and incomplete.  It can be fixed as below.
Let $\alpha$ be an ordinal, that is, $\alpha$ is a transitive set (every element of it is a subset of it) such that the relation $\in$ on $\alpha$ is a well-order (i.e., a (strict) total order for which every nonempty subset has a minimal element).
At the end of the proof of I.8.6 we have $x\in y\in z\in\alpha$ with $x,y,z\in\alpha$.  We have to show that $x\in z$. Because $\in$ is a total order, we either have (1) $z=x$, or (2) $z\in x$, or (3) $x\in z$.  Case (1) is impossible because that would give a cycle $x\in y\in x$ with no minimal element for the set $\{x,y\}$.  Case (2) is impossible because that would give a cycle $x\in y\in z\in x$ with no minimal element for the set $\{x,y,z\}$.  So case (3) $x\in z$ must hold.
Your example $A=\{\emptyset, \{\emptyset\},\{\{\emptyset\}\}\}$ is a transitive set, but the $\in$ relation on $A$ is not a well-order, as it is not even a total order: the elements $\emptyset$ and $\{\{\emptyset\}\}$ are not comparable, as neither is a member of the other.  As you correctly observed, $A$ being a transitive set is not enough to conclude that the relation $\in$ on $A$ is a transitive relation.
