TITLE: Is there a transitive set that is non empty and doesn't contain the empty set?
QUESTION [5 upvotes]: Sorry for being so naive and because this is maybe a silly question but all the examples I can think of contains the empty set, and it's not clear to me whether this makes sense at least intuitively. 
For example, let's supose that we have a set $A\neq \emptyset$ that is transitive. Then there's $x\in A$. But then every element $x_{i}$ in $x$ is element of $A$.  Then every element $x_{j}$ of $x_{i}$ is in $A$. And so on.  This processes might be infinite and hopefully $A$ doesn't have the empty set...

REPLY [7 votes]: Assuming the Axiom of Foundation, no there isn't.  For example, assume $t \neq \emptyset$ is transitive.  Let $x$ be an $\in$-minimal element of $t$.  Then $x \cap t = \emptyset$. But since $t$ is transitive, $x \subseteq t$ so that $x = x \cap t = \emptyset$.
Edit 1: As you point out, possibly that chain of elements could be infinite, but the Axiom of Foundation says that $\in$ is well-founded and that is equivalent to saying that there is no descending chain $ \cdots \in x_1 \in x_0$.
Edit 2: As Andres points out below not assuming Foundation it is consistent that there is a set $A = \{ A \}$ i.e. $A \in A$ so that $A \neq \emptyset$.  But then $A$ is transitive because $\forall y \in A (y \subseteq A)$ ($A \in A$ and surely $A \subseteq A$). Relating this to my first edit, we would have a descending chain $ \cdots \in A \in A \in A$.