I have to prove the following
The set $S = \{x \in X: x$ is transitive and $\forall \emptyset \neq z \subset x, z$ has an $\in$-minimal element$\}$ is an inductive set if $X$ is an inductive set
I'm given the definition that
$t$ is $\in$-minimal in $z$ if $\nexists s \in z$ such that $s \in t$.
I've already proved that $\emptyset \in S$ and that if $x \in S$, $x \cup \{x\}$ is transitive. But I need to show that all nonempty subsets of $x\cup\{x\}$ contain an $\in$-minimal element.
My argument goes like this. Allow $z \subset x\cup\{x\}$. If $x \notin z$, then $z$ contains an $\in$-minimal element since $z\subset x$ and $x \in S$. Now assume that $x \in z$ and $z$ is not the singleton. I must show that there is an $\in$-minimal element.
Find $t \in z\setminus \{x\}$ such that $t$ is $\in$-minimal. I only need to show that $x \notin t$ to show that $t$ is a $\in$-minimal element of $z$ but I dont know how. Any hints?
Edit: If I assume $x \in t$, then it would imply that $x \in t \in x$, but I'm not sure how to arrive at contradiction.