Showing that a Transitive Set of Transitive Sets is an Ordinal

Question

My definition of an ordinal is a transitive set that's well ordered by $\in$. Let $\alpha$ be a transitive set all of whose elements are transitive sets. Since every element of $\alpha$ is transitive, the $\in$ relation among the elements of $\alpha$ is a transitive relation. By Foundation, $\in$ is an irreflexive relation, so since $\in$ is a transitive, irreflexive relation, it makes $\alpha$ into a partially ordered set.

If I can show that $\alpha$ is totally ordered, then well-orderedness follows from there being no infinite descending chain of $\ni$'s. But I'm having trouble showing that any two elements of $\alpha$ are comparable.

Let $x, y$ be distinct elements of $\alpha$. Since $\alpha$ is transitive, $x, y$ are also distinct subsets of $\alpha$. Here's where I get stuck. How do I show that $x\in y$ or $y\in x$ (or perhaps $x\subsetneq y$ or $y\subsetneq x$)?

Answer 1

I suppose you already know some facts about ordinals, such as:
(1) No set contains all the ordinals.
(2) Every transitive set of ordinals is an ordinal.

Let $x$ be any transitive set of transitive sets. Let $\alpha$ be the least ordinal such that $\alpha\notin x$. If $x\subseteq\alpha$ then $x$ is an ordinal by (2); so we may assume that $x\not\subseteq\alpha$, i.e., $x\setminus\alpha\ne\emptyset$. By the Axiom of Foundation, we can choose $y\in x\setminus\alpha$ so that $y\cap( x\setminus\alpha)=\emptyset$. Since $x$ is transitive, it follows that $y\subseteq x$ and $y\setminus\alpha=\emptyset$, i.e., $y\subseteq\alpha$. But $y$ is transitive (since $y\in x$), so it follows by (2) that $y$ is an ordinal.

Now we get a contradiction if we try to compare the ordinals $y$ and $\alpha$. We know that $y\notin\alpha$ because $y$ was chosen from $x\setminus\alpha$. But if $\alpha=y$ or $\alpha\in y$, then $\alpha\in x$ (since $y$ is an element of the transitive set $x$), contradicting the assumption that $\alpha\notin x$.

Answer 2

Say that $x$ and $y$ in $\alpha$ are incomparable if they are distinct and neither is an element of the other. If $\alpha$ contains incomparable elements, let $x\in\alpha$ be $\in$-minimal among elements incomparable with at least one element of $\alpha$, and let $y\in\alpha$ be $\in$-minimal among elements incomparable with $x$. Let $z\in y$; then $z$ is comparable with $x$, i.e., $z\in x$, $z=x$, or $x\in z$. In the last two cases $x\in y$, which is impossible, so $z\in x$. But $z$ was an arbitrary element of $y$, so $y\subseteq x$.

Now let $z\in x$; $z$ is comparable with $y$ by the $\in$-minimality of $x$. If $y\in z$ or $y=z$, then $y\in x$, which is impossible, so $z\in y$; and $z$ was arbitrary, so $x\subseteq y$, and hence $x=y$, contradicting the incomparability of $x$ and $y$. It follows that $\in$ totally orders $\alpha$.