Is this a way to define an ordinal $\lambda$?

Question

A set $\lambda$ is by definition an ordinal if

1. $\forall x\in\lambda : x\subseteq \lambda$ and
2. $\lambda$ is well-ordered by $\in$ .

May I equally define an ordinal as a set $\lambda$ having property (1.) and the property

3. $\lambda$ is well-ordered by $\subseteq$ ?

A reformulation of the question is this:

Let $\lambda$ be a set satisfying property (1.). Does $\lambda$ then satisfies property (2.) if and only if it satisfies property (3.)?

Answer 1

Yes, this is true (though in (3) you should really say $\subset$ well-orders $\lambda$, or that $\subseteq$ nonstrictly well-orders $\lambda$). First, note that (2) implies (3) even without assuming (1). For suppose $\lambda$ satisfies (2) and $x,y\in \lambda$. Then since $\in$ is a transitive relation on $\lambda$, $x\in y$ implies $x\subseteq y$, but in fact by Regularity it implies $x\subset y$. Conversely, if $x\subset y$, then we cannot have $x=y$ or $y\in x$ (since in the latter case, we would have $y\subset x$), so since $\in$ is a total order we must have $x\in y$. Thus $x\in y$ iff $x\subset y$, so $\subset$ is also a well-ordering of $\lambda$.

Conversely, suppose $\lambda$ satisfies (1) and (3). We will show by induction on $y\in\lambda$ (with respect to the well-ordering $\subset$) that $x\subset y$ iff $x\in y$ for all $x,y\in\lambda$. So suppose that $x\subset z$ iff $x\in z$ for all $x\in\lambda$ and all $z\in\lambda$ such that $z\subset y$. First, suppose $x\in\lambda$ and $x\in y$. If $x\not\subset y$, then $y\subseteq x$. This implies $x\in x$, contradicting Regularity. Conversely, suppose $x\in\lambda$ and $x\subset y$. If there exists a $z\in\lambda$ such that $x\subset z\subset y$, then by the induction hypothesis, $x\in z$, so $x\in y$. So we may assume that no such $z$ exists; i.e., that $y$ is the successor of $x$. Since $x\subset y$, there exists some element $a\in y$ such that $a\not\in x$. By (1), we have $a\in \lambda$, and so by the induction hypothesis, $a\not\in x$ means $x\subseteq a$. If $x=a$, we get $x\in y$, so we're done. If $x\subset a$, then because $y$ is the successor of $x$, we have $y\subseteq a$. But $a\in y$, so this implies $a\in a$, contradicting Regularity.