Request for an example from the definition of Ordinal I have a request for an example. I have the following definition of an ordinal: 
A set $\alpha$ is said to be an $\textit{ordinal}$ if
1.) $\alpha$ is well-ordered by $\in$, and
2.) $\beta\in\alpha\rightarrow\beta\subseteq\alpha$
My question is, is there such thing as a set where 1 holds and 2 doesn't? As far as I can tell it seems that satisfying 1 would make 2 a given. Am I correct or no?
 A: Sure: let $0 = \emptyset$, $1 = \{0\}, 2 = \{0,1\}$ — the usual definitions of these integers in set theory. Then the set $A = \{0,2\}$ is well-ordered by $\in$, but it doesn't satisfy condition 2. because $1\in 2\in A$ but $1\notin A$.
A set $X$ satisfying $(\forall x\in X)\,x\subseteq X$ is said to be transitive. So a (von Neumann) ordinal, an $\alpha$ as in your definition, is just a transitive set well-ordered by $\in$. 
As Noah Schweber points out in his answer, any set of ordinals $A$ is well-ordered by $\in$; so a set $A$ will fail to be an ordinal only when for some $\alpha\in A$, the set omits some $\xi < \alpha$ (here, "<" means precisely "$\in$"). 
Although the circularity involved makes the following a crummy definition, nevertheless it's true:

A set $A$ is an ordinal iff all of its members are ordinals, and it's downward-closed with respect to the ordinal ordering — that is, $\alpha < \beta \in A \implies \alpha\in A$, which can also be written $(\forall \beta\in A)\,[0.\beta)\subseteq A$.

Nothing deep there: if $\beta$ is an ordinal, then $\beta = [0,\beta)$.
A: Sure - fix any ordinal $\alpha$, and look at the set $\{\alpha\}$. If $\alpha=0$, then this is an ordinal (specifically, $1$), but otherwise this isn't - e.g., $\{17\}$. However, $\{\alpha\}$ is always (trivially) well-ordered by "$\in$." More generally, any set of ordinals is always well-ordered by "$\in$."
