$X$ is ordinal if and only if X is transitive and well-ordered under $\in$.
Let $X$ be a non-empty set of ordinals. What are the necessary and sufficient conditions for $X$ to be an ordinal?
I found that the answer is quite simple in case $X$ is finite.
Assume $X$ has $n$ elements $a_1,a_2,\cdots,a_n$. First of all, $X$ is well-ordered under $\in$, so WLOG we can safely assume that $a_1\in a_2\in a_3\in\cdots\in a_n$, hence $a_i\in a_n$ for all $0<i<n$. It is easily to prove that $a_1=\emptyset$. As a result, $X$ is ordinal $\implies$ $\emptyset\in X$ and $\exists c\in X,X\setminus\{c\}\subsetneq c$. One can easily verify that $\emptyset\in X$ and $\exists c\in X,X\setminus\{c\}\subsetneq c\implies X$ is ordinal.
Please help me in case $X$ is infinite!
A set $X$ of ordinals is itself an ordinal if and only if $X$ is transitive, i.e. every element of $X$ is a subset of $X$.
Since, for every ordinal $\alpha$, $\alpha$ is transitive this is clearly necessary.
But it is also sufficient: A set $Y$ is an ordinal iff it is transitive an strictly totally ordered by $\in$. Since $X$ is a set of ordinals, it is strictly totally ordered by $\in$ and by our requirement it is transitive as well.