I have some trouble understanding a statement for ordinals.
First of all, the definition I'm using is this: We say that $\alpha$ is an ordinal when $\alpha$ is a transitive set (that is, $\cup\alpha\subset\alpha$) and well-ordered with the $\in$ ordering.
The statement says that if $\alpha,\beta$ are ordinals such that $(\alpha,\in)\cong(\beta,\in)$, then $\alpha=\beta$.
The proof starts as following: Let $f:\alpha\to\beta$ be an isomorphism and $f^{-1}$ its inverse. Then for all $x,y\in\alpha$ we have that $x\in y\implies f(x)\in f(y)$. Also, for all $x,y\in\beta$ we have that $x\in y\implies f^{-1}(x)\in f^{-1}(y)$. These two yield that for all $x,y\in\alpha$ it is $x\in y\iff f(x)\in f(y)$. Denote this statement by $(*)$.
Up to this point everything is clear. Now we want to prove that for all $x$ it is $f(x)=x$ and we do this by using transfinite induction. The proof says at this point that "obviously, $\min\alpha=\emptyset$, therefore by $(*)$, $f(\emptyset)=\emptyset$". I really can't see why $\emptyset$ is an element of $\alpha$.
I really need some clarification on this.
$OR$, the class of the ordinals, is well ordered by the $\in$ and $\emptyset$ is an ordinal. So we should have $\emptyset \in \alpha$ (assuming $\alpha \neq \emptyset$).
Let us verify that $\emptyset \in \alpha$ from the definition of ordinal: Let $\sigma = min(\alpha,\in)$, suppose for a contradiction that there is $\tau \in \sigma $. Since $\alpha$ is transitive it implies $\tau \in \alpha $ which contradict the minimality of $\sigma$. Hence $\sigma = \emptyset$.