prove that for an ordinal $\alpha \notin \alpha$

Question

I understand bits of this proof but I don't understand what is meant by this proof contradicts the asymmetry of $(\alpha,\in_\alpha)$

Thanks

Answer 1

Observe $\in_\alpha$ is a well-order in $\alpha$.

This means that $\in_\alpha$ is a partial order and trichotomous and well-founded.

I am not sure what proof you are referring to, but if you show that any of the above properties do not hold, then a contradiction has taken place.

If $\alpha \in \alpha$, then one can derive a contradiction. Therefore it cannot be the case that $\alpha\in\alpha$.

Answer 2

Recall that a well-ordering is in particular a strict linear ordering. Namely, irreflexive, transitive, and total.

Since $(\alpha,\in)$ is a well-ordering, it means that for every $\beta\in\alpha$, $\beta\notin\beta$. Note that $\in$ plays the role of both the conventional membership, and the linear ordering.

So if $\alpha\in\alpha$, it implies that $\alpha\notin\alpha$. Of course, this is a contradiction. So it must be that $\alpha\notin\alpha$.