Set membership as a relation on a particular set

Question

I am well aware that set membership does not form a relation over all sets. Indeed, it would form a proper class of ordered pairs of sets. However, if we are already given a particular set, say $\alpha$, does $$< : = \{(\mu, \pi) \in \alpha \times \alpha \mid \mu \in \alpha\}$$ not form a set? If so couldn’t we just have that relation be defined on all ordinals? Many set theory books mention that universal membership is a proper class and so we only think of it as a pseudo relation, but make no mention of that question when considering a particular fixed set. So then each ordinal $\beta$ is actually a well-ordered set, yet the class of all ordinals $\Omega$ is still only pseudo ordered?

Answer 1

Yes, this is a consequence of the separation axiom scheme: for $\alpha$ a set (ordinal or otherwise), the class of pairs $(\beta,\gamma)\in\alpha\times\alpha$ with $\beta\in\gamma$ is a set. Just apply separation to the set $\alpha\times\alpha$ using the formula "$x\in y$."

So in particular, yes, every ordinal is genuinely a well-ordered (by "$\in$") set.