In Jech's Set Theory he claims (p. 20, new millennium edition):
If $C$ is a nonempty class of ordinals, then $\cap C$ is an ordinal, $\cap C \in C$ and $\cap C = \inf C$
Jech does not give a proof, stating that it is routine. I'm finding it difficult to prove that $\cap C \in C$ if $C$ is a proper class.
I know how to do this if $C$ is a set, but if $C$ is a proper class I see no way except by induction over all first-order formulae. Am I right in thinking this, or is there a more direct method?
Pick some element $c$ of $C$. Because the ordinals are totally ordered by inclusion, the intersection over the non-empty set $(c+1)\cap C$ is the same as the intersection over the class $C$. Now apply the set version and you're done.