Talk:Union of Subset of Ordinals is Ordinal

Set or class?
I believe we define an ordinal to be a set, and $\On$ to be the class of all ordinals, so that $\On$ is not an ordinal, so $\bigcup \On = \On$ is not an ordinal (note: if $n$ is an ordinal, so is $n^+$, so $n^+ \ni n$. Thus $n \in n^+ \in \On$, so $n \in \bigcup \On$.). Thus it seems, from this perspective, that we require a set of ordinals, rather than a class of them, in the premise to this theorem. On the other hand, Kelley, if I remember correctly, defines an ordinal as a class, so that $\On$ is the greatest ordinal, and the only ordinal which is not a set. --Dfeuer (talk) 00:04, 5 April 2013 (UTC)