Upper bound of set of ordinals This is probably a naive question but, given a set of ordinals, is every upper bound of this set necessarily an ordinal?
 A: Intuitively yes, but that is because the order you are thinking of only compares ordinals.  Any set that is not an ordinal is not compared in this order, so cannot provide an upper bound.  If you use the order of set inclusion the answer is no.  $\emptyset$ is a fine ordinal, $\{\{\emptyset\}\}$ is not an ordinal in the usual encoding, yet $\emptyset \subset \{\{\emptyset\}\}$ so $\{\{\emptyset\}\}$ is an upper bound to $\{\emptyset\}$, which is a set of ordinal(s).
A: Given a set of ordinals $S$ there are many sets $y$ such that $\forall x\in S\mid x\in y$. An example is $y=S$, and $S$ is not necessarily an ordinal. In a weird sense these sets $y$ are "upper bounds" for the set $S$, but we don't consider sets that are not ordinals as part of the order of ordinals. So when the other answer given here says "Intuitively yes, but that is because the order you are thinking of only compares ordinals", to me this is not "intuitively" but "formally", as the order of ordinals only contains ordinals, and an upper bound for a set of ordinals must be part of the order of ordinals (i.e. an ordinal).
A: Is every upper bound of a set of rational numbers is a rational number? The answer is "yes" if you're only considering rational upper bounds, and "no" if you allow arbitrary real numbers. So when you say upper bound, you need to tell us what is the domain from which upper bounds are allowed to be taken.
Implicitly, this question seems to assume the answer to that is "the ordinals", in which case, of course. Every object you're allowed to compare is an ordinal, so given a set of ordinals, and upper bound is an ordinal, pretty much by definition.
