What's the least class of ordinals closed under successor and the limits of omega-sequences? What is the smallest class $S$ of ordinals that contains $0$, closed under successor and the limits of omega-sequences, i.e.,
$$
\forall i \in \omega [\alpha_i \in S] \Rightarrow \sup_{i} \alpha_i \in S.
$$
Will it be an initial segment of the class of all ordinals?  Or will it contain all ordinals of cofinality $\aleph_0$?  What other ordinals does $S$ contain?
Background: I've been reading Chapter 6 of Proofs and Computations by Schwichtenberg and Wainer, where they explain the generalized recursion theory in the abstract setting, in the style of Scott/Plotkin.  There, they give a type-theoretic definition of ordinals: the type of ordinals is
$$
\mathbf O := \mu_\xi(\xi, \xi\rightarrow\xi, (\mathbf N\rightarrow\xi)\rightarrow\xi).
$$
where $\mathbf N$ is the type of naturals. I assume this consists of three constructors, each of which corresponds to zero, successor and limits, respectively, and I believe $\mathbf O$ corresponds to the class $S$ defined above.
 A: This gives you exactly the set of all countable ordinals (assuming the axiom of countable choice).  Indeed, you can prove by induction that $S$ contains all countable ordinals (every countable limit ordinal is the supremum of all the ordinals below it, which form a countable set).  On the other hand, any successor of a countable ordinal is countable and any limit of a countable collection of countable ordinals is countable, since a countable union of countable sets is countable.
As Noah commented, this argument uses the axiom of countable choice to conclude that a countable union of countable sets is countable, and in $ZF$ it is actually possible for $\omega_1$ to be a countable union of countable sets, so $S$ would contain some uncountable ordinals.  However, even in $ZF$, $S$ will always be an initial segment of the ordinals.  Indeed, if $\beta$ is the least ordinal not in $S$, then let $S'$ be the set of ordinals less than $\beta$.  If $\alpha>\beta$, then $\alpha$ cannot be the sup of any collection of ordinals in $S'$, since $\beta$ is a smaller upper bound for any such collection.  Similarly, $\alpha$ cannot be the successor of any ordinal in $S'$.  Thus $S'$ is also closed under your operations, so $S'$ must be all of $S$.
I don't know whether it's consistent with $ZF$ for $S$ to be all of the ordinals.
