Let S be the set of all ordinaldefinable real numbers and let A(S) be the statement that S is denumerably infinite. If ZFC is consistent, it has been proved (by A. Levy) that it remains consistent if we adjoin A(S) to it as a new axiomso let us assume that this has been done. Now S has a natural wellordering inherited from the wellordering of the class of all ordinaldefinable sets. Let O(S) be the ordinal number of this natural wellordering of S. My question is, how does the size of O(S)which must be very large indeedcompare with other large denumerable ordinal numbers, such as the smallest nonrecursive ordinal number (in the sense of CHURCH/KLEENE) or some of the socalled "admissible" ordinal numbers?
The class HOD consisting of the hereditarily ordinal definable sets is transitive proper class inner model of ZFC, and your assumption A(S) amounts to the assertion that $\mathbb{R}^{\text{HOD}}$, the set of reals of this model, is a countable set in the ambient settheoretic universe V. The shortest enumeration of this set in HOD, however, has the order type of the continuum of HOD, which we may denote $\frak{c}^{\rm \text{HOD}}$, and this would seem to be (one natural interpretation of) the ordinal to which you are referring.
Although your assumption means that this ordinal is countable in the ambient universe $V$, we may also consider it inside the inner model HOD, where it is simply the continuum. Thus, from the perspective of HOD, the ordinal is very large indeed by the standards that you have mentioned. In particular, HOD thinks that it is much larger than $\omega_1^{ck}$, the first noncomputable (nonrecursive) ordinal, and also that it is admisssible, and a limit of admissibles and a limit of limits of admissibles, and so on. In particular, it is a cardinal in the constructible universe L, since it is a cardinal in HOD, which contains L.
And once we know that HOD thinks those things, then it follows that our universe V also agrees, since those properties about this ordinal are absolute from HOD to V, since HOD computes the L hierarchy the same as we do in V.
Finally, let me point out that the forcing method allows us enormous flexibility in controlling the precise value of this particular ordinal. Namely, suppose that V is any model of ZFC and that $\delta$ is any cardinal of $V$ with uncountable cofinality and above the continuum. First, by a theorem of Solovay, there is a forcing extension $V[G]$ in which the continuum becomes exactly $\delta$; one simply adds $\delta$ many Cohen reals. Next, by a theorem of MacAloon, there is a further extension $V[G][H]$ not adding any reals in which the reals of $V[G]$ become ordinaldefinable in $V[G][H]$; one may do this by forcing to code the individual reals into the GCH pattern high above $\delta$. Finally, we may perform collapse forcing to build a model $V[G][H][K]$ in which $\delta$ becomes countable. If the reals of $V[G]$ were coded into the GCH pattern up high, this last step will preserve those reals as ordinal definable, and it will create no new ordinal definable sets since it is almost homogeneous forcing. Thus, the $\mathbb{R}^{\text{HOD}}$ of $V[G][H][K]$ is the same as $\mathbb{R}^{V[G]}$ and it will have cardinality $\delta$ there. Thus, we have arranged your ordinal to be exactly $\delta$ in $V[G][H][K]$, a model satisfying your assertion $A(S)$. But $\delta$ was an arbitrary cardinal with uncountable cofinality above the continuum in $V$.

1$\begingroup$ In particular, the argument of my last paragraph shows that your ordinal can be any uncountable regular cardinal of $L$, in the sense that for any such cardinal there is a forcing extension of $L$ in which that cardinal is your countable ordinal, and these cardinals can of course become very large by the measures you have mentioned. $\endgroup$ – Joel David Hamkins Jun 12 '11 at 1:13

$\begingroup$ Thanks alot for your answer. Much of of what you present goes far beyond my knowledge of the subject. I am fascinated by large countable ordinals (above the recursive ordinals) and whether there are any limits on how large they can be made to be while still remaining countable. I wonder if, given any specific preassigned countable ordinal Z, it is cosistent with ZFC+A(S) to assume that O(S)>Z. $\endgroup$ – Garabed Gulbenkian Jun 14 '11 at 17:36

1$\begingroup$ The method of forcing shows that any ordinal at all (indeed, any set) can be made countable in a forcing extension of the universe. In this sense, there is no a priori limit to the countable ordinals, and one will not be able to identify any internal property of an ordinal that reveals it as necessarily uncountable. Meanwhile, if you are interested in definability issues in set theory, you might enjoy my recent paper (arxiv.org/abs/1105.4597), and particulalry the discussion there of what we call the "math tea" argument. $\endgroup$ – Joel David Hamkins Jun 15 '11 at 10:22