
The least ordinal not in any transitive model of ZFC can also be described as the supremum of the heights of transitive models of ZFC. It is natural here to consider the class S consisting of all ordinals λ for which there is a transitive model of ZFC of height λ. Thus, the ordinal of your title, when it exists, is simply the supremum of the countable members of S. There are a number of relatively easy observations:
If M is a transitive model of ZFC, then so is L^{M}, the constructible universe as constructed inside M, and these two models have the same height. Thus, one could equivalently consider only models of ZFC + V = L.
It is relatively consistent with ZFC that S is empty, that is, that there are no transitive models of ZFC. For example, the least element of S is the least α such that L_{α} is a model of ZFC. This is sometimes called the minimal model of ZFC, though of course it refers to the minimal transitive model. It is contained as a subclass of all other transitive models of ZFC. The minimal model has no transitive models inside it, and so it believes S to be empty.
The least element of S (and many subsequent elements) is Δ^{1}_{2} definable in V. This is because one can say: a real codes that ordinal iff it codes a wellordered relation and there is a model of that order type satisfying that no smaller ordinal is in S (a Σ^{1}_{2} property), also iff every wellfounded model of ZFC has ordinal height at least the order type of the ordinal coded by z (a Π^{1}_{2} property).
If S has any uncountable elements, then it is unbounded in ω_{1}. The reason is that if L_{β} satisfies ZFC and β is uncountable, then we may form increasingly large countable elementary substructures of L_{β}, whose Mostowski collapses will give rise to increasingly large countable ordinals in S.
In particular, if there are any large cardinals, such as an inaccessible cardinal, then S will have many countable members.
If 0^{#} exists, then every cardinal is a member of S. This is because when 0^{#} exists, then every cardinal κ is an Lindiscernible, and so L_{κ} is a model of ZFC. Thus, under 0^{#}, the class S contains a proper class club, and contains a club in every cardinal.
S is not closed. For example, the supremum of the first ω many elements of S cannot be a member of S. The reason is that if α_{n} is the n^{th} element of S, and λ = sup_{n} α_{n}, then there would be a definable cofinal ω sequence in L_{λ}, contrary to the Replacement axiom.
S contains members of every infinite cardinality less than its supremum. If β is in S, then we may form elementary substructures of L_{β} of any smaller cardinality, and the Mostowski collapses of these structures will give rise to smaller ordinals in S.
If β is any particular element of S, the we may chop off the universe at β and consider the model L_{β}. Below β, the model L_{β} calculates S that same as we do. Thus, if β is a limit point of S, then L_{β} will believe that S is a proper class. If β is a successor element of S, then L_{β} will believe that S is bounded. Indeed, if β is the α^{th} element of S, then in any case, L_{β} believes that there are α many elements of S.
If S is bounded, then we may go to a forcing extension V[G] which collapses cardinals, so that the supremum of S is now a countable ordinal. The forcing does not affect whether any L_{α} satisfies ZFC, and thus does not affect S.
Reading your question again, I see that perhaps you meant to consider a fixed M, rather than letting M vary over all transitive models. In this case, you will want to look at finestructural properties of this particular ordinal. Of course, it exhibits many closure properties, since any construction from below that can be carried out in ZFC can be carried out inside M, and therefore will not reach up to ht(M).


answered Feb 25 '10 at 13:33

