Unary Predicate Symbol Research Articles

An axiomatization for a class of two-cardinal models

In this note we give a simple recursive axiomatization for the class of structures of type (ℶω ℵ0). This solves a problem of Vaught which is Problem 13 in the book [1] of Chang and Keisler. The same technique is used to get a recursive axiomatization for the class of κ-like structures where κ is strongly ω-inaccessible.Let us fix throughout some recursive first-order language L, and until further notice let us suppose that included in L is a distinguished unary predicate symbol U. For cardinals κ and λ with κ ≥ λ ≥ ℵ0, we say the structure has type (κ, λ) if card(A)= κ and card . Let K(κ, λ) be the class of all structures of type (κ, λ). For each ordinal α define 2ακby 20κ = κ, and 2ακ= ⋃ {2λ: λ = 2βκ for some β < α} when α > 0. Let Vaught proved the following theorem in [7].Theorem (Vaught). Suppose a is a sentence such that for each n < ω there are κ, λ with κ > 2λn and a model of σ of type (κ, λ). Then whenever κ ≥ λ ≥ ℵ0, the sentence σ has a model of type (κ, λ).

Effectiveness and Vaught’s gap $\omega $ two-cardinal theorem

We consider functions f with the property that whenever a is a sentence in L,,,,, then f(a) f(a) model, then a admits all types. A question of Barwise is answered by showing that no such f is recursive, and that the least such f is not co-r.e. Barwise proves in [1] an effective version of Vaught's gap X two-cardinal theorem [7] for a class of sentences which have a certain syntactic form. He then asks (Problem 2.12 of [1]) the following question concerning the possibility of extending this result to the set of all sentences: What can be said about the effectiveness of a function f (and, in particular, the least J) such that for any sentence a, if a has a gap > f (a) model, then a admits all types? Note that we have reformulated Barwise's question in a way that best fits our answer. First, some notation and definitions. Let e be a sufficiently rich, recursive, first-order language which has among its symbols a distinguished unary predicate symbol U. We will deal throughout with this language P. We let E denote the set of all 6-sentences. A type is a pair (K, X) of infinite cardinals such that K >? X, and we say that the X-structure W = (A, U,... ) has type (K, X) if card (A) = K and card (U) = X. A sentence a admits the type (K, X) if it has a model of type (K, X). For each n n structure iff K > Dn(X) We can now state Vaught's gap X theorem. VAUGHT'S THEOREM. If a has a gap > n model for each n > X, then a admits all types. Barwise defines, for each n < t, a class of formulas which he calls 3V( )mod U formulas. By induction on n, we define when 4 is a V(n)-mod U formula: 0 is a V(?)-mod U formula iff it is universal, and 4 is a V(n+')-mod U formula iff it is in the form Vx (3y, E U) (3Y2 E U)... (3ymEe U)01, where 41 is a V(n)-mod U formula. Then, a formula is a Vg(n)-mod U formula if it is obtained from a V(n)-mod U formula by means of existential quantification. Received by the editors December 23, 1975. AMS (MOS) subject classifications (1970). Primary 02H05, 02G05.

A note on the two cardinal problem

In this note we prove a theorem concerning the two cardinal problem (see [1 ], [4], [6], [7] for reference and for some of the standard notation); this result has been referred to in [6, p. 311] and [7, (3.7) ]. The problem, first proposed by Vaught, is as follows. Let T be a first-order theory and let U be a unary predicate symbol in the language of T. T is said to admit the pair a, ( of cardinals if there exists a model M=(A, U, S, * * * ) of T such that |A| =a and I U|I =j3. Suppose T admits a pair a, ( where a >l >?w. Then what other pairs of cardinals y, 5 must T admit? The following theorem gives a partial answer.