On Existence Proofs of Hanf Numbers

This paper refines some results of Barwise [1] as well as answering the open question posed at the end of [1] about the Hanf number of , positively. We conclude by showing that the existence of a Hanf bound for Y.,., cannot be proved in the natural formally intuitionistic set theories with bounded predicates decidable of [3], [4] and [5]. All notation not explained below is taken from [1]. In the Appendix, we give the axioms of ZFO, ZF1, and T in full. We remark that an important point about the axiom of foundation was not emphasized in [1]. This axiom was intended to be the axiom scheme (Vx)((Vy e x)(A(y)) -> A(x)) -> (Vx)(A(x)), where y does not occur in A = A(x), instead of the more customary (Vx)(Vy)(y e x -> (3z e x)(Vw e z) (w O x)). This is of no consequence in the presence of full separation, but is vital when considering ZFo and the T below, for with the customary form of foundation, these cannot even prove the existence of R., In [1], a proof of the following is sketched. THEOREM A. ZFo V "the Hanf number of c-logic exists." 2 But ZF1F "the Hanf number of each 2.cA exists." In [1], the question of whether T proves "the Hanf number of .F, exists" is raised. We will show that T proves "the Hanf number of .2,, exists for all infinite cardinals K." A related argument similarly shows that T proves X1(91)-separation. Standard arguments show that T proves A,(9P)-separation. It is easy to see how one can show in Tthat cardinal successor and exponentiation are defined for all cardinals, and that every set is in one-to-one correspondence with a cardinal. These two consequences of T will be used in Theorem 1 below. THEOREM 1. T "for all infinite cardinals K, the Hanf number of ?Tx exists." T E1(?)-separation. PROOF. As noted in [1, ?2], if cp E SKx has a model of power > A, then 'p has a model of power A A) is A1(69). Let us call a functionf special just in case (1) the domain off is some ordinal a, (2) for each P < a, f(p) is the least infinite cardinal A with Q(A) and f(y) < A, all