[FOM] Higher Order Set Theory [Ackermann Set Theory]

[Robert M. Solovay]

Mr. Taranovsky's claims about the weakened form of SP are not correct.
Details follow below.


On Fri, 11 Mar 2005, Dmytro Taranovsky wrote:

>
> Being an elementary substructure implies that V(alpha) satisfies correct
> statements with parameters in V(alpha).  If we only allow formulas without
> parameters, then the statement would be stronger than ZFC + there is an
> (infinite) ordinal k such that there are at least k inaccessible cardinals and
> L_k satisfies ZFC, but weaker than ZFC + there are omega_1 inaccessible
> cardinals.
>
	First let me call SP- the variant of SP where we only require that
"there exists a strongly inaccessible kappa and an alpha > kappa such that
V(alpha) and V(kappa) are elementarily equivalent [satisfy the same
sentences without parameters]".

	ZFC + "there exist aleph_2 strongly inaccessible cardinals" proves
the consistency of ZFC + SP-; ZFC + SP- proves the consistency of ZFC +
"the order type of the class of inaccessible cardinals is greater than
aleph_1". Clearly the second claim refutes the last claim of the quoted
passage from Taranovsky.

	First work in ZFC + "there exists aleph_2 inaccessible cardinals".
Then in L, there are at least aleph_2 inaccessibles. But in L, GCH holds
and we can use the Shipman argument to prove the consistency of ZFC + SP-.
Since arithmetical statements are absolute from L to V, this completes the
proof of Con(ZFC + SP-).

	Now, towards our second claim, work in ZFC + SP-. Clearly SP-
relativizes to L. So without loss of generality, we work from here on out
in the theory ZFC + V=L + SP-. Let kappa be inaccessible and alpha > kappa
such that V(kappa) and V(alpha) are elementarily equivalent.

	We seek to show that ZFC + "the order type of the inaccessibles is
greater than aleph_1" holds in L(kappa). If there are at least aleph_2
inaccessibles in V(kappa) we are done. So, wlog, assume that there is an
ordinal eta < aleph_2 which is the order-type of the strong inaccessibles
in V(kappa). Similarly, let eta* be the order-type of the strong
inaccessibles in V(alpha). Then eta < eta* since kappa is strongly
inaccessible in V(alpha).

	We first argue that eta is not < aleph_1. If it were, let X be the
eta^{th} subset of omega. Clearly eta* has a similar definition in
V(alpha) and using the elementary equivalence of V(kappa) and V(alpha) we
conclude first that X = X* (X* is the eta*^{th} subset of aleph_1) in
V(alpha))  and then that eta = eta*. Contradiction.

	Similarly if eta = aleph_1 in V(kappa) then eta* = aleph_1 in
V(alpha). Since both V(kappa) and V(alpha) both correctly compute aleph_1,
we again have eta = eta* which is absurd.

	The upshot is that eta > aleph_1 as was to be shown.

	--Bob Solovay