[FOM] What is second order ZFC?

Harry Deutsch hdeutsch at ilstu.edu
Thu Sep 5 19:28:17 EDT 2013

Here's a stab at answering Harvey's question as asked.  No. Ch is not an axiom of ZFC.  No, it is not an axiom of ZFC second order, since this is not axiomatizable.  That's just part of the test, and I'm hardly an expert.  Harry
On Sep 5, 2013, at 12:31 PM, Cole Leahy wrote:

> On Tue, 3 Sep 2013, Martin Dowd wrote:
> Suppose at stage alpha one adds the subsets which are second order definable. Call the result L_2.  Clearly L\subseteq L_2\subseteq V, so if V=L all three are equal. Obvious questions include the following. Is L_2 a model of ZFC? ... Is it consistent that L_2=neq V? ... Is L_2=L? ... What kind of sets might be in L_2-L? Is CH true in L_2?
> If memory serves, Myhill and Scott showed in their "Ordinal Definability" (1971) that L_2 = HOD follows from AC. We can therefore answer your questions by noting that ZF + L != HOD != V and ZFC + V = HOD + ~CH are consistent relative to ZF. (These facts are apparently due to McAloon.) I'm not sure whether AC can be added in the first case.
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