We develop an account of logical consequence for the second-order language of set theory in the spirit of Boolos's plural interpretation of monadic second-order logic. There is little doubt that a second-order axiomatization of Zermelo-Fraenkel set the- ory plus the axiom of choice (ZFC) is desirable. One advantage of such an ax- iomatization is that it permits us to express the principles underlying the first-order schemata of separation and replacement. Another is its almost-categoricity: M is a model of second-order ZFC if and only if it is isomorphic to a model of the form � Vκ, ∈∩ (Vκ × Vκ)� ,f orκ a strongly inaccessible ordinal. We obtain similar benefits when we allow for the existence of Urelemente. The axioms of second-order ZFC with Urelemente (ZFCU) are not able to specify the structure of the universe up to isomorphism, but McGee has recently shown that, pro- vided one takes the range of its quantifiers to be unrestricted, the addition of an axiom that states that the Urelemente form a set to the axioms of ZFCU will characterize the structure of the universe of pure sets up to isomorphism. 1 In sum, there is much to be gained from the ability to employ second-order quantification in the context of set theory. What is much more controversial is that we can, with a clear conscience, develop set theory within a second-order language. The standard interpretation of second- order quantification takes second-order variables to range over the sets of individuals which first-order variables range over. This interpretation may be convenient for the development of second-order arithmetic but it will not do for the purpose of develop- ing set theory in a second-order language. The reason is not difficult to state. When we do set theory, we take our first-order variables to range over all sets. But if we take our second-order variables to range over sets of sets in the range of the first-order vari- ables, then second-order comprehension will fail. A simple instance of second-order comprehension such as ∃X∀y (Xy←→ y / ∈ y) will be false on account of Russell's
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