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I don't see why the lack of a complete effective proof system for it should be a problem. I do in fact consider the power set of the naturals to exist, so I'm

• danharaj 8y ago
  It matters because it means second order arithmetic isn't a finite object in the same way first order effective theories are where you might have "infinitely" many axioms but they are packaged up in a finite object, an algorithm. The whole point of a formal system, its formality, is that it doesn't require any semantic notions to describe it. Second order arithmetic with its full model is no such thing. It is inherently infinitary and therefore not formal at all. And it's a treacherous object. C.f. Richard's paradox which demonstrates that "subset of the integers" is by no means a naive idea.
  • drdeca 8y ago
    Isn't it true that the only countable model of the natural numbers (including ordering) which is computable is the standard model of the naturals?