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How is the Axiom 'V=L' related to Second-Order ZFC? Since it is consistent relative to first-order ZFC, is it consistent relative to second-order ZFC as well? Since 'V=L' is a first-order axiom, is there a second-order analogue to 'V=L' one must consider in order to make the question "Is [V=L] consistent relative to second-order ZFC" sensible?

Thomas Benjamin
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    Usually when people say "second-order ZFC", they mean second-order ZFC with standard semantics, which is a non-recursive theory (because second-order logic with standard semantics doesn't have a recursive proof system). Is that what you're talking about? If so, then the answer is quite interesting: if the continuum hypothesis (or V = L) is true, then it "follows" from second-order ZFC, whereas if it's false, then it's negation "follows". (I'm using "follows" in a semantic sense). But there's not really a way to tell which of these is true. See here: http://mathoverflow.net/a/78083 – Keshav Srinivasan Nov 03 '13 at 15:27
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    By the way, as mentioned in that Math.Overflow answer, you don't even need to go to a theory as powerful as second-order ZFC to have this phenomenon. The continuum hypothesis can be written as a statement of third-order arithmetic, and if it's true, then it "follows" from third-order arithmetic (with standard semantics), and if it's false, then its negation "follows". – Keshav Srinivasan Nov 03 '13 at 16:25
  • @ keshav: yes, that is exactly what I am talking about. The motivation for asking the question is to see whether the following theorem of Paul Cohen applies in the second-order case (found on pages 108-9 of his book SET THEORY AND THE CONTINUUM HYPOTHESIS, Dover Edition): "From ZF+SM or indeed from any axiom system containing ZF which is consistent with V=L, one cannot prove the existence of an uncountable standard model in which AC is true and CH false, nor even one in which AC holds and which contains non-constructible real numbers." Although Cohen certainly intended the theorem – Thomas Benjamin Nov 04 '13 at 07:10
  • to apply to the first-order case, if one studies the proof (found on page 109), there seems nothing in that proof limiting it to apply only to the first-order case. If that is the case then (since second-order ZFC is Categorical and thus essentially has only one uncountable model) if (as in the first-order case) second-order ZFC has L as its smallest inner model (since by the fact that second-order ZFC is categorical and therefore can be deemed to contain 'all' the ordinals), one has, by Cohen's proof, a 'ready-made' continuum, that is, the continuum of the constructible reals. – Thomas Benjamin Nov 04 '13 at 07:31
  • In fact, by the categoricity of second-order ZFC, L (whatever form of it is the correct form for second-order ZFC) would seem to be THE model for second-order ZFC (and L could again be deemed to be "the rewarder of the patient"). – Thomas Benjamin Nov 04 '13 at 07:42
  • @ThomasBenjmain Second-order ZFC is not actually categorical. It can be made categorical, however, if you add axioms concerning exactly what inaccessible cardinals exist. See here: http://math.stackexchange.com/a/317763/71829 – Keshav Srinivasan Nov 04 '13 at 08:02
  • @keshav: Very interesting. It would seem, then, that for the first inaccessible cardinal k, V_k=L could hold, or am I misunderstanding Asaf's answer? – Thomas Benjamin Nov 05 '13 at 05:02

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