Timeline for A question about Second-Order ZFC and the Continuum Hypothesis
Current License: CC BY-SA 3.0
11 events
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| Oct 20, 2011 at 19:39 | comment | added | Garabed Gulbenkian | Andreas, I will make this following last attempt to get around your counter-examples. "If T is a consistent and recursively axiomatizable sub-theory (in my sense) of ZFC2 then either (1) T contains no proof of of CH and no proof of not-CH or (2) T contains a proof of one of these sentences such that all the sentences which occur in that proof are true sentences of pure second-order logic. If this seems vague or trivial or still doesn't work, I give up! | |
| Oct 19, 2011 at 20:18 | comment | added | Garabed Gulbenkian | Andreas, I agree completely. That is why I asked my original question about sub-theories of ZFC2 with a recursively enumerable set of (logical and non-logical) axioms. Wouldn't it look a bit like cheating to include either CH or not-CH among the axioms of such a sub-theory- and how would you decide which to include? Incidentally, I apologize for my mixed-up and incorrect impressions of what I thought Rogers' book was stating about second-order arithmetic. | |
| Oct 18, 2011 at 20:52 | comment | added | Andreas Blass | @Garabed: If you identify "logical axiom of ZFC2" with "true sentence of pure second-order logic", then, by Kreisel's observation, either CH is one of those axioms, or not-CH is one of those axioms. So that axiom, whichever one it is, gives you a singleton subset axiomatizing a subtheory of ZFC2 that trivially decides CH. (Note that CH is expressible in pure second-order logic; the formulation would begin with "for every Dedekind-complete ordered field ....") | |
| Oct 18, 2011 at 18:30 | comment | added | Garabed Gulbenkian | I would consider the set L of the logical axioms of ZFC2 to be the same as the true sentences of pure second-order logic which according to the last sentence on page 391 of ROGERS' book do not even "constitute an analytic set". This is also, I think, what Kreisel conceived L to be. | |
| Oct 17, 2011 at 15:25 | comment | added | Andreas Blass | @Noah: The power set axiom doesn't need a second-order extension; the first-order version already says that there's a set containing all subsets of any given set. Where "second-order" comes in is making sure that "all subsets" means what it should, not "all subsets that happen to be in the model". That job is done by the second-order version of separation axioms (a.k.a. subset axioms, Aussonderung, or comprehension). And these follow from second-order replacement just as in the first-order situation. | |
| Oct 17, 2011 at 15:21 | comment | added | Andreas Blass | @Garabed: Looking up "second-order arithmetic" in the index of Rogers's book, I come to Section 16.2, where I see nothing about "logical axioms of second-order arithmetic". A footnote on page 389 gives a recursive set of axioms and says that this is not what's under discussion. The rest of the section is about truth (not axiomatics) of statements of second-order arithmetic. It's shown (among other things) that this notion of truth is not in the analytical hierarchy. So I still have no clear idea what you mean by axioms of ZFC2. | |
| Oct 17, 2011 at 15:02 | comment | added | Garabed Gulbenkian | In the last chapter of the book "Theory of Recursive Functions and Effective Computability" by Hartley Rogers, it is stated that the set L of logical axioms of Second-order Arithmetic (and thus no doubt of ZFC2 also) is not only non-recursively enumerable but is of high rank in the Analytical Hierarchy of Descriptive Set Theory. | |
| Oct 14, 2011 at 20:54 | comment | added | Noah Schweber | Yes, you definitely need to clarify what you mean by $ZFC_2$ here. My gut instinct is what I said below: there's no good way to get what you want to be true. (Also, @Andreas: is second-order replacement enough to get second-order powerset? Because I'd think we'd definitely want second-order powerset.) | |
| Oct 14, 2011 at 19:02 | comment | added | Andreas Blass | Since you want the question to refer to the sets of axioms, I'm curious what you take to be the set of axioms of ZFC2; in other words, what set should T be a subset of? I suppose the set-theoretic axioms should be those of first-order ZFC except that the replacement schema is replaced with the obvious single second-order axiom. But what should the logical axioms be? | |
| Oct 14, 2011 at 18:15 | comment | added | Garabed Gulbenkian | In response to your answer, Andreas, I should have explained more clearly what I mean by saying that a theory A is a sub-theory of a theory B. I mean that the set of axioms (both logical and non-logical) of A is a subset of the set of axioms of B. How do you know that either CH or its negation are among the axioms of ZFC2 since the set of axioms of ZFC2 is not recursively enumerable. Perhaps neither CH nor its negation is an axiom of ZFC2. | |
| Oct 13, 2011 at 22:40 | history | answered | Andreas Blass | CC BY-SA 3.0 |