Are there any known statements that are provably independent of $ZF + V=L$? A similar question was asked here but focusing on "interesting" statements and all examples of statements given in that thread are not provably indepedent of $ZF + V=L$, they all raise the consistency strength bar. For example, the claim that "there exists an inaccessible" is independent of $ZF + V=L$, is really just an assumption. Because of Gödel's second incompleteness theorem, we cannot prove this. It is well possible that $ZFC$ proves "there is no inaccessible". The same holds for $Con(ZF)$ or "there is a set model of ZF". Those are assumed to be independent of $ZF + V=L$, but this cannot be proved without large cardinal assumptions.

So my question is: Is there any known (not necessarily "interesting" or "natural") statement $\phi$ and an elementary proof of $Con(ZF) => Con(ZF + V=L + \phi) \wedge Con(ZF + V=L + \neg\phi)$? Or is there at least a metamathematical argument that such statements should exists? (Contrast this with the situation of $ZFC$ and $CH$!)

And if not: Might $ZF + V=L$ be complete in a weak sense: There is no statement provably independent of it?

What is known about this?

  • $\begingroup$ Does the answer to your question not depend on the exact system in which you allow the proof of the independence to happen? $\endgroup$ – Mariano Suárez-Álvarez Mar 13 '10 at 14:50
  • 3
    $\begingroup$ If ZF turns out to be inconsistent then no statement is provably independent of ZF+V=L because every statement is provable assuming it :P . Hence no known statements are provably independent of ZF+V=L ;-) $\endgroup$ – Kevin Buzzard Mar 13 '10 at 15:46
  • 3
    $\begingroup$ But Kevin, the OP asked for a proof of the implication Con(ZF) implies Con(ZF+ psi) + Con(ZF + not-psi). If Con(ZF) fails, then this implication is vacuously true. $\endgroup$ – Joel David Hamkins Mar 13 '10 at 17:24
  • $\begingroup$ @Joel: Serves me right for not reading the question. I feel I should delete my comment but perhaps I'll leave it and just say here that of course you're right. Let me run another one past you though while you're here though: how about letting psi be "Con(ZF+V=L)"? $\endgroup$ – Kevin Buzzard Mar 13 '10 at 17:52
  • 1
    $\begingroup$ Yes, that statement is what Goedel uses in the 2nd Incompleteness theorem, and it is provably equivalent to "I am not provable". The situation in general is complicated by the need for omega-consistency in Goedel's version of the Incompletness theorem, to get both sides of the indpendence. Thus, one should use the Rosser sentence instead. For example, some consistent theories T do prove their own inconsistency. e.g. ZFC+not-Con(ZFC), and so one cannot always say that Con(T) is independent of T, even when T is consistent, although Con(T) is not probable. $\endgroup$ – Joel David Hamkins Mar 13 '10 at 18:01

The Incompleteness theorem provides exactly the requested independence. (But did I sense in your question that perhaps you thought otherwise?)

The Goedel Incompleteness theorem says that if T is any consistent theory interpreting elementary arithmetic with a computable list of axioms, then T is incomplete. Goedel provided a statement σ, such as the "I am not provable" statement, which is provably equivalent to Con(T), or if you prefer, the Rosser sentence, "there is no proof of me without a shorter proof of my negation", such that T does not prove σ and T does not prove ¬σ.

This establishes Con(T) implies Con(T + σ) and Con(T + ¬σ), as you requested. [Edit: one really needs to use the Rosser sentence to make the full conclusion here.]

In particular, this applies to the theory T = ZFC+ V=L, since this theory interprets arithmetic and has a computable list of axioms. Thus, this theory, if consistent, is incomplete, using the corresponding sentences σ above. Since it is also known (by another theorem of Goedel) that Con(ZF) is equivalent to Con(ZFC + V=L). This establishes the requrested implication:

  • Con(ZF) implies Con(ZFC + V=L + σ) and Con(ZFC + V=L + ¬σ)

The Incompleteness theorem can be proved in a very weak theory, much weaker than ZFC or even PA, and this implication is similarly provable in a very weak theory (PA suffices).

One cannot provably omit the assumption Con(ZF) of the implication, since the conclusion itself implies that assumption. That is, the existence of an independent statement over a theory directly implies the consistency of the theory. So since we cannot prove the consistency of the theory outright, we cannot prove the existence of any independent statements. But in your question, you only asked for relative consistency (as you should to avoid this triviality), and this is precisely the quesstion that the Incompleteness theorem answers.

  • 1
    $\begingroup$ Thank you! But i'm still somewhat confused: I don't think that Con(T) implies Con(T + Con(T)), otherwise the theory T + Con(T) could prove it's own consistency. So one cannot take Con(ZF + V=L) as an example for a statement provably independent of ZF + V=L. Or am i missing something? $\endgroup$ – user4607 Mar 13 '10 at 18:20
  • $\begingroup$ In this general case, one must use the Rosser sentence. The issue you mention is exactly the difference between the Goedel Incompleteness theorem and the Goedel-Rosser Incompleteness theorem. The statement sigma you want is: "For any proof of me, there is a shorter proof of my negation". $\endgroup$ – Joel David Hamkins Mar 13 '10 at 18:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.