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The statement A = "There exists a well-ordering of the reals" is independent of ZF. My understanding is that the statement B = "There exists an explicit well-ordering of the reals" is also independent of ZF, yet this seems counterinutitive to me, because of the following line of reasoning: If B were true, then A would be provable in ZF, which is impossible since A is independent of ZF, so B must be false. This line of reasoning seems perfectly clear to me, and I see no reason why it cannot be carried out in ZF itself. But if the line of reasoning could be carried out in ZF, then that would mean that ZF implies that not B, contradicting the claim that B is independent of ZF.

So can anyone clarify how exactly ZF+B is consistent?

Any help would be greatly appreciated.

Thank You in Advance.

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See related question on definable well-orderings of the reals mathoverflow.net/questions/6593/… –  Joel David Hamkins Sep 6 '12 at 17:19

4 Answers 4

Here is the error in your reasoning "If B were true, then A would be provable in ZF, which is impossible since A is independent of ZF, so B must be false." The statement "B is independent of ZF" means that there are models of ZF where it is true (V=L) and models where it is false(almost everything else). You are right in saying that in a model where B is true, then A is true but that does not mean that A is provable from ZF. There are also models where A is true and B is false (i.e. if we assume choice that will give us a well ordering, but that does not mean that it is definable.)

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It sounds like you're taking the word "explicit" (which has no precise mathematical meaning) to mean "can be proved to exist by ZF". This is problematic because a theory (e.g., ZF) proves statements. It does not construct objects, or prove that a particular object exists, although it may prove statements asserting the existence of objects with certain properties.

For example, although every model of "ZF + $V=L$" has a definable wellordering of its reals, this does not help us construct wellorderings of the reals in models of "ZF + $V \ne L$". The most serious obstacle in this case is that the two models we are considering could have different sets of reals. So although ZF does prove that $L$ satisfies "there is a definable wellordering of the reals," this only gives us a definable wellordering of the reals of $L$. It is consistent that $L$ does not contain all the reals, and even that it contains only countably many reals.

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From Gödel's proof of the consistency of AC: There is an explicit subset $A \subseteq \mathbb R$ and an explicit well-ordering defined on $A$. It is consistent with ZF (and even with ZFC) that $A = \mathbb R$. But of course it is not provable in ZFC that $A = \mathbb R$.

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Just to clarify, you are using "explicit" to mean "definable", right? –  Trevor Wilson Sep 6 '12 at 16:50

I suspect that the root of your confusion is an ambiguity in the word "explicit". Your statement B could mean, as Gerald Edgar suggests in a comment and as others seem to have assumed, "there exists a definable relation that well-orders the reals." This is consistent and does not imply the ZF-provability of statement A. But B could also mean "There is a definable relation that can be proved, in ZF, to well-order the reals", so that "explicit" involves both definability and provability. This version of B (when sufficiently precisely formulated) is not consistent with ZF. I think you had the latter version of B in mind when you used B to infer provability of A.

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I think your second formulation actually is consistent with ZF, since it is consistent with ZF that $\neg\text{Con}(\text{ZF})$, and in such a model, proofs from ZF are easy to come by. That is, if there are any models of ZF at all, then there are models of ZF that have a definable well-ordering of the reals such that, in that model, there is a proof from ZF that this definition does define a well-ordering of the reals. –  Joel David Hamkins Sep 6 '12 at 20:27
    
Joel, you're right. In fact, it seems that the second formulation of B is actually equivalent (in PA or appropriate weaker theories) to $\neg$Con(ZF). –  Andreas Blass Sep 6 '12 at 21:42

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