5
$\begingroup$

I am familiar with the concept that a statement can be proven indepent such as in the case of the continuum hypothesis where both ZFC+CH and ZFC+(CH is false) are both proven consistent, but I would like to know if any current publications show that ZFC can show the following to be true of a statement T:

[ZFC+T is cconsistent and ZFC+(T is false) is consistent] is independent

(I apologize at how informally the question is posed; I've just started to read the work of Cohen and Godel and have been unable to find any text refering to this problem)

$\endgroup$
1
  • 3
    $\begingroup$ To nitpick, ZFC+CH is not proven to be consistent. It was proven that if ZFC is consistent then ZFC+CH is consistent (and same for ZFC+~CH). It is unknown whether ZFC is consistent or not, and by Godel's incompleteness we will probably never know as well. $\endgroup$
    – Asaf Karagila ♦
    Commented Jul 17, 2011 at 18:10

1 Answer 1

Reset to default
14
$\begingroup$

Statements of the form Con(ZFC + T) are never provable in ZFC, because they imply Con(ZFC). Similarly, ZFC can not prove any statement of the form "T is not provable in ZFC", because such statements also imply Con(ZFC). Both of these facts are consequences of Gödel's incompleteness theorems.

On the other hand if ZFC + T is consistent, then it is also impossible for ZFC to prove that "ZFC+ T is inconsistent". Because ZFC is $\omega$-consistent, if ZFC proves some theory is inconsistent then the theory really is inconsistent.

In short, ZFC cannot prove "ZFC + T is consistent" for any T and cannot disprove that statement for any T such that ZFC + T is actually consistent.

$\endgroup$
3
  • 3
    $\begingroup$ Of course you assume ZFC is actually consistent for your answer... $\endgroup$
    – GEdgar
    Commented Jul 17, 2011 at 18:41
  • 5
    $\begingroup$ Of course; everyone who works with ZFC assumes it is consistent (in fact as I point out it is $\omega$-consistent). If ZFC is inconsistent then it proves every statement in its language and the question itself is of no interest. At the same time, I refer always to ZFC, not to ZFC + Con(ZFC). $\endgroup$
    – Carl Mummert
    Commented Jul 17, 2011 at 18:50
  • 4
    $\begingroup$ I agree that it is always important to remark that ZFC cannot be proven consistent and we always have to assume that, it is not always necessary to mention that - just like we do not start every answer, paper, whatnot with "Assume first order logic is consistent..." and so on. Lately it feels like every question in set theory gets a comment about this... :-) $\endgroup$
    – Asaf Karagila ♦
    Commented Jul 17, 2011 at 18:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .