1
$\begingroup$

Is there any statement unprovable in ZFC without the Continuum Hypothesis and provable with it, which is not (unprovable in ZFC with the "weakened continuum hypothesis" and provable with it):

In which the weakened continuum hypothesis is as follows: "The existence of any set whose cardinality is strictly between that of the integers and the real numbers, is independent of ZFC".

I am probing at the notion of whether the continuum hypothesis is an "ungrounded statement". In particular I suspect that proofs which depend upon the continuum hypothesis are, in reality depending on the weaker property of its "truth or otherwise" being independent of ZFC, rather than depending upon its "truth or otherwise" itself.

Sorry - an edit: when I ask for an unprovable/provable statement, I mean some meaningful statement about the numbers such as the existence of a solution to some diophantine equation.

$\endgroup$
  • $\begingroup$ I don't understand at all what is your "weakened CH". Can you perhaps explain more clearly? $\endgroup$ – Asaf Karagila Mar 17 '17 at 16:45
  • $\begingroup$ @AsafKaragila here's where you tell me it's nonsense. I'm trying to understand if some tangible consequence for the numbers can be derived from "CH" which cannot be derived from "CH is independent of ZFC"... or substitute for ZFC any nice theory about the numbers. $\endgroup$ – user334732 Mar 17 '17 at 18:15
6
$\begingroup$

Yes. The Continuum Hypothesis is an example of such a statement.


Some other (less pithy) thoughts:

I'm curious what motivated your question. Do you have any example of a statement provable from ZFC + CH (but not from ZFC), where you think you can prove it from ZFC + WCH instead?

Do I understand correctly that your weakened continuum hypothesis (WCH) is the statement "CH is independent of ZFC"? If so, I think WCH is not a good name. "Weakened" suggests that CH is stronger that WCH over ZFC, i.e. that ZFC + CH proves WCH. But it doesn't.

Why? ZFC + WCH proves Con(ZFC), because if ZFC is inconsistent, then it proves everything, and in particular it proves CH. And ZFC proves Con(ZFC)$\rightarrow$Con(ZFC + CH). So if ZFC + CH proved WCH, it would prove Con(ZFC), so it would prove Con(ZFC + CH), which contradicts Gödel's theorem.

Also, since ZFC + Con(ZFC) proves that CH is independent of ZFC (thanks to Gödel and Cohen), WCH is actually equivalent to Con(ZFC) over ZFC. So it really has nothing to do with the real CH.

$\endgroup$
  • $\begingroup$ Sorry Alex, I meant to say a meaningful, grounded statement about the numbers such as the nonexistence of a solution to some diophantine equation. I will correct. With regards to "weaker", I agree they are incomparable under ZFC but it was not my intention to compare them under ZFC. $\endgroup$ – user334732 Mar 17 '17 at 15:21
  • $\begingroup$ Let me adjust my question to you, then: Do you have any example of a "meaningful, grounded statement about the numbers" which is provable from ZFC + CH but not from ZFC? $\endgroup$ – Alex Kruckman Mar 17 '17 at 15:33
  • $\begingroup$ I think so... just searching for it. $\endgroup$ – user334732 Mar 17 '17 at 15:34
  • 1
    $\begingroup$ Indeed, you'll have trouble finding such a statement. ZFC + GCH has various conservativity properties over ZFC, e.g. it's $\Pi^2_1$-conservative and $\Pi^1_4$-conservative. This means, in particular, that if a statement with quantifiers which range over the natural numbers (and possibly some limited level of quantification over sets of natural numbers) is provable in ZFC + GCH, it's already provable in ZFC. $\endgroup$ – Alex Kruckman Mar 17 '17 at 15:44
  • 1
    $\begingroup$ I think the classic reference is Platek's "Eliminating the Continuum Hypothesis" projecteuclid.org/euclid.jsl/1183736771 $\endgroup$ – Alex Kruckman Mar 17 '17 at 15:47

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.