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Which extension of $\sf ZFC$ prove that $$ {\sf ZFC} \not \vdash \exists x \, ( \operatorname {CH}(x) \land x \neq \emptyset \land x \neq 1)$$

Where $\operatorname {CH}(x) \iff \neg \exists \kappa \, (|x| < \kappa < |P(x)|) $

In English: $\sf ZFC$ doesn't prove the continuum hypothesis of any set other than the empty set and the singleton of the empty set.

I know that "$\sf ZFC + CH$ fails everywhere", can prove that, which is too strong. But I'm asking if this can be proved in a much less strong theory, some theory whose consistency strength is just a little above $\sf ZFC$.

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    $\begingroup$ Complete failure of GCH has large cardinal consistency strength. The precise answer was already given in an answer to a question of yours from three years ago: math.stackexchange.com/q/3186630/127263 $\endgroup$
    – Wojowu
    Commented Apr 24, 2022 at 20:32
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    $\begingroup$ Also, it's not true that ZFC+GCH fails everywhere proves that. If it did, then this theory would prove its own consistency, contradicting Godel's theorems. $\endgroup$
    – Wojowu
    Commented Apr 24, 2022 at 20:57
  • $\begingroup$ @Wojowu Those comments should be answers. $\endgroup$
    – Noah Schweber
    Commented Apr 25, 2022 at 0:15

1 Answer 1

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The statement you propose is equivalent to consistency of GCH failing for all (infinite) cardinals. This problem is well-studied. Of course the theory ZFC+Con(ZFC+"GCH fails everywhere") gives an answer to your problem, but this is completely tautological. It is standard in such situations to calibrate the consistency strength against the large cardinal hierarchy: as is discussed in this answer, the precise equiconsistency statement is not easy to state in ZFC, but it amounts to existence of stationarily many measurable cardinals of high Mitchell rank. It is between measurable cardinals and strong cardinals.

Let me also mention that your claim in the last paragraph is false: ZFC+"GCH fails everywhere" doesn't prove Con(ZFC+"GCH fails everywhere"), by Godel's incompleteness.

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  • $\begingroup$ What is the proof of the first line in this answer? $\endgroup$
    – Zuhair Al-Johar
    Commented Apr 25, 2022 at 11:22
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    $\begingroup$ @ZuhairAl-Johar It should be obvious. ZFC doesn't prove X iff ZFC + not X is consistent. $\endgroup$
    – Wojowu
    Commented Apr 25, 2022 at 11:28

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