These sentences are usually of two kinds. The first kind are actually theorems of ZFC asserting the existence of various cardinal numbers, and their negations are inconsistent with ZFC. The second kind are the so called "large cardinal axioms" and if A is such a sentence, it cannot be proved (in ZFC) that the consistency of ZFC implies the consistency of ZFC+A. My question is whether any examples of such sentences are known which can be proved (in ZFC) to be undecidable in ZFC-assuming that ZFC is consistent? More precisely, does there exist a formula $C(x)$ in the language of first order ZFC - containing $x$ as its one and only free variable - such that the following three statements are provable in ZFC? (1) If ZFC is consistent, then so is ZFC+"There exists an $x$ such that $C(x)$" (2) If ZFC is consistent, then so is "ZFC+"There does not exist an $x$ such that $C(x)$" (3) For all $y$ and $z$, $C(y)$ and $C(z)$ implies that there exists an injective mapping of $y$ onto $z$.

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    $\begingroup$ How about "$x$ is an injection from $\mathbb{R}$ into $\omega_1$"? $\endgroup$ Apr 24, 2014 at 20:00
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    $\begingroup$ François, it seems that great minds think alike... $\endgroup$ Apr 24, 2014 at 20:07
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    $\begingroup$ You should use line breaks more often. $\endgroup$
    – Asaf Karagila
    Apr 25, 2014 at 1:37
  • $\begingroup$ Francois, you have really pricked my baloon! Your example satisfies all three of my conditions. But the mathematical object whose existence your example shows to be undecidable in ZFC is not a cardinal number. It is a set of dijoint ordered pairs whose cardinal number is that of the continuum-and this cardinal number can be proved to exist in ZFC. So, as Joel states, my definition does not capture the notion of a cardinal number whose existence is undecidable in ZFC. $\endgroup$ Apr 26, 2014 at 18:40

1 Answer 1


I'm not sure your requirements capture the property you are intending, because the formula $$C(x)\qquad\iff\qquad (x=0\text{ and the CH holds })$$ has all the properties you seek. Any model of ZFC+CH has $C(0)$, and a model of not CH has no $x$ with $C(x)$. and $C(x),C(y)$ implies $x=y=0$.


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