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The continuum hypothesis (CH) states there does not exist $A \subset \mathcal{P}(\mathbb{N})$ whose cardinality lies between $|\mathbb{N}|$ and $|\mathcal{P}(\mathbb{N})|$.

For each model of ZFC, CH is either true or false. In particular, for the "true model" of $\mathcal{P}(\mathbb{N})$, CH is either true or false.

It was shown ZFC neither proves nor disproves CH. This was done by showing there exists a model of ZFC where CH holds, and another model where CH fails. This seems partly like a defect of first order logic. In particular, first order logic is outrageously bad at specifying intended models (for us $\mathcal{P}(\mathbb{N})$), which leads to my question.

$\textbf{Question}$: Is there any hope of an extension of first order logic (which can accurately specify $\mathcal{P}(\mathbb{N})$ and omit unintended models) resolving the continuum hypothesis? For instance, one might wonder if the models Cohen and Godel made also work for $\mathscr{L}_{\omega_1 \omega}$ and $\mathscr{L}_Q$.

Extensions of first order logic tend not to be complete, so even if $\mathcal{P}(\mathbb{N})$ is accurately specified, it isn't given either CH or not CH is provable.

Stephen Harrison
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  • IIRC, the Continuum hypothesis was proven by Paul Cohen to be kind of like the parallel postulate of euclidean geometry. You can have models of set theory where the continuum hypothesis holds (like euclidean geometry) and you can have models of set theory where the continuum hypothesis fails (like non-euclidean geometry). So you have to turn to philosophy with regards to whether set theory is in some metaphysical sense better with (or without) the continuum hypothessis. – Fomalhaut Jan 02 '24 at 05:19
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    Second-order ZFC actually does decide CH. The problem is that interpreting second-order logic requires set theoretical assumptions, and so all we get is that CH is a consequence of ZFC$_2$ if and only if it is true. – spaceisdarkgreen Jan 02 '24 at 05:21
  • @spaceisdarkgreen any good source to read up this? – Ajin Shaji Jose Jan 02 '24 at 05:29
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    More precisely, models of ZFC$2$ are all isomorphic to $V\kappa$ for some inaccessible cardinal $\kappa,$ so for any statements quantifying over only small sets, all models just agree with $V$ on the truth value. The things that remain independent are questions like 'how many inaccessible cardinals are there?' – spaceisdarkgreen Jan 02 '24 at 05:30
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    @AjinShajiJose I don't have a reference handy... the keyword "quasi-categoricity" will probably get you somewhere. This goes back to Zermelo in the 1930s iirc. – spaceisdarkgreen Jan 02 '24 at 05:31
  • @spaceisdarkgreen Are you saying $\text{ZFC}_2 \vdash$ CH iff CH is true, or $\text{ZFC}_2 \models$ CH iff CH is true? – Stephen Harrison Jan 02 '24 at 05:56
  • @StephenHarrison The second (and moreover, ZFC$_2\models \lnot$CH iff CH is false). – spaceisdarkgreen Jan 02 '24 at 06:04

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