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Let $FA_{\aleph_1}$(cardinal preserving proper forcings) be the forcing axiom: if $\mathbb{P}$ is a cardinal preserving proper forcing notion and $(D_\xi)_{\xi<\omega_1}$ are dense subsets of $\mathbb{P},$ then there exists a filter $G \subseteq \mathbb{P}$ which meets all $D_\xi$'s, $\xi < \omega_1$.

Does this forcing axiom decide CH. More precisely,

Question. Assuming the existence of large cardinals, is $FA_{\aleph_1}$(cardinal preserving proper forcings) consistent with large values of the continuum?

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  • $\begingroup$ If mean, it implies MA (because every ccc forcing is proper and cardinal preserving), so presumably it decides $\lnot\sf CH$. Since it follows from $\sf PFA$, it's compatible with $2^{\aleph_0}=\aleph_2$. So by "large" do you mean larger than $\aleph_2$? $\endgroup$
    – Asaf Karagila ♦
    Commented Oct 28, 2020 at 13:23
  • $\begingroup$ @AsafKaragila. Yes, sure, i mean given any $\kappa,$ can we get it with the continuum above $\kappa$?. Th motivation comes from the fact that to show PFA decides CH, we use forcings which collapse the continuum!. $\endgroup$
    – Mohammad Golshani
    Commented Oct 28, 2020 at 14:10
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    $\begingroup$ By "decide CH", do you mean just whether CH is true or false? Because hitting $\aleph_1$ dense subsets of Cohen forcing implies that CH is false, right? But deciding the value of the continuum is another matter. $\endgroup$
    – Todd Eisworth
    Commented Oct 28, 2020 at 15:14
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    $\begingroup$ I mean determine its value, like what PFA does, namely it implies the continuum is $\aleph_2$. $\endgroup$
    – Mohammad Golshani
    Commented Oct 28, 2020 at 15:41
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    $\begingroup$ If it determines the value, it has to be $\aleph_2$. So I guess your question can be clarified by asking specifically if this forcing axiom is consistent with a larger continuum. This is a question I'd ask David Aspero, not MathOverflow, to be honest. I'll alert him to this. $\endgroup$
    – Asaf Karagila ♦
    Commented Oct 28, 2020 at 19:23

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