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We know that if $\kappa>2^{\aleph_0}$ and $\kappa^{<\kappa}=\kappa$, then there is a c.c.c. forcing which forces $\sf MA+2^{\aleph_0}=\kappa$. Traditionally, we even start with $\sf GCH$, which simplifies even further the cardinal arithmetic, although going through the proof it seems excessive.

Consider the following scenario.

You wake up in a universe; to your left there is a regular cardinal $\kappa$, not smaller than the continuum, which itself is greater than $\aleph_1$. Do you want to try and force $\sf MA+2^{\aleph_0}=\kappa$.

Can we always do it? Can we always do it with a c.c.c. forcing? The latter clearly has a negative answer. If, say, $2^{\aleph_1}=\kappa^+$, then we must either add $\kappa^+$ new reals (which we don't want) or collapse $\kappa^+$ (which we cannot do), since $\sf MA$ implies that $2^{\aleph_0}=2^{\aleph_1}$.

So, what and when can we do things, and how can we do them?

Suppose that $\kappa\geq2^{\aleph_0}$ is a regular cardinal.

  1. Can we always force $\sf MA+2^{\aleph_0}=\kappa$ with a proper forcing? (E.g. $\operatorname{Add}(\kappa,1)*\Bbb P_\kappa$, where $\Bbb P_\kappa$ is the standard iteration. What if $\kappa=2^{\aleph_0}$ already in the ground model?)

  2. Under which assumptions can we just use a c.c.c. forcing? In particular, does $\sf ZFC$ prove that there is always a c.c.c. forcing that forces $\sf MA$, or do things like "For all $\kappa\geq\omega$, $2^\kappa=\kappa^{++}$" get in the way?

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    $\begingroup$ Some classic scenario, for example, would be to have forced, directly or otherwise, a bunch of Cohen reals over a model of MA... $\endgroup$
    – Asaf Karagila ♦
    Commented Dec 1, 2022 at 16:22
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    $\begingroup$ Since MA implies that $2^{\lambda} = 2^{\aleph_0}$ for every infinite $\lambda < 2^{\aleph_0}$, I don't see how you can force MA without collapsing cardinals, assuming your universe satisfies $\forall \kappa \text{ regular } \kappa^{<\kappa} > \kappa$, (for example, this is the case in a model in which GCH fails everywhere and there are no strongly inaccessible cardinals). $\endgroup$
    – Yair Hayut
    Commented Dec 1, 2022 at 21:11
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    $\begingroup$ I figured that's probably going to be the case. But presumably there's a broader and better condition to answer (2)? Or is it really just $\kappa^{<\kappa}=\kappa$? $\endgroup$
    – Asaf Karagila ♦
    Commented Dec 1, 2022 at 21:55
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    $\begingroup$ Is it known whether it is ever possible to force MA without adding reals? I think Paul Larson has a model where, after you force with a Suslin tree, there are no Suslin trees left. I'm not sure if that can stretch to full MA... $\endgroup$
    – Miha Habič
    Commented Dec 3, 2022 at 18:01
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    $\begingroup$ @MihaHabič That should be possible: Start with a model in which e.g. $2^\omega=\omega_3$, $\mathrm{MA}_{\omega_1}$ holds and $2^{\omega_2}$ is large, so MA fails. Force with $\mathrm{Add}(\omega_2, 1)$. In the extension, $2^\omega=\omega_2$ and $\mathrm{MA}_{\omega_1}$ still holds since both models have the same subsets of $\omega_1$, so full MA holds. In some sense this is the only way this can happen: If $V$ and $V[G]$ have the same reals and MA holds in $V[G]$ then $V$, $V[G]$ must have the same bounded subsets of $(2^\omega)^{V[G]}$. So if MA fails in $V$ then $(2^\omega)^V$ is larger. $\endgroup$
    – Andreas Lietz
    Commented Dec 6, 2022 at 14:43

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