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Recall that

$\mathfrak{p}=\min\{|F|: F$ is a subfamily of $[\omega]^{\omega}$ with the sfip which has no infinite pseudo-intersection $\}$.

The cardinal $\mathfrak{q}_0$ defined as the smallest cardinality of a subset of $\mathbf{R}$ which is not a $Q$-space.

Q1. Is it true that $\mathit{MA}(\omega_1)$ iff $\omega_1<\mathfrak{p}$ ?

Q2. Is there a model of set theory in which $\omega_1=\mathfrak{q}_0$ and a non-$\mathit{MA}(\omega_1)$ ?

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    $\begingroup$ KP Hart has answered your questions here, but let me add that a good resource for lots of further information about related topics is David Fremlin's book "Consequences of Martin's Axiom". $\endgroup$
    – Andreas Blass
    Commented Jan 15, 2023 at 16:09
  • $\begingroup$ @‍KPHart's answer referenced by @AndreasBlass. $\endgroup$
    – LSpice
    Commented Jan 15, 2023 at 16:56

1 Answer 1

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Q1: No, see Between Martin's Axiom and Souslin's Hypothesis by Kunen and Tall. Note: Bell proved in The combinatorial principle $P(\mathfrak{c})$ that $\mathfrak{p}>\aleph_1$ is equivalent to $\mathsf{MA}(\aleph_1)$ for $\sigma$-centered partial orders.

Q2: $\mathsf{MA}(\aleph_1)$, even $\mathfrak{p}>\aleph_1$, implies $\mathfrak{q_0}>\aleph_1$, see page 162 of Internal Cohen Extensions by Martin and Solovay. The principle $\mathsf{S}_\aleph$ is proved on page 154, using a $\sigma$-centered partial order. So, every model with $\mathfrak{q}_0=\aleph_1$ satisfies $\neg\mathsf{MA}(\aleph_1)$ as well, for example every model of $\mathsf{CH}$ will do.

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    $\begingroup$ Superb answer. It would be nice if you could add a referece for your second answer $\endgroup$
    – Boaz Tsaban
    Commented Jan 15, 2023 at 11:04
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    $\begingroup$ Thanks for your answer. Yes, a link to the second answer would be very helpful. $\endgroup$
    – Alexander Osipov
    Commented Jan 15, 2023 at 11:26
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    $\begingroup$ I added two more references, to make the picture complete. $\endgroup$
    – KP Hart
    Commented Jan 15, 2023 at 15:55
  • $\begingroup$ Many thanks! Very helpful references ! $\endgroup$
    – Alexander Osipov
    Commented Jan 15, 2023 at 18:33

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