Is $\mathfrak p=\omega_1$ equivalent to the existence of a Hausdorff gap without infinite pseudointersection?

Question

Given two set $A,B$ we write $A\subset^* B$ if the complement $A\setminus B$ is infinite.

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A Hausdorff gap is a transfinite family $\langle A_\alpha,B_\alpha\rangle_{\alpha\in\omega_1}$ of infinite subsets of $\omega$ satisfying the following two properties:

$\bullet$ $A_\alpha\subset^* A_\beta\subset^* B_\beta\subset^* B_\alpha$ for any countable ordinals $\alpha\le\beta$;

$\bullet$ for any infinite subset $I\subset \omega$ there exists a countable ordinal $\alpha$ such that $A_\alpha\not\subset^* I$ or $I\not\subset^* B_\alpha$.

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A in infinite subset $I\subset\omega$ is called a pseudointersection of a Hausdorff gap $\langle A_\alpha,B_\alpha\rangle_{\alpha\in\omega_1}$ if $I\subset^* (B_\alpha\setminus A_\alpha)$ for any $\alpha\in\omega_1$.

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The definition of the small uncountable cardinal $\mathfrak p$ implies that under $\mathfrak p>\omega_1$ each Hausdorff gap has an infinite pseudointersection.

What does happen under $\mathfrak p=\omega_1$?

Problem. Is $\mathfrak p=\omega_1$ equivalent to the existence of a Hausdorff gap without infinite pseudointersection?

Answer 1

Yes. This is a result due to Nyikos and Vaughan from 1983, appearing the paper

Nyikos, Peter J.; Vaughan, Jerry E., On first countable, countably compact spaces. I: ((\omega_ 1,\omega^*_ 1))-gaps, Trans. Am. Math. Soc. 279, 463-469 (1983). ZBL0542.54004.

Hausdorff gaps without a pseudo-intersection are called tight gaps, and Theorem 1.2 of the paper shows that these exist if and only if $\mathfrak{p}=\omega_1$.