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

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

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 $$I\not\subset^* B_\alpha$$.

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 $$\alpha\in\omega_1$$.

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?

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 $$\mathfrak{p}=\omega_1$$.