Hausdorff gaps and $\mathfrak{p}=\mathfrak{t}$

Recently Malliaris and Shelah proved that $$\mathfrak{p}=\mathfrak{t}$$ (see: http://math.uchicago.edu/~mem/Malliaris-Shelah-CST.pdf). Their results are far more general, however for the specify context of this question what they showed is that if $$G$$ is an ultrafilter on $$\omega$$, then the ultra product $$\omega^\omega/G$$ is a linear order with no $$(\kappa,\lambda)$$-gaps with $$\kappa,\lambda<\mathfrak{t}$$.

They can obtain the desired equality combining it with a result of Shelah on the existence of a certain type of small gap (i.e. with both sequences in the gap of size smaller than $$\mathfrak{t}$$) in structures of the form $$\omega^\omega/G$$ in the case $$\mathfrak{p}<\mathfrak{t}$$.

Recall that

• $$\mathfrak{t}$$ is the least size of a decreasing chain in $$P(\omega)/Fin$$ without a positive lower bound in the boolean algebra $$P(\omega)/Fin$$.

• $$(f_\alpha:\alpha<\kappa,g_\beta:\beta<\lambda)$$ is a $$\kappa,\lambda$$ pregap in a partial order $$(X,<^*)$$ if $$f_\alpha<^*f_\beta<^*g_\eta<^*g_\gamma$$ for all $$\alpha<\beta<\kappa$$ and $$\gamma<\eta<\lambda$$.

• A pregap $$(f_\alpha:\alpha<\kappa,g_\beta:\beta<\lambda)$$ on $$(X,<^*)$$ is a gap if for no $$f\in X$$ we have that $$f_\alpha<^*f<^*g_\gamma$$ for all $$\alpha<\kappa$$ and $$\gamma<\lambda$$.

• A gap $$(f_\alpha:\alpha<\omega_1,g_\beta:\beta<\omega_1)$$ on the partial order $$(\omega^\omega,<^*)$$ where $$<^*$$ is eventual domination is an Hausdorff gap if for all $$\beta<\omega_1$$ and $$n<\omega$$

$$\{\alpha<\beta:\forall m>n \: f_\alpha(m) is finite.

Malliaris and Shelah's result has also the following consequence on Hausdorff gaps of which I was not aware and which I ask if it had a proof prior to their result (may be even without extra set theoretic assumptions).

Assume $$\mathfrak{t}>\omega_1$$, $$(f_\alpha:\alpha<\omega_1,g_\beta:\beta<\omega_1)$$ is an Hausdorff gap.

Let $$X\subset P(\omega)$$ be the set of $$A$$ infinite subsets of $$\omega$$ such that $$(f_\alpha\restriction A:\alpha<\omega_1,g_\beta\restriction A:\beta<\omega_1)$$ is no longer a gap in the partial order $$(\omega^A,<^*)$$.

Then $$X$$ is open dense in the Ellentuck topology.

Peter Nyikos and Jerry Vaughan in this paper prove this result, although they don't quite state it that way, and they work in the partial order $(P(\omega),\subseteq^*)$ rather than $(\omega^\omega,<^*)$.

For this partial order, a gap is defined in nearly exactly the same way: just replace the symbol $<^*$ with the symbol $\subseteq^*$ in the definition you gave. Given a gap $(F_\alpha,G_\alpha: \alpha < \omega_1)$, they say that a set $E$ is beside the gap if $(F_\alpha \cap E,G_\alpha \cap E: \alpha < \omega_1)$ is no longer a gap in $(P(E),\subseteq^*)$. The analogy with the situation in your question is (I hope) obvious. They say that a gap is tight if there is no set $E$ beside the gap. They prove (theorem 1.2 of their paper) that

Theorem: There is a tight $(\omega_1,\omega_1)$-gap if and only if $\mathfrak{t} = \aleph_1$.

We'll show in a minute (after proving a lemma below) that this actually implies your statement regarding the Ellentuck topology.

This shows that your proposition (or, at least, a version of it for a different partial order) does depend on the assumption $\mathfrak{t} > \aleph_1$.

As for the proposition in the form you gave it, I don't know whether it had a proof prior to Malliaris and Shelah's work. But it does have a fairly simple proof that doesn't depend on their work. I'll sketch that out now.

Recall that the Ellentuck topology has a basis of sets of the form $$[s,A] = \{s \cup B : B \in [A]^\omega\},$$ where $s \in [\omega]^{<\omega}$ and $A \in [\omega]^{\omega}$.

Lemma: Suppose $X$ is a subset of $2^\omega$ such that

$\qquad (1)$ if $A \in X$ and $B \subseteq^* A$, then $B \in X$.

$\qquad (2)$ if $B \in [\omega]^{\omega}$, then there is some infinite $A \in X$ such that $A \subseteq B$.

Then $X$ is open dense in the Ellentuck topology. (To paraphrase: If you're open dense in the partial order $([\omega]^{\omega},\subseteq^*)$, then you're open dense in the Ellentuck topology.)

Proof:

To see that $X$ is open: Suppose $A \in X$. By $(1)$, $[\emptyset,A] \subseteq X$, so $A$ is in the interior of $X$.

To see that $X$ is dense: Let $[s,A]$ be a basic open neighborhood in the Ellentuck topology. By $(2)$, there is some infinite $B \in X$ with $B \subseteq A$. By $(1)$, $s \cup B \in X$. But $s \cup B \in [s,A]$, so $X \cap [s,A] \neq \emptyset$.

QED

Note: this lemma shows that (in the terminology of Nyikos and Vaughan given above) if $\mathfrak{t} > \aleph_1$ and we have a gap $(F_\alpha,G_\alpha:\alpha<\omega_1)$, then $$\{E \in [\omega]^{\omega} : E \text{ is beside the gap}\}$$ is open dense in the Ellentuck topology. Condition $(1)$ is obvious. Condition $(2)$ follows from the aforementioned theorem of Nyikos and Vaughan, since otherwise their theorem I quoted would be false for the gap $(F_\alpha \cap B, G_\alpha \cap B: \alpha < \omega_1)$ (which is a gap in the partial order $(P(B),\subseteq^*)$).

Now let's check that the set $X$ mentioned in your question satisfies the conditions of this lemma.

Condition $(1)$ is fairly obvious.

For condition $(2)$, fix $B \in [\omega]^{\omega}$. For each $\alpha < \omega_1$, let $$C_\alpha = \{(m,n) : m \in B \text{ and } f_\alpha(m) < n < g_\alpha(m)\}.$$ Using the definition of a pregap, it's fairly easy to see that $\langle C_\alpha : \alpha < \omega_1 \rangle$ is a decreasing chain in $P(\omega \times \omega)/fin$. Since $\mathfrak{t} > \aleph_1$, there is some infinite $C \subseteq \omega \times \omega$ such that $C \subseteq^* C_\alpha$ for every $\alpha$. Let $$A = \{m : (m,n) \in C \text{ for some } n\},$$ $$h = \{(m,n) : m \in A \text{ and } n = \min \{k : (m,k) \in C\}\}.$$ Clearly $h$ is a function from $A$ to $\omega$, and, for every $\alpha$, $f_\alpha \upharpoonright A <^* h <^* g_\alpha \upharpoonright A$. We know that $A$ must be infinite, because $C$ is infinite and, for each $m \in A$, $C \cap \{m\} \times \omega$ is finite.

Notice that this proof doesn't depend on the gap being Hausdorff, so you can eliminate that assumption from the statement of the proposition.