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

Question

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 $n<\omega$

  $\{\alpha<\beta:\forall m>n \: f_\alpha(m)<g_\beta(m)\}$ 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.

Answer 1

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.