Game versions of the tower number $\mathfrak t$

Question

Let us recall that $\mathfrak t$ is the smallest cardinal $\kappa$ for which there exists a family $(T_\alpha)_{\alpha\in\kappa}$ of infinite subsets of $\omega$ such that

$\bullet$ for any ordinals $\alpha\le\beta<\lambda$ we have $T_\beta\subseteq^* T_\alpha$;

$\bullet$ for any infinite set $I\subseteq \omega$ there exists $\alpha\in\kappa$ such that $I\not\subseteq^* T_\alpha$.

Here for two sets $A,B$ we write $A\subseteq^* B$ iff $A\setminus B$ is finite.

Now given an infinite ordinal $\kappa$ let us consider the tower game of length $\kappa$ played by two players $I$ and $J$ who construct two transfinite sequences of sets $(I_\alpha)_{\alpha\le\kappa}$ and $(J_\alpha)_{\alpha\le\kappa}$ by the following rules.

The player $I$ starts the game selecting a countable inifinite set $I_0$ and the player $J$ answers with an infinite set $J_0\subseteq I_0$. At the $\alpha$-th inning the player $I$ selects an infinite set $I_\alpha$ such that $I_\alpha\subseteq^* J_\beta$ for every ordinal $\beta<\alpha$. If such an infinite set $I_\alpha$ does not exist, then the player $I$ is forced to put $I_\alpha=\emptyset$. The player $J$ answers by selecting a subset $J_\alpha\subseteq I_\alpha$ of cardinality $|J_\alpha|=|I_\alpha|$. At the end of the game, the player $I$ is declared a winner if $I_\kappa\ne\emptyset$. If $I_\kappa=\emptyset$, then the player $J$ wins the game.

Let $\mathfrak t_I$ (resp. $\mathfrak t_J$) be the smallest ordinal $\kappa$ for which the player $I$ has no winning strategy (resp. the player $J$ has a winning strategy) in the tower game of length $\kappa$.

It is easy to see that $\mathfrak t\le \mathfrak t_I\le\mathfrak t_J\le\mathfrak c^+$.

Problem 1. Is $\mathfrak t_I\le\mathfrak c$? $\mathfrak t_J\le\mathfrak c$?

Problem 2. Is the strict inequality $\mathfrak t<\mathfrak t_J$ (resp. $\mathfrak t<\mathfrak t_I$ or $\mathfrak t_I<\mathfrak t_J$) consistent?

Answer 1

For problem 1: yes, $\mathfrak{t}_I \leq \mathfrak{t}_J \leq \mathfrak{c}$.

To force the game to end by step $\mathfrak{c}$, player $J$ can begin with an enumeration $\langle A_\alpha :\, \alpha < \mathfrak{c} \rangle$ of all subsets of $\omega$. On move $\alpha$, player $J$ selects a set $J_\alpha$ that is either contained in or disjoint from $A_\alpha$. By round $\mathfrak{c}$, this means that no infinite set $I$ can be contained in all the $J_\alpha$'s: there is some $\alpha < \mathfrak{c}$ where $A_\alpha$ "splits" $I$, meaning that both $A_\alpha \cap I$ and $I \setminus A_\alpha$ are infinite, and at that stage player $J$ played to ensure that $I$ is not contained in $J_\alpha$.

In fact, this argument shows that $\mathfrak{t}_I \leq \mathfrak{t}_J \leq \mathfrak{s}$, where $\mathfrak{s}$ denotes the splitting number. (A different version of the argument could improve this to $\mathfrak{h}$, actually.) This suggests that for problem 2, we should look at a model where $\mathfrak{t} < \mathfrak{s}$ (or $\mathfrak{h}$). The Mathias model is a natural candidate.