Game versions of the tower number $\mathfrak t$

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?

1 Answer

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.

• $\mathfrak h$ is a very $\mathfrak t$ight upper bound for $\mathfrak t_J$. Thank you. Commented Oct 20, 2021 at 6:18