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 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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.