The Banach-Mazur game on a poset $\mathbb P$ is the $\omega$-length game where the players alternate choosing a descending sequence $a_0 \geq b_0 \geq a_1 \geq b_1 \geq \dots$. Player II wins when the sequence has a lower bound.
A strategy is a function $\sigma : \mathbb P^{<\omega} \to \mathbb P$ such that for each linearly ordered $\vec x \in \mathbb P^{<\omega}$, $\sigma(\vec x) \leq \min(\vec x)$.
A tactic is a descending function $\tau : \mathbb P \to \mathbb P$. (Meaning $\tau(p) \leq p$ for all $p$. Thanks to Joel for the better terminology.)
The Banach-Mazur game on $\mathbb P$ is $\omega$-strategically closed when there is a strategy $\sigma$ such that II wins whenever II plays according to $\sigma$, meaning if the sequence of plays so far is $\vec x$ and it is II’s turn, then II plays $\sigma(\vec x)$. The game is $\omega$-tactically-closed when II has a winning tactic, meaning II wins when they always play $\tau(a)$, where $a$ is the last move by I.
These notions have generalizations to games of length longer than $\omega$. The separation of tactical and strategic closure for games of uncountable length has been studied by Yoshinobu. The situation is different because the games involve limit stages. However, I don't know whether these notions have been separated for games of length $\omega$.
Question: Suppose $\mathbb P$ is $\omega$-strategically-closed. Is it $\omega$-tactically-closed?
Best Answer
Consistently, there is a $\sigma$-strategically closed forcing which is not $\sigma$-tactically closed. Such a forcing is constructed (through forcing) by Jech-Shelah in
Jech, Thomas; Shelah, Saharon, On countably closed complete Boolean algebras, J. Symb. Log. 61, No. 4, 1380-1386 (1996). ZBL0871.06008.
Precisely, they show that in a generic extension, there is a (seperative atomless) $\sigma$-strategically closed forcing $(\mathbb P,\leq)$ so that the Boolean completion $\mathbb B(\mathbb P)$ of $\mathbb P$ does not contain a dense $\sigma$-closed subset. Their argument really proves the stronger statement that $\mathbb P$ is not $\sigma$-tactically closed.
After constructing $\mathbb P$, their proof is roughly as follows: Let $\mu\colon \mathbb P\rightarrow \mathbb B(\mathbb P)$ be a dense embedding. Suppose toward a contradiction that $D$ is a dense $\sigma$-closed subset of $\mathbb B(\mathbb P)$. Then define $\prec$ on $\mathbb P$ by $q\prec p$ iff $\exists d\in D$ with $\mu(q)<d<\mu(p)$. Jech-Shelah now go on and argue that $\prec$ cannot possibly exist on $\mathbb P$, but they use only the following properties of $\prec$:
Jech-Shelah mention 1.-4. explicitly and use 5. implicitly. Now, if $\tau$ was a winning tactic for II in the Banach-Mazur game on $\mathbb P$, there would be a different way to define $\prec$ with properties 1.-5.: let $q\prec p$ iff $q < \tau(p)$. (In fact, for an atomless partial order $\mathbb Q$, the existence of such a $\prec$ on $\mathbb Q$ is equivalent to $\mathbb Q$ being $\sigma$-tactically closed). So $\mathbb P$ is not $\sigma$-tactically closed.