# Can I win this variant of the Banach-Mazur Game?

Suppose I play the following game against the Opponent. My moves are rational numbers $$p_i$$ and the Opponent's moves are real numbers $$\epsilon_i>0$$.

On turn $$n+1$$ the past move sequence is $$p_1,\epsilon_1,\ldots, p_{n}, \epsilon_{n}$$. I select a point $$p_{n+1}\in \mathbb Q \cap (p_{n}-\epsilon_{n},p_{n}+\epsilon_{n})$$ and the Opponent selects some $$\epsilon_{n+1} >0$$.

I win provided the sequence $$p_i$$ tends to an irrational number.

Does anyone know if I have a winning strategy? Certainly I don't if the rational and irrational numbers are swapped, as the Opponent can just enumerate the rationals and select the intervals small enough to exclude each rational in turn.

I imagine I can always win. But that's only because this looks like the Banach-Mazur game, which I can win if the target set is the irrationals.

The motivation behind this is I want to recursively build a set of homeomorphisms $$F_1,F_2,\ldots: \mathbb R \to \mathbb R$$ such that $$F(x) =\displaystyle \lim_{n \to \infty} F_n \circ F_{n-1} \circ \ldots \circ F_1 (x)$$ is a well-defined homeomorphism and such that some fixed $$p \in \mathbb Q$$ is sent to an irrational number.

I have conditions under which the limit exists. Namely I have to ensure the next $$\max d(x,F_n(x)) < \epsilon_n$$. Unfortunately $$\epsilon_n$$ are not known in advance. Each $$\epsilon_n$$ is determined by $$F_1,\ldots, F_{n-1}$$. This leads to the above game where $$p_n = F_n \circ F_{n-1} \circ \ldots F_1(p)$$.

## 1 Answer

Sure you can win. Let's enumerate rationals as $$q_n, n = 1, 2,\ldots$$. Also we can WLOG assume that $$\varepsilon_{i+1} \le \frac{\varepsilon_i}{20000}$$. We will make it so that $$d(q_n, p_{n+1}) \ge 10\varepsilon_{n+1}$$. Then for $$m > n$$ we have $$d(q_n, p_m) \ge 10\varepsilon_{n+1} - \varepsilon_{n + 2} - \ldots \ge \varepsilon_{n+1}$$.

For $$p_{n+1}$$ the possible values are an interval of length $$\varepsilon_n \ge 20000\varepsilon_{n+1}$$. Therefore there exists at least one admissible value of $$p_{n+1}$$ such that $$d(q_n, p_{n+1})\ge 10\varepsilon_{n+1}$$ whatever the value of $$q_n$$ is.

Thus, $$p_n$$ can not converge to any rational $$q$$ since if $$q = q_n$$ for some $$n$$ then $$d(q, p_m) \ge \varepsilon_{n+1} > 0$$ for $$m > n$$. On the other hand since $$\varepsilon_{i+1} \le \frac{\varepsilon_i}{20000}$$ the sequence $$p_n$$ is Cauchy and thus converge for some $$q\in \mathbb{R}$$. Therefore $$q\in \mathbb{R}\backslash \mathbb{Q}$$.

• Cool, thanks. It's not important to my problem, but any ideas what happens if we replace $\mathbb Q$ with some arbitrary comeagre set, or if this game can be reduced to Banach-Mazur or something more well-known? – Daron Dec 10 '19 at 10:25