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

Question

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 $p_n = F_n \circ F_{n-1} \circ \ldots F_1(p)$.

Answer 1

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 $q\in \mathbb{R}\backslash \mathbb{Q}$.