Alice and Bob are playing a game. The rules are as follows: First Alice chooses a compact interval $A_1\subset\mathbb R$ (in this question, intervals contain more than one points, they are not allowed to choose single-point sets,) then Bob chooses a compact interval $B_1\subset A_1$ with $|B_1|\leq\frac12|A_1|$ (here $|I|$ means the length of the compact interval $I$.) After that, Alice chooses a compact interval $A_2\subset B_1$ with $|A_2|\leq\frac12|B_1|$ and so on. Hence they get a sequence of nested compact intervals $$A_1\supset B_1\supset A_2\supset B_2\supset \cdots\supset A_n\supset B_n\supset \cdots,$$ with $\lim_{n\to\infty} |A_n|=\lim_{n\to\infty} |B_n|=0$. By nested interval theorem, there is a unique $x\in\mathbb R$ such that $$\{x\}=\left(\bigcap_{n=1}^\infty A_n\right) \bigcap \left(\bigcap_{n=1}^\infty B_n\right).$$ If $x$ is rational, then Alice wins; if $x$ is irrational, then Bob wins. Decide whether there exists a winning strategy for either of them.

Background and thoughts

This problem is from my imagination.

It is clear that for any $x\in\mathbb R$, whether $x$ is rational or not, we can always find a sequence of nested intervals $I_n=\left[x-\frac1n,x+\frac1n\right]$ such that $\{x\}=\cap_n I_n$. The key point is that, Alice and Bob are both very very clever, and they are free to change the center of the intervals to any numbers they want. For example, if Alice choose $A_1=[-1,1]$, which is centered at $0\in\mathbb Q$, then Bob can find $B_1\subset A_1$ such that $B_1$ is centered at an irrational number; and then Alice can find $A_2\subset B_1$ such that $A_2$ is centered at a rational number; and so on... I cannot determine who will win in this kind of procedure. It seems that Bob will win, because irrational numbers are much more than rational numbers. But I can't figure out a rigorous proof.

Any help would be appreciated!

  • $\begingroup$ Is a single point not a compact interval? $\endgroup$
    – ItsMe
    Aug 1, 2022 at 11:46
  • 2
    $\begingroup$ Look up the Banach-Mazur game $\endgroup$ Aug 1, 2022 at 11:48
  • $\begingroup$ @ItsMe Thanks for your comment! Strictly speaking, a single point is not a interval, as I learned from high school. OK. Now I know that makes things confusing. Let me edit. $\endgroup$
    – Feng
    Aug 1, 2022 at 11:48
  • $\begingroup$ @AlessandroCodenotti Thanks! Interesting game! It seems that this wikipedia page has already answered my question. $\endgroup$
    – Feng
    Aug 1, 2022 at 11:55

1 Answer 1


Almost one day past, but nobody has written an answer. So I decide to write an answer here to close this qeustion, using the hint given by @Alessandro Codenotti.

This kind of game is generally called Banach–Mazur game. My proof follows this wikipedia page. We prove that Bob has a winning strategy, as I conjectured in OP. Note that all intervals we will mention have positive lengths.

Since $\mathbb Q$ is a countable set, we can Index the elements of $\mathbb Q$ as a sequence: $\mathbb Q=\{x_1, x_2, \cdots\}$. Suppose Alice has chosen a compact interval $A_1$ with $|A_1|>0$, then the interior $(A_1)^\circ$ of $A_1$ is a non-empty open interval. Hence Bob can choose a compact interval $B_1$ such that $$B_1\subset (A_1)^\circ\setminus\{x_1\}\subset A_1,\qquad 0<|B_1|\leq\frac12|A_1|.$$ Then Alice choose any compact interval $A_2\subset B_1$ with $0<|A_2|<\frac12|B_1|$ and, in a similar fashion, Bob can choose $B_2\subset (A_2)^\circ\setminus\{x_1\}\subset A_2$ with $0<|B_n|\leq\frac12|A_n|$. Continuing in this way, each point $x_n$ will be excluded by the set $B_n$, hence $$\mathbb Q\cap\left(\bigcap_{n=1}^\infty B_n\right)=\emptyset.$$ Therefore, Bob wins!


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