How to win this Rational versus Irrational game?

Question

Suppose this game can continue indefinetely. One of the players has the rationals, the other the irrationals. The one with the rationals chooses a subinterval I1 = [a1 , b1] of [0, 1] (a1 e b1 may be rational or irrational) with length b1 - a1 <= 1 = 1/1. On the next move, his opponent chooses a subinterval I2 = [a2, b2] of I1 with length less or equal to 1/2. On the n-th move, it's chosen a subinterval I_n = [a_n, b_n] of I_(n-1) with length b_n - a_n <= 1/n. So, 2 sequences a_n e b_n are constructed in [0,1], such that a_n < b_n for every n and lim (a_n - b_n) = 0 Since the reals are a complete set, a_n and b_n converge to a same x in [0,1] (ie, x is the only element common to all of the I_n's). If x is rational, the player with the rationals wins, and vice-versa. Show that there exists an strategy that, if followed by the player that has the irrationals, assures his victory no matter the moves of his opponent are.

Answer 1

Let p_1, p_2, ,,, be an enumeration of the rational (which we know are countable in number). For the first move of the irrational player, choose an interval that fails to contain p_1. In general, at the n^th move, choose an interval that does not contain p_n. Since every rational has been excluded, x is irrational.

Answer 2

I'm assuming each subinterval must be of positive length, else the game ends on the first play in the obvious way.

Under that assumption, as defined, victory is never reached at any finite stage of the game, because it is based on the limit of an infinite series.

I think what you are asking for here is, to exhibit an algorithm, which, if followed by the player with the irrationals, will lead to a sequence converging to an irrational number regardless of the strategy employed by the player with the rationals. If that is the case, I think the strategy given by mathematician is one that works.

Answer 3

by working hard