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I got this from Thomson et al.'s freely available "Elementary Real Analysis" p.356.

They introduce Baire's category theorem through a game where, given two players (A) and (B)

Player (A) is given a subset $A$ of $\mathbb{R}$, and player (B) is given the complementary set $B = \mathbb{R} \backslash A$. Player (A) first selects a closed interval $I_1 \subset \mathbb{R}$; then player (B) chooses a closed interval $I_2 \subset I_1$. The players alternate moves, a move consisting of selecting a closed interval inside the previously chosen interval.

The play of the game thus determines a descending sequence of closed intervals \begin{align} I_1 \supset I_2 \supset \ldots I_n \supset \ldots\end{align} where player (A) chooses those with odd index and player (B) those with even index. If \begin{align} A \: \bigcap_n^{\infty} I_n \neq \emptyset \end{align} then player (A) wins; otherwise player (B) wins.

Then they argue that if player (A) is dealt the irrational set and (B) is dealt the rational set, (A) always has a strategy to win.

I'm confused in several ways by this argument. One confusion is about the term "closed interval". Does, for example, the closed interval [1,1] count as an interval? Because if that's the case, can't whoever has the first turn just end the game then and there without regard to whether he has the rationals or irrationals? Say, if (A) received the rationals, he can just pick [0.5,0.5]. Game over. (But I'm guessing probably not, because the game wouldn't happen)

If that is not the case, and an interval has to be defined $[a,b]$ s.t. $a < b$, isn't it always true that for any $I_{2n+1} = [a_{2n+1}, b_{2n+1}]$ for the odd numbered turns of (A), and where (A) has the irrationals, there exists some $q \in [a_{2n+1},b_{2n+1}]$ s.t. $q$ is rational? Because $\mathbb{Q}$ is also dense on the real line. Or is this argument relying on the fact that a countable intersection of closed sets is always closed? And that it can converge to a single point. But if it converges to a single point, it might be a closed set, but is it still a closed interval? And does that still count as winning? And can't (A) still play this same game even if he were dealt the rationals if all that is needed is that his choice of intervals converge to a single point?

I've been thinking about this for a few days now and none of the ways I've approached it convince me that their discussion of the game is true (though I do trust that it is true, since the authors are mathematicians and I'm not), so any help would be appreciated.

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1 Answer 1

up vote 2 down vote accepted

A closed interval is of the form $[a,b]$, $[a, \infty)$ or $(-\infty,b]$.

Since $A$ deals first, he can start with a bounded interval, so let's assume that all intervals $I_{n}$ are bounded (I leave it to you to describe a winning strategy for $B$ if $A$ is silly enough to always deal an unbounded interval). The only really interesting case is if the length of the interval converges to zero, for if not, the intersection will be a non-trivial interval, which will then contain an irrational number. So, if $|I_{n}| \to 0$ then there will be a unique real number $x \in \bigcap_{n} I_{n}$ by the priniple of nested intervals and the question becomes: can $B$ force this number to be rational so that $A$ loses? The answer is no. The reason is that the decimal expansion of a rational number is eventually periodic, so $A$ only has to take care to break any possible period of the decimal expansion of $x$ at each step and this is easily achieved by choosing an interval of sufficiently small length.

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Does this mean the outcome of the game still stands (i.e. the holder of irrationals $A$ wins) even if $A$ starts play after $B$? It sounds like it would. –  JasonMond Feb 9 '11 at 16:50
    
@user6690: Yes, no matter what $B$ does, $A$ can force the intersection of the intervals to contain an irrational number, even if $B$ starts. –  t.b. Feb 9 '11 at 16:54

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