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Topic: A rational X irrational game
Replies: 10   Last Post: Mar 20, 2004 5:52 PM

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Posts: 72
Registered: 12/13/04
Re: A rational X irrational game
Posted: Mar 20, 2004 7:41 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply (Artur) wrote in message news:<>...
> David C. Ullrich <> wrote in message news:<>...
> > On 19 Mar 2004 07:30:09 -0800, (Artur) wrote:
> >

> > >Hello
> > >I've been thinking about the following problem (or puzzle) but hasn't
> > >come to a conclusion yet. Maybe someone can give a hint.
> > >
> > >A and B are playing the following game: On the real line, A chooses a
> > >closed interval I1 of length 0<L1<=1. Then, B chooses a closed
> > >interval I2, of length 0<L2<=1/2, contained in I1. Then A, in turn,
> > >chooses a closed interval I3, of length 0<L3<=1/3, contained in I2,
> > >and so on. We know there's one, and only one, real number x that
> > >belongs to all of the intervals I1, I2, I3.... If x is rational, A
> > >wins the game, and if x is irrational, then B wins. We are asked to
> > >find an strategy that B should follow so that he will certainly win
> > >the game no matter how B chooses his intervals.
> > >
> > >Since the rationals are countable and the irrationals are not, it
> > >really seems that there is such a strategy that assures B will ever
> > >win, but I couldn't find it so far.

> >
> > Hint: B doesn't have to force a win on the first move. On
> > the first move B prevents one way he can lose, on the
> > next move he prevents another method of losing...
> >

> Ummm... based on your hint, this may work:
> Once the interval I1 has been choosen by A, B enumerates the rationals
> in I1. Let {x_1, x_2....} be such enumeration. Then, B chooses I2 in
> such a way that x_1 is not in I2 (this is always possible, because the
> length of I2, though positive, is allowed to be arbitrarily small). In
> the next move, A chooses an I3 contained in I2. And, in his turn, B
> chooses an I4 that doesn't contain x_2 .
> Carrying on with this method of choices, B ensures the game will
> generate a sequence {Im} of closed intervals such that no x_n belongs
> to all of the Im's. Therefore, the element x, common to all Im's, is
> not covered by the enumeration {x_1, x_2....}, which implies x must be
> irrational. And B wins!
> Artur

The important point here is that the rationals are countable and the
irrationals are not. This same strategy works if we replace the words
rational and irrational by algebraic and transcendental, because the
algebraics are countable and the trascendental are not. Keeping the
other rules, it seems we get really complicated puzzles if we replace
the game's winning criterion by the following:

a)Suppose the sets Da and Db, Da with measure zero but uncountable,
both dense in R, form a partition of R. A wins if x is in Da and B
wins if x is in Db.

b) A wins if x is Liouville and B wins otherwise.


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