Hacker News Reader

I don't follow why Bob can play any legal move when s_n is not possible. The proposition is that, if S is countable, Bob has a winning strategy. Suppose S = (s1, s2, s3), which is finite and therefore countable. On turn 1 Alice picks s1. Since s1 is already taken, Bob picks any legal move, which means any number in [0, 1]. Suppose he picks s3. Then, Alice picks s2 and wins?

View on HN

They must continue playing. There is no termination condition for the game. They always continue playing an infinite number of steps.

Alice picks s2. Then Bob picks some number less than s3, so the sequence does not converge to s3. Then Alice picks some number larger than s2, so the sequence does not converge to s2. Any number that is picked after a finite number of steps, cannot be the value that the sequence converges to.

Ok, that makes more sense. The part I still don't get is how we can pick a number x such that s2 < x < s3 without directly assuming that [0, 1] is uncountable.

Even the rational numbers have that property (the Archimedean property): for any distinct rationals s2 and s3, there exists a rational x such that s2 < x < s3.

Proof: x = (s2 + s3) / 2.

This is easily extended to show that there is an arbitrary number of such x.

Thanks, that makes sense!

Even some subsets have that property, like eg all the finite decimal fractions.

Reply on news.ycombinator.com