I have a question regarding the proof of theorem 6.2 which states that,

Thm 6.1: There is a strategy in which is sure to win iff is of first category

The game played is this: there is a set $B \subseteq I_0$, where $I_0$ is a closed interval (all intervals here are required to have positive length). The first player $(A)$ chooses a closed interval $I_1 \subseteq I_0$, and then they alternate forever, choosing a closed subinterval $I_{k+1}$ of the previous interval $I_k$. Player (A) wins if the intersection $\bigcap_k I_k$ is disjoint from $B$, otherwise player (B) wins.

The proof is given here below and my question is regarding the second line:

How can we know that given $f_1$, it is possible to define a sequence of closed intervals $J_i$ contained in $I_0^o$ such that the intervals $K_i$ are disjoint? and also, how do we make sure that the union of their interior is dense??

Thank you, Shir

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1 Answer 1


They explain it in the very next sentence. They take all intervals with rational endpoints, and plug them in as A's move. The K's are then B's move according to the strategy. The interiors of the K's are all disjoint because the J's are chosen to disjoint from all previous K and each K lies in the corresponding J.

This is dense because if it weren't, there would be an interval with rational endpoints not intersecting the K's, but since we added ALL such intervals as we reached them, this is impossible.

  • $\begingroup$ Sorry! I didn't continue reading because I didn't understand and didn't think it is going to be explained :| Thank you! $\endgroup$
    – topsi
    Jan 9, 2014 at 5:14

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