Let $n<\sqrt{5\cdot2^{11}}$, and say we have a 2 colouring on ${1,2,...,n}$. Prove that there exists an arithmetic progression of length 10 that isn't monochromatic.
This was given to us in an exam question in the context of positional games. It stated, either prove this using the fact that the second player does not have a winning strategy in the strong game $(X,F)$, and using the Erdos Selfridge theorem for the maker-breaker game.
My first idea was to prove the breaker had a win for the following game; Player 1 (maker) and Player 2 (breaker) take it in turns to colour the integers ${1,2,...,n}$ until they are all coloured. If there is an arithmetic progression of length 10 once all $n$ elements are coloured, then the breaker wins. Otherwise, the maker wins.
Doing this, it is clear that any winning set in the family $A \in F$ satisfies $\mid A \mid = 10$, so we can turn the erdos selfridge theorem from;
$$ \sum_{A \in F} 2^{-\mid A \mid} < \frac{1}{2}$$
To;
$$\mid F \mid< 2^{9}$$
So it suffices to prove for out given n, F satisfies this. However from here I became very confused. Any hints?