This is an exercise in chapter 2 of Probability with Martingales by David Williams.
Question: Let $\mathcal{A}$ be the set of all maps $\alpha : \mathbb{N} \rightarrow \mathbb{N}$ such that $\alpha(1) < \alpha(2) < \dots $,. For $\alpha \in \mathcal{A}$, let
$$F_\alpha = \{ \omega : \frac{\# (k \le n : \omega_{\alpha_{(k)}} = H)}{n} \rightarrow 0.5 \} .$$
Prove that
$$\bigcap_{\alpha \in \mathcal{A}} F_\alpha = \emptyset.$$
It is not too clear to me what can be done. Would appreciate some hints.
The text gives you a hint for this: "For any given $\omega$ find an $\alpha \ldots$". To extend this hint a little:
Let $\omega\in\Omega$, with $\omega=(\omega_1,\omega_2,\ldots)$ and with infinitely many Heads. Then consider a map $\alpha_\omega:\mathbb{N}\to\mathbb{N}$ defined by
$$\alpha_\omega(n) = j,\qquad\text{where $\omega_j$ is the $n^{th}$ Head in sequence $\omega$.}$$
Show that $\alpha_\omega\in \mathcal{A}$ and that $\omega\notin F_{\alpha_{\omega}}$. Also show that any $\omega\in\Omega$ with finitely many Heads will not belong to any $F_{\alpha}$ set. The result follows.