Almost surely statement in Williams book. In Probability with Martingales (Williams) I came across the following proposition

and then they give the following contradictory example

Could someone please explain how it can be so? Also why is that in the truth set they use $\rightarrow \frac{1}{2}$ and not $= \frac{1}{2}$ when comparing each outcome?
Thank you.
 A: This statement says that for every $\omega$ there exists some $\alpha$ in $\mathcal A$ such that $\omega\notin F_\alpha$. To see that this holds, consider $N_\omega=\{n\in\mathbb N\mid\omega_n=H\}$. 


*

*If $N_\omega$ is infinite, order it as $N_\omega=\{\nu_\omega(k)\mid k\in\mathbb N\}$ and consider $\alpha$ such that $\alpha(k)=\nu_\omega(k)$ for every $k$. Then $\omega_{\alpha(k)}=H$ for every $k$ hence $\frac1n\#\{k\leqslant n\mid\omega_{\alpha(k)}=H\}=1$ for every $n$, in particular $\omega\notin F_\alpha$. 

*If $N_\omega$ is finite, consider $\alpha$ such that $\alpha(k)=k$ for every $k$. Then $\omega_{\alpha(k)}=H$ for finitely many $k$ hence $\frac1n\#\{k\leqslant n\mid\omega_{\alpha(k)}=H\}\to0$ when $n\to\infty$, in particular $\omega\notin F_\alpha$. 
One sees that $\alpha$ such that $\omega\notin F_\alpha$ depends on $\omega$.
And this does not contradict the proposition in the book because $\mathcal A$ is uncountable. To see this, note that $\mathcal A$ is in bijection with $\mathbb N^\mathbb N$ through the application $\alpha\mapsto\beta$ defined by $\beta(1)=\alpha(1)$ and $\beta(n+1)=\alpha(n+1)-\alpha(n)$ for every $n$ in $\mathbb N$. Since $\mathbb N^\mathbb N$ is uncountable, this proves the claim  
