A question in the book 'probability and martingales' Our teacher uses this book as a textbook and in $2.3, there is a statement that about which is a model of tossing a coin
$$F=\sigma({w\in\Omega:w_n=W}:n\in \textbf{N}, W\in \{H,T\})\\
w=(w_1,w_2,...), w_n \in \{H,T\}$$
and it says that $F \ne P(\Omega)$ which is the power set of the set of $w$, I want to prove it but I have no ideas.
$\sigma(M)$ denotes the generated $\sigma$-algebra of M, i. e. the cut of all $\sigma$-algebras containing M:
$$ \sigma(M) = \bigcap_{\substack{\mathfrak{A}\text{ $\sigma$-algebra} \\ M \in \mathfrak{A}}} \mathfrak{A}$$
$\Omega$ is the set of each $w$.
 A: $\Omega=\{H,T\}^\mathbb N$
$\mathcal F=\sigma\left(\left\{\left\{\omega\in\Omega\mid\omega_n=W\right\}\mid n\in\mathbb N,W\in\{H,T\}\right\}\right)$
Let $d$ be the metric defined on $\Omega^2$ by
$$
\forall x,y\in\Omega,\quad d(x,y)=\sum_{n\in\mathbb N}\frac{1}{2^n}\vert x_n-y_n\vert
$$
If for all $k\in\mathbb N$, $\omega^{(k)}\in\Omega$ and $\omega\in\Omega$, then $d(\omega^{(k)},\omega)\to0\iff\forall n\in\mathbb N,\omega^{(k)}_n\to\omega_n$. Then we have
$$
\mathcal F=\mathcal B(\Omega),
$$
where $\mathcal B(\Omega)$ denotes the Borel $\sigma$-algebra of $\Omega$ with respect to the metric $d$. Indeed, $\{\omega\in\Omega\mid\omega_n=W\}$ is a closed subset of $\Omega$, so $\mathcal F\subset\mathcal B(\Omega)$. Conversely, let $A$ be a closed subset of $\Omega$. Let
$$
D=\{\omega\in\Omega\mid\exists N\in\mathbb N,\quad\forall N\ge n,\quad\omega_n=H\}
$$
Then $D$ is a countable subset of $\Omega$ such that $\overline D=\Omega$, so since $A$ is closed, we have
$$
A=\{\omega\in\Omega\mid d(\omega,D)=0\}=\{\omega\in\Omega\mid\inf_{y\in D}d(\omega,y)=0\}
$$
For all $y\in D$, $\omega\in\Omega\mapsto d(\omega,y)$ is $\mathcal F$-measurable. Since $D$ is countable, $\omega\in\Omega\mapsto\inf_{y\in D}d(\omega,y)$ is $\mathcal F$-measurable. So $A\in\mathcal F$. Hence $\mathcal B(\Omega)\subset\mathcal F$.
Let $f:\Omega\to[0,1]$ be defined by
$$
\forall\omega\in\Omega,\quad f(\omega)=\sum_{n\in\mathbb N}\frac{1}{2^n}\omega_n
$$
Then $f$ is one-to-one and onto, $f$ is $(\mathcal B(\Omega),\mathcal B([0,1]))$-measurable and $f^{-1}$ is $(\mathcal B([0,1]),\mathcal B(\Omega))$-measurable.
Let $V$ be a subset of $[0,1]$ such that $V\notin\mathcal B([0,1])$. Then $f^{-1}(V)\in\mathcal P(\Omega)$. However, if we had $f^{-1}(V)\in\mathcal B(\Omega)$, then we would have $V=f(f^{-1}(V))\in\mathcal B([0,1])$, which is not. Therefore,
$$
V\in\mathcal P(\Omega)\backslash\mathcal F
$$
