$\sigma$-Algebra generated 
I have a question : 
Let $Z=\lbrace 0,1\rbrace ^{\mathbb{N}}$. Why does the sets $E_1,E_2,...$, where $E_k=\lbrace (n_i)\in Z; n_k=1 \rbrace $ generate $\mathcal{B}(Z)$? 
How can I prove it?
I find that they use :

$\quad$Proposition $\mathbf{8.1.5.}$
$\quad$Let $X_1$, $X_2$, $\ldots$ be a finite or infinite sequence of separable metrizable spaces. Then $\mathscr{B}(\Pi_nX_n)=\Pi_n\mathscr{B}(X_n)$.

but I don't understand why ?
Thank you
 A: *

*We can give a complete characterization of $\mathcal B(Z)$.

*Show that the $\sigma$-algebra generated by the $E_k$ is $$\left\{\prod_{n=0}^{+\infty}S_n, S_n\subset \{0,1\}\right\}.$$

A: Proof. Let $S$ be a minimal $\sigma$-algebra of subsets of $Z=\{0,1\}^N$ generated by the family $\{E_k: k \in N\}$, i.e., $S=\sigma(\{E_k: k \in N\})$. We have to show that $S={\cal{B}}(Z)$.  Since $\{E_k: k \in N \} \subseteq {\cal{B}}(Z)$ we claim that $$S=\sigma(\{E_k: k \in N\})\subseteq \sigma( {\cal{B}}(Z))={\cal{B}}(Z).$$ We have to show the validity of the converse inclusion.
Note that $${\cal{B}}(Z)=\sigma (\{ Y \times \prod_{i>n}\{ 0,1 \}_i : Y \subseteq 
\{0,1\}^n~\& ~n \in N  \}).$$ Let us show that $$(\omega_1,\cdots,\omega_n)\times \prod_{i>n}\{ 0,1 \}_i \in S$$ for $n \in N$ and $(\omega_1,\cdots,\omega_n)\in \{ 0,1 \}^n.$
Since $E_k \in S$ for $k \in N$ we deduce that $$Z \setminus E_k=\{(n_i)_{i \in N}:(n_i)_{i \in N} \in Z ~\&~n_k=0~\} \in {\cal{B}}(Z).$$ For $1 \le k \le n$ we set $A_k=E_k$ if $\omega_k=1$ and $A_k=Z \setminus E_k$ if $\omega_k=0$. Then   $$(\omega_1,\cdots,\omega_n)\times \prod_{i>n}\{ 0,1 \}_i= \cap_{k=1}^n A_k \in S. $$
Now if $Y \subseteq \{0,1\}^n$ then $Y=\cup_{(\omega_1,\cdots,\omega_n)\in Y}\{(\omega_1,\cdots,\omega_n)\}$. Since $S$ is closed under taking finite unions we deduce that $ Y \times \prod_{i>n}\{ 0,1 \}_i= \cup_{(\omega_1,\cdots,\omega_n) \in Y }(\omega_1,\cdots,\omega_n)\times \prod_{i>n}\{ 0,1 \}_i \in S$. 
Thus $$\{ Y \times \prod_{i>n}\{ 0,1 \}_i : Y \subseteq \{0,1\}^n~\& ~n \in N  \} \subseteq S$$ and we deduce that $${\cal{B}}(Z)=\sigma (\{ Y \times \prod_{i>n}\{ 0,1 \}_i : Y \subseteq \{0,1\}^n~\& ~n \in N  \}) \subseteq  \sigma(S)=S.$$
This ends the proof.
A: Let $\mathcal{S}$ be the $\sigma$-algebra generated by the $E_k$'s.
Each $E_k$ is open, so $\mathcal{S} \subseteq \mathcal{B}(Z)$.
Furthermore, the $E_k$'s form a sub-basis for the product toplogy, so $\mathcal{B}(Z) \subseteq \mathcal{S}$.
