Confusion regarding an equality on product set We have $\{ X_{\alpha} \}_{\alpha \in A} $ collection of nonempty sets. $X = \prod_{\alpha} X_{\alpha} $ and $\pi_{\alpha} : X \to X_{\alpha }$ is coordinate maps. Given the collection 
$$ \mathcal{F} = \{ \pi_{\alpha}^{-1}( E_{\alpha} ) : E_{\alpha} \in \mathcal{M}_{\alpha} , \alpha \in A \} $$
where $\mathcal{M}_{\alpha} $ is a $\sigma-$algebra on $X_{\alpha}$. The product sigma algebra on $X$ is the sigma algebra generated by $\mathcal{F} $.
Why does it follow that 
$$ \prod_{\alpha \in A} E_{\alpha} = \bigcap_{\alpha \in A} \pi_{\alpha}^{-1}(E_{\alpha} ) $$
?? Ive been trying to understand this for hours, and I cant get it. 

This is just about set theory, and has nothing to do with sigma algebras and such.
Let $(x_\alpha)_{\alpha \in A}\in \prod_{\alpha \in A}E_\alpha$. So: $$\pi_\alpha((x_\alpha)_{\alpha \in A}) = x_\alpha \in E_\alpha \implies (x_\alpha)_{\alpha \in A} \in \pi_\alpha^{-1}(E_\alpha).$$Since $\alpha \in A$ is arbitrary, this gives $\prod_{\alpha \in A}E_\alpha \subset \bigcap_{\alpha \in A}\pi_\alpha^{-1}(E_\alpha)$.
On the other hand, take $(x_\alpha)_{\alpha \in A} \in \bigcap_{\alpha \in A}\pi_\alpha^{-1}(E_\alpha)$. So given any $\alpha \in A$, we have that: $$x_\alpha = \pi_\alpha((x_\alpha)_{\alpha \in A}) \in E_\alpha \implies (x_\alpha)_{\alpha \in A}\in \prod_{\alpha \in A}E_\alpha,$$by the definition of  the inverse images. This gives $\bigcap_{\alpha \in A}\pi_\alpha^{-1}(E_\alpha)\subset \prod_{\alpha \in A}E_\alpha$. Done.
