I am reading Folland's "Real Analysis" on my own, and I was hoping that someone could tell me if I have understood the following proof correctly.
Proposition: If $A$ is countable, then $\otimes_{\alpha \in A} \ \mathcal{M}_\alpha$ is the $\sigma$-algebra generated by $$\left\{\prod_{\alpha \in A} E_{\alpha} : E_{\alpha} \in \mathcal{M}_{\alpha} \right\}.$$
Previously, the product $\sigma$-algebra on a set $X = \prod_{\alpha \in A} X_\alpha$ where $\mathcal{M}_\alpha$ is a $\sigma$-algebra on each $X_\alpha$ was said to be the $\sigma$-algebra generated by $$\left\{{\pi_\alpha^{-1}}(E_\alpha) : E_\alpha \in \mathcal{M}_\alpha, \alpha \in A\right\}.$$
The proof uses the following lemma:
Lemma: Let $\mathcal{M}(\mathcal{E})$ be the $\sigma$-algebra generated by $\mathcal{E}$. If $\mathcal{E} \subset \mathcal{M}(\mathcal{F})$, then $\mathcal{M}(\mathcal{E}) \subset \mathcal{M}(\mathcal{F})$.
Proof of proposition (from Folland):
If $E_\alpha \in \mathcal{M}_\alpha$, then $\pi_\alpha^{-1}(E_\alpha) = \prod_{\beta \in A} E_{\beta}$ where $E_\beta = X_\beta$ for $\beta \neq \alpha.$
Question 1: Is this because for a countable product the map $\pi_\alpha^{-1}(E_\alpha)$ takes in a set like $E_\alpha$ and maps it back to $X_1 \times X_2 \times \cdots \times E_\alpha \times X_{\alpha+1} \times \cdots$ ? How does this map work for uncountable products?
On the other hand, $$\prod_{\alpha \in A} E_\alpha = \cap_{\alpha \in A} \pi_\alpha^{-1}(E_\alpha).$$
This statement makes sense to me if the coordinate map behaves the way I asked about above.
The result therefore follows from the lemma.
Question 2: Is this because on one hand, $$\{\prod_{\alpha \in A} E_\alpha \} = \left\{\cap_{\alpha \in A} \pi_\alpha^{-1}(E_\alpha)\right\} \subseteq \left\{{\pi_\alpha^{-1}}(E_\alpha) : E_\alpha \in \mathcal{M}_\alpha, \alpha \in A\right\}$$ which is in turn a subset of the $\sigma$-algebra generated by $\{{\pi_\alpha^{-1}}(E_\alpha) : E_\alpha \in \mathcal{M}_\alpha, \alpha \in A\}$? Like, in general, $\mathcal{M}(\mathcal{E})$ contains all elements of $\mathcal{E}$? Therefore using the lemma, $$\mathcal{M}(\prod_{\alpha \in A} E_\alpha) \subset \mathcal{M}(\{{\pi_\alpha^{-1}}(E_\alpha) : E_\alpha \in \mathcal{M}_\alpha, \alpha \in A\}).$$
On the other hand,
$$\left\{\pi_\alpha^{-1}(E_\alpha)\right\} =\left\{ \prod_{\beta \in A} E_\beta \text{ | $E_\beta = X_\beta$ for $\beta \neq \alpha$}\right\} \subset \mathcal{M}(\{\prod_{\alpha \in A} E_{\alpha} : E_{\alpha} \in \mathcal{M}_{\alpha} \}),$$ so we can apply the lemma again. For this part, I believe the subset inclusion hold because $X_\alpha$ is necessarily in $M_\alpha$.
Thanks in advance for any help!