Is my understanding of product sigma algebra (or topology) correct? Let $(E_i, \mathcal{B}_i)$ be measurable (or topological) spaces, where $i \in I$ is an index set, possibly infinite. Their product sigma algebra (or product topology) $\mathcal{B}$ on $E= \prod_{i \in I} E_i$ is defined to be the coarsest one that can make the projections $\pi_i: E \to E_i$ measurable (or continuous).
Many sources said the following is an equivalent definition:
$$\mathcal{B}=\sigma \text{ or }\tau\left(\left\{\text{$\prod_{i    \in I}B_i$,  where
    $B_i \in \mathcal{B}_i, B_i=E_i$ for all    but a finite
    number of $i \in I$}\right\}\right),$$
where $\sigma \text{ and }\tau$ mean taking the smallest sigma algebra and taking the smallest topology. Honestly I don't quite understand why this is the coarsest sigma algebra (or topology) that make the projections measurable (or continuous).
Following is what I think is the coarsest one that can make the projections measurable
$$\mathcal{B}=\sigma \text{ or }\tau\left(\left\{\text{$\prod_{i    \in I}B_i$,  where
    $B_i \in \mathcal{B}_i, B_i=E_i$ at least for all    but one $i \in I$}\right\}\right),$$
because $\pi^{-1}_k (E_k) = \text{$\prod_{i    \in I}B_i$,  where
    $B_i=E_i$ for all   $i \neq k$}$. So I was wondering if the two equations for $\mathcal{B}$ are the same?
Thanks and regards!

For the comments: I retract my error for the definition of measurability. Sorry.
For the two things generating the same sigma algebra (or topology, which is similar):
We use $\langle - \rangle$ to denote the smallest sigma algebra containing the thing in the middle. We want to show that 
$$(1) \hspace{5mm}\langle \prod_{i} B_i \rangle$$ 
where $B_i \in \mathcal{B}_i$, and $B_i = E_i$ for all but finitely many $i$s, is the same as
$$(2) \hspace{5mm}\langle \prod_{i} B_i \rangle$$ 
where $B_i \in \mathcal{B}_i$, and $B_i = E_i$ for all but one $i$.
It is clear that $(2) \subset (1)$, since the generating collection in (2) is a subset of that of (1).
On the other hand, $(2)$ contains $\prod_{i} B_i$ where $B_i \in \mathcal{B}_i$, and $B_i = E_i$ for all but finitely many $i$s, since it is the (finite) intersection of the generators. For example, 
$$B_1 \times B_2 = (B_1 \times E_2) \cap (E_1 \times B_2)$$
So $(2) \supset (1)$. Therefore $(2) = (1)$.
