Equivalence of the Box and product topologies on a finite product This is a very basic question but am having trouble showing it so will really appreciate any help with this, thanks in advance. 
Let $(X_i,\mathscr{T}_i)$, $i=1,...,n$ for some $n\in\{2,3...\}$ and $X=\prod\limits_{i=1}^nX_i$.  Then the box topology $\mathscr{T}^*$ on $X$ is generated by the basis $\mathscr{B}^*:=\{\prod\limits_{i=1}^n U_i|U_i\in\mathscr{T}_i\}$ and the product topology $\mathscr{T}'$ is generated by the subbasis $\mathscr{S}':=\{\pi^{-1}_i(U_i)|U_i\in\mathscr{T}_i\}$. Now let $\mathscr{B}'$ be the basis of $\mathscr{T}'$, so the set of of finite intersections of $\mathscr{S}'$. So it seems to me that $\mathscr{B}^*\subset\mathscr{B}'$, (as $V_i\times ...\times V_n=\bigcap\limits_{i=1}^n\pi_i^{-1}(V_i)$) so $\mathscr{T}^*\subset\mathscr{T}'$ but I cannot see why the reverse inclusion is true. Also why does this inclusion fail for infinite products as I read that the box topology is finer.
So any assistance and corrections of any mistakes will be greatly appreciated. Thanks.
 A: Let $S$ be a finite intersection from the subbase $\mathcal{S}'$, so there is some $N$ and $I = i_1, \ldots, i_N \in \{1, \ldots, n\}$ and non-empty open sets $U_{i_1} \subseteq X_{i_1}, \ldots, U_{i_N} \subseteq X_{i_1}$ such that
$S = \cap_{j=1}^N \pi_{i_j}^{-1}[U_{i_j}]$. We can  collect sets from the same space together and some indices might not appear, so set $J_i = \{j \in \{1,\ldots, M\}: i_j = i \}$ for $i \in \{1,\ldots,n\}$ and $K = \{i \in \{1,\ldots, n\}: J_i =\emptyset\}$, all indices of spaces that are not used.
Then
$$S  =\cap_{i \notin K} (\pi_{i}^{-1}[\cap_{j \in J_i} U_{i_j}]) = \prod_{i=1}^n U_i \text{, where } U_i = X_i \text{ for } i \in K \text{ and } U_i = \cap_{j \in J_i} U_{i_j} \text{ for } i \notin K$$
so $S \in \mathcal{B}^\ast$. And indeed, as you said, $S = \prod_i U_i \in \mathcal{B}^\ast$ then $S = \cap_{i=1}^n \pi_i^{-1}[U_i]$ which is in the base generated by $\mathcal{S}'$. So the standard base for the box topology, is exactly the same as the base generated by the subbase for the product topology, so they generate the same topology.
EDIT: made more slick:
Let $\mathcal{B}'$ be the base generated by $\mathcal{S}'$. Note that the box base
$\mathcal{B}^\ast$ is closed under finite intersections as
$$\prod_i U_i \cap \prod_i V_i = \prod_i (U_i \cap V_i)$$
and also note that $(\pi_k)^{-1}[U] = \prod_i U_i \in \mathcal{B}^\ast$, where $U_i = X_i$ for $i \neq k$ and $U_k = U$, all of which are open if $U$ is.
The last remarks hows that $\mathcal{S}' \subseteq \mathcal{B}^\ast$ and as the latter is closed under finite intersections, $\mathcal{B}' \subseteq \mathcal{B}^\ast$. And the old
$$\prod_i U_i = \cap_i \pi_i^{-1}[U_i] \in \mathcal{B}'$$
shows that $\mathcal{B}^\ast \subseteq \mathcal{B}'$, hence equality of the bases is ensured, and the equality of the topologies as well.
For infinite products, the product topology is strictly coarser (for non-trivial spaces): because any proper box, $U = \prod_{i \in I} U_i$, where all $\emptyset \neq U_i \nsubseteq X_i$ are open, is not open in the product topology: if it were, pick $p_i \in U_i$, so that $p =(p_i) \in U$, and suppose there are finitely many $i_1, \ldots, i_N$ and open sets $O_{i_1},\ldots, O_{i_N} $ such that $p \in S = \cap_{i=1}^N (\pi_{i_j})^{-1}[O_{i_j}] \subseteq U$. But $I$ being infinite, we have $j \in I$ and $j \notin \{i_1,\ldots,i_N\}$, and pick $q_j \in X_j \setminus U_j$. Then $p'$ defined by $p'_i = p_i , i \in I\setminus \{j\}, p'_j = q_j$ obeys $p' \notin U$, but $p' \in S$ which cannot be. The intuition is easy: $S$ only sets conditions on finitely many coordinates (as a finite intersection of single "conditions"), while the box needs control over all infinitely many, so the esentially finite boxes of the product topology always stick out of the infinite boxes at the missing coordinates. The above makes this more formal.
