The product of the $A_j$ is the set $\prod_{j\in J} A_j$:=$\{ f: J \rightarrow \bigcup A_j \ | f(j) \in A_j \ for \ each \ j \in J \}$ We usually represent $f$ as $(a_j)=f(j) \in A_j$. For each $k \in J$ the k-th projection is the function $ \prod_{k} : \prod_{j \in J} A_j \rightarrow A_k$ defined as $\prod_k ((a_j)_{j \in J})=a_k$.
We assume that $X_j$, $j \in J$ are top. spaces The product topology on $\prod_{j \in J} X_i$ is the coarsest topology that makes all projections $\prod_j$ continuous.
The set $B:=\{\prod U_j \ | \ U_j \subseteq X_j \}$ is a basis for a topology on $\prod X_j$. Topology generated by this base is called the box topology on $\prod X_j$
TASK
Determine necessary and sufficient conditions, in terms of the topological spaces $X_j$ and the index set $J$, for the box topology to be different from the product topology on $\prod X_j$
Things I know:
1.In product topology $\prod_{j}^{-1} (U_j)$ is oppen in $\prod X_j$ for every $j\in J$ and every open set $U_j \subseteq X_j$
2.If $J$ is finite then the topology $t^{box} = t^{product}$. In general $t^{prod} \subseteq t^{box}$. The box top. is finer than the prod. top. so also for $\prod X_j$ equipped with the box top. all projections $\prod_j$ are continuous.
The box topology is different from the product topology if and only if $J$ is infinite and infinitely many of the $X_j$ have a topology that is not the trivial topology. I.e., there is an infinite subset of $J$, call it $K$ such that there exists $\varnothing \subsetneq U_k\subsetneq X_k$ with $U_k$ open in $X_k$ for all $k\in K$.
Proof. First suppose that $J$ is infinite and infinitely many of the $X_j$ have a topology that is not the trivial topology. Then let $K$ be an infinite subset of $J$ such that for all $k$ in $K$, $X_k$ does not have the trivial topology, and choose a family of nonempty open subsets $U_k\subsetneq X_k$ for each $k\in K$. Then let $$U=\prod_{j\in J} \begin{cases}U_j & j\in K\\ X_j &\text{otherwise.}\end{cases}$$ Clearly $U$ is open in the box topology, so to show that the box topology is different than the product topology, it suffices to show that this is not open in the product topology. For this, we note that $U$ doesn't contain any basic open subset, because the basic open subsets in the product topology are of the form $$V=\bigcap_{\ell\in L}\pi_\ell^{-1}(U_\ell)$$ for $L$ a finite subset of $J$, $U_\ell$ open in $X_\ell$, and since $L$ is finite, there will be some $k\in K\setminus L$, and $\pi_k(V)=X_k \not\subseteq U_k=\pi_k(U)$, so $V\not\subseteq U$. Thus $U$ cannot be open in the product topology.
Conversely, suppose that $J$ is not infinite, or $J$ is infinite and only finitely many of the $X_j$ have a topology that is not the trivial topology. Now we want to show that the box topology is the same as the product topology. First suppose that all the spaces have the trivial topology, then since any product with a null set is the null set, the box topology and the product topology on a product of spaces all of which have the trivial topology are both themselves the trivial topology.
Now let $K\subset J$ be the subset of indices whose corresponding spaces have the trivial topology. Let $L$ be the (finite) complement of $K$. Then we can factor the product as sets as $$\prod_{j\in J} X_j = \prod_{k\in K} X_k \times \prod_{\ell\in L} X_\ell.$$ Now if we use subscripts $b$ and $p$ to denote whether we're putting the box or product topology on the set product respectively, we see that $$\left(\prod_{j\in J} X_j\right)_b = \left(\prod_{k\in K} X_k\right)_b \times \left(\prod_{\ell\in L} X_\ell\right)_b,$$ and $$\left(\prod_{j\in J} X_j\right)_p = \left(\prod_{k\in K} X_k\right)_p \times \left(\prod_{\ell\in L} X_\ell\right)_p.$$ However, the first product is a product of spaces with the trivial topology, so $$\left(\prod_{k\in K} X_k\right)_p=\left(\prod_{k\in K} X_k\right)_b,$$ and the second product is finite, so $$\left(\prod_{\ell\in L} X_\ell\right)_p=\left(\prod_{\ell\in L} X_\ell\right)_b$$ as well. But that implies that $$\left(\prod_{j\in J} X_j\right)_p=\left(\prod_{k\in K} X_k\right)_p \times \left(\prod_{\ell\in L} X_\ell\right)_p=\left(\prod_{k\in K} X_k\right)_b \times \left(\prod_{\ell\in L} X_\ell\right)_b=\left(\prod_{j\in J} X_j\right)_b.$$ Thus the box and product topologies are the same in this case.