Understanding Theorem 19.2 in Munkres 
Theorem 19.2 in Munkres Suppose the topology on each space $X_{\alpha}$ is given by a basis $\mathcal{B}_{\alpha}$. The collection of all sets of the form $$\prod_{\alpha \in J} B_{\alpha}$$ where $B_{\alpha} \in \mathcal{B}_{\alpha}$ for each ${\alpha}$, will serve as a basis for the box topology on $\prod_{\alpha \in J} X_{\alpha}$

The problem I'm having with understanding the theorem is due to the use of the indices. 

From what I understand the above theorem is saying that the collection of all possible sets formed by:
Taking one arbitrary basis element from each arbitrary basis $\mathcal{B}_{\alpha}$, and then taking the product of each of those basis elements 
forms a basis for the box topology $\prod_{\alpha \in J} X_{\alpha}$. 

Is my understanding correct? 
I apologize in advance if this question is somewhat vague.
 A: Yes, your interpretation is correct.  The open sets in the box topology on $\displaystyle \prod_{\alpha \in J} X_\alpha$ will be unions of basis elements.  The basis elements are of the form $\displaystyle \prod_{\alpha \in J} B_\alpha$, where each $B_\alpha$ is some basis element for $X_\alpha$.
Perhaps more simply, the basis for the box topology can be defined in terms of open sets.  That is, $\displaystyle \prod_{\alpha \in J} X_\alpha$ has as a basis $\displaystyle \left\{ \prod_{\alpha \in J} U_\alpha \ \Big| \ U_\alpha \text{ is open in } X_\alpha \right\}$.

As a side note, we typically choose the product topology over the box topology in practice; I'm sure this'll come up shortly in Munkres.  The product topology is defined as we defined the box topology in the second paragraph above, but with the added stipulation that $U_\alpha = X_\alpha$ for all but finitely many $\alpha \in J$.  Though this seemingly more complicated, the product topology satisfies more "desirable" things.  Some examples off the top of my head include:


*

*The product of compact spaces will be compact.  

*A function $f: Y \rightarrow \displaystyle \prod_{\alpha \in J} X_\alpha$ is continuous $\iff$ each $(\pi_\alpha \circ f): Y \rightarrow X_\alpha$ is continuous, where $\pi_\alpha$ denotes the projection map from the product space onto $X_\alpha$.
Lastly, note that the product topology and the box topology are the same thing unless we are looking at the product of infinitely many $X_\alpha$.
