Alternative definition of product topology In James Munkres's book, product topology on any many sets is defined based on projection mapping which is based on J tuple. This definition seems to be not very intuitive.
Why isn't it defined as:
The product topology on $\prod X_\alpha$ has as subbasis of all sets of the form 
 $\prod U_\alpha$ where $U_\alpha$ is open in $X_\alpha$ for each $\alpha$ and $U_\alpha$ equals $X_\alpha$ except for a single $\alpha$.
 A: That's equivalent to the usual definition.
The topology with a certain subset as a subbasis is the topology with the fewest open sets where all elements of that subset is open. 
The product topology is the topology with the fewest open sets where every projection is continuous. That is: the preimage of any open set under the projection must be open. That is precisely the definition you give. 
A: Equivalently, it's the topology generated by sets of the form $\prod U_i$, where $U_i=X_i$ for all but finitely many $i$.  This can be contrasted with the "box topology", where any number of $U_i\ne X_i$.  
The product topology makes the product $X=\prod X_i$ a "categorical product".
Also, the product topology satisfies a universal property, namely that any continuous map from another toplogical space to $X$ "factors through" the projections.
A: It may be easier to understand the definition, as well as its equivalences, in a slightly more general context.
Suppose that for each $j\in J$ we have a topological space $X_j$ and a function $f_j\colon X\to X_j$, where $X$ is just a set. In this scenario we can ask about a topology over $X$ in which each function $f_j$ becomes continuous. One obvious answer could be the discrete topology over $X$. However, this is not good enough, since every function $f\colon X\to Y$ is continuous when $X$ is discrete (and $Y$ is a topological space).
The issue here is that the discrete topology is "too big". So, we could ask about a smaller topology. Actually, let us ask ourselves about the smallest topology over $X$ such that $f_j\colon X\to X_j$ is continuous - when $X$ is endowed with such topology.
Certainly there is such topology, since the intersection of a nonempty family of topologies over $X$ is a topology. Let us call it $\tau$.
Now, if $\sigma$ is another topology over $X$ such that $f_j\colon (X,\sigma)\to X_j$ is continuous for each $j\in J$, then $f_j^{-1}[U]\in\tau$ whenever $U\subseteq X_j$ is an open set. After some reasoning, it is not hard to see that this shows us that the topology generated by the subbasis $\mathcal{B}=\{f_j^{-1}[U]:j\in J$ and $U\subseteq X_j$ is an open set$\}$ is the least topology over $X$ such that each $f_j\colon X\to X_j$ is continuous. In other words, $\mathcal{B}$ is a subbasis for the topology $\tau$.
So, it is equivalent to define $\tau$ as the topology generated by the subsets $f_j^{-1}[U]$, where $U\subseteq X_j$ is an open set.
Note that your question is a particular case of the above reasoning: just take $X=\prod_{j\in J}X_j$ and let $f_j\colon X\to X_j$ be the $j$-th projection.
