What is the intuition behind product topology? Let $X$ denote the cartesian product of the sets $X_i,i\in \mathbb N$ where $(X_i,\tau_i)$ are topological spaces. Define product topology by defining $\mathcal S=\bigcup\limits_{i\in \mathbb N}\mathcal S_i$ as a subbasis where $\mathcal S_i=\{p^{-1}(U_i):U_i\in \tau_i\}$.
My question is what is the intuition behind this product topology? 
I am lacking the motivation behind this definition, can someone provide me an example to illustrate the motivation in a better way. I know that for finite case, it is same as box topology. But I am looking for motivation for infinite case and want to know why this has to be the natural way to define a topology on cartesian product. I also want to know why it is the smallest topology such that $p_i$ , the projection maps are continuous. I think it is so because smallest topology containing $\mathcal A$ is obtained by taking subbasis $\mathcal{S=A}\cup \{X\}$.
 A: The product topology is indeed chosen to be the minimal topology on $\prod_{i \in I} X_i$ that makes all projections continuous. This minimality also ensures that it interacts well with other constructions of new spaces fom old ones, like the subspace topology, and it also obeys some nice criteria to check continuity of maps, see my long post on the general theory here if you want to know all the details. It's the most "natural" choice of topology, and the one that fits best into more modern developments like category theory.
Another reason to pick this topology over the other sort of natural one, the box topology, is that it simply has nicer properties and so is more useful in practice: the space $\Bbb R^\omega$ (of infinite sequences of reals) in the product topology is metrisable, separable and connected, while in the box topology it is none of those; not even first countable e.g. So infinite box products are nice counterexamples in some cases but the product topology has the properties that are desirable: preserving compactness, connectedness etc etc. Maybe not intuition, but it is motivation.
A: The intuition is precisely that - we want projections to be continuous and there is a minimal way of making it happen.
As for why it's smallest. Let $\tau ^\times$ be the product topology and $\tau$ some topology such that all $p_j$ are continuous w.r.t $\tau$. Let $U\in \tau ^\times$, then $B\subseteq U$ for some $B= \bigcap _{k=1}^n p_{i_k}^{-1}(V_{i_k})$. By assumption, the preimages are in $\tau$, hence also $B\in\tau$, which implies $U\in\tau.$
