Understanding definition of subbasis of product topology

Question

According to Munkres, the product topology has a subbasis which the union of all $S_i$ such that:

$$S_i =\{\pi_i^{-1}(U): U \text{ is open in }X_i \}$$

I am quite certain that I am misunderstanding this definition. To me this looks like $S_i$ contains all Cartesian products such that the i'th element is open in $X_i$. However, that would make the union of all $S_i$ the power set of the product space. It looks to me like Munkres, and other resources as well, are treating this definition as one where $S_i$ comprises all sets of the form

$$\dots X_{i-3}\times X_{i-2} \times X_{i-1} \times U \times X_{i+1} \times X_{i+2} \times \dots$$

Where U is an open set in $X_i$. All other sets in this product are the entire space $X_j$. Why is this the case? Why, for instance, doesn't $S_i$ contain this set:

$$\dots U_{i-3}\times U_{i-2} \times U_{i-1} \times U \times U_{i+1} \times U_{i+2} \times \dots$$

Where $U_j$ may or may not be open in $X_j$ for $j \neq i$?

Answer 1

But your first formula is exactly what $\pi_i^{-1}[U]$ means: the only condition for a point to be in that set, is that the $i$-th coordinate of that point is in $U$; all other coordinates are completely free.

It contains the other set as a subset, but topologies aren't closed under subsets, so that doesn't mean anything. It can be written as $\bigcap_{i \in I} \pi_i^{-1}[U_i]$, but that will be an infinite intersection, and topologies are only guaranteed to be closed under finite intersections.

Answer 2

You are misunderstanding $\pi_i^{-1}(U)$. This is the set which has $X_j$ in the $j^{th}$ slot (for $j\neq i$) and $U$ in the $i^{th}$ slot.

For example, in $\mathbb R^3$, one has $\pi_1^{-1}(A) = A\times{\mathbb R}\times{\mathbb R}$.

In other words, $\pi_i^{-1}(U)$ is the single set of all points in the product which project into $U$ under the action of $\pi_i$.