Confusion about the definition of open sets in a product topology. Munkres' "Topology" says:

Say $X=X_1\times X_2\times\dots$ The subbasis for this topology is $X_1\times X_2\times\dots A_i\times X_{i+1}\dots$. 

Say we want the open set $A_1\times A_2\times\dots$. We will then have to have an infinite intersection of subbasis sets. Is the infinite intersection of subbasis sets an open set?
Thanks!  
 A: In general, when generating a topology from a subbasis $\mathcal{S}$, the first step is to construct a basis $\mathcal{B}$, and the way to do this is to let $\mathcal{B}$ consist of all finite intersections of sets in $\mathcal{S}$.  
The basic idea is that you want the eventual topology generated to be the coarsest topology such that every set in $\mathcal{S}$ is open: since open sets are closed under finite intersections, we know that every finite intersection of sets in $\mathcal{S}$ must be open in the eventual topology. It turns out that by forming $\mathcal{B}$ in the manner above we do get a basis, and we know how to generate a topology from a basis: the open sets are arbitrary unions from the basis, and this is the coarsest topology in which all the basis sets are open.
If you do instead take arbitrary intersections of the form you describe, you will end up with what is called the box topology.  This topology differs greatly from the product topology.  For instance, if you start with infinitely many copies of the two point discrete space $X_i = \{ 0 , 1 \}$, the product topology $\prod_{i \in \mathbb{N}} X_i$ is a compact space (actually, this space is homeomorphic to the Cantor set).  However, if you take the box topology, $\Box_{i \in \mathbb{N}} X_i$  you end with with a discrete space of size continuum, which is not compact.
The product topology also has the following very nice property which fails for box topologies.

Suppose that $X_i$ ($i \in I$) are topological spaces, $Y$ is a topological space, and for each $i \in I$ we have a function $f_i : Y \to X_i$.  Then the function $f : Y \to \prod_{i \in I} X_i$ defined by $f(y) = \langle f_i(y) \rangle_{i \in I}$ is continuous iff each $f_i$ is.

