Can an infinite-dimensional product space have the defining subbase?

Question

The following is an excerpt from Schaum's Outlines: General Topology

The class of subsets of a product space $X= \prod X_i$ of the form $$\pi_{j_0}^{-1}[G_{j_0}]=\prod\{X_i: i \neq j_0\} \times G_{j_0}$$ where $G_{j_0}$ is an open subset of the coordinate space $X_{j_0}$ is a subbase and is called the defining subbase for the product topology.

This means that we take intersections of the $\pi_{j_0}^{-1}$'s to form a base for the product topology. If the product space $X = \prod X_i$ is infinite-dimensional, then wouldn't we need an infinite number of intersections, thus breaking the definition of the defining subbase? I suspect that the product space must be finite-dimensional for the defining subbase to yield open subsets after taking intersections.

Answer 1

In the product topology, you are only allowed to have all but a finite number of the coordinates be anything but the whole space. If you want to allow arbitrary cross products of open sets to be open, that is called the box topology. We don't use that one because it doesn't preserve the properties we would like to preserve.

Since we are only allowed a finite number of coordinates with non-whole space open sets, this works as as a sub-basis as a finite intersection generates our basic open sets.

Answer 2

The family $$\{\pi_{j}^{-1}[O_j]\mid j \in J, O \subseteq X_j \text{ open }\}$$ always forms a subbase for $\prod_{j \in J} X_j$, regardless of the size of $J$. This is just the definition of the product topology and follows also from its description as being the smallest topology that makes all maps $\pi_j$ continuous.

To get the generated base, as always for any subbase, we indeed take finite intersections of subbase elements and so a standard basic set can only "depend" on finitely many coordinates $j$, indeed. This might seem strange but is an essential feature in the proving of properties of infinite products: open sets are always "large".