Product topology for $\mathbb{R}^{\mathbb{N}}$ and generating subbasis

Question

The product topology on $\mathbb{R}^{\mathbb{N}}$ is defined to be the topology generated by the subbasis:

$$S_n := \{ \pi_n ^{-1} (U) : U \subseteq \mathbb{R} \text{ open}\} $$

where $\pi_n$ are the projection maps from the product space to the component space. Now I learned the product topology as the topology with least number of open sets such that the projection maps are continuous (although I never understood the "why" for this definition). However, I don't quite get why the subbasis above would generate the product topology (I actually don't even know how can a subbasis "directly" generate a topology since I always though that a topology is generated by the union of basis elements, so first we would need to construct the basis from the subbasis and then get the topology).

Answer 1

The product topology on $\mathbb{R}^{\mathbb{N}}$ is defined to be the topology generated by the subbasis: $$S_n := \{ \pi_n ^{-1} (U) : U \subseteq \mathbb{R} \text{ open}\} $$

Although you're saying the correct thing, this notation is not correct. You want to take into the subbasis the sets $\pi_n ^{-1}(U)$ for all $n$'s. So the right notation would be $$\mathcal S := \{ \pi_n ^{-1} (U) : U \subseteq \mathbb{R} \text{ open}, n\in\mathbb N\}.$$ (The main thing I object to in the notation you wrote down is that it looks as if the subbasis depends on $n$.)

the topology with least number of open sets such that the projection maps are continuous (although I never understood the "why" for this definition)

Well, if you want $\pi_n$ to be continuous, then $\pi_n^{-1}(U)$ has to be open for every open set $U$. This is basically the definition of continuity.

So you want the topology which contains all such sets, i.e., a topology which contains $\mathcal S$. Saying that $\mathcal S$ is a subbasis is simply saying that you are taking the smallest topology which contains $\mathcal S$.

I actually don't even know how can a subbasis "directly" generate a topology since I always though that a topology is generated by the union of basis elements

If you have a subbasis $\mathcal S$, then the corresponding basis is obtained as set of all finite intersections of elements from $\mathcal S$. $$\mathcal B=\{S_1\cap \dots \cap S_k; k\in\mathbb N, S_i\in\mathcal S \text{ for }i=1,2,\dots,k\}.$$ Here, if you work with projections, the basis for the product topology on $\prod\limits_{i\in I} X_i$ can be written simply in this way $$\mathcal B=\{\pi_{i_1}^{-1}(U_1) \cap \pi_{i_2}^{-1}(U_2) \cap \dots \cap \pi_{i_k}^{-1}(U_k); k\in\mathbb N, U_j\text{ is open in }X_{i_j}\text{ for }i=1,2,\dots,k\}.$$

In you case the index set is $\mathbb N$ and each $X_n$ is equal to $\mathbb R$. So you're dealing with sequences of real numbers. A reasonable ways to visualize the basic open sets is that you have chosen finitely many coordinates $i_1,\dots,i_k$ and for each of those coordinates you've chosen some open sets. And you're taking only the sequences where the $i_k$-th term belongs to the chosen open set $U_k$. (On coordinates outside of the set $\{i_1,\dots,i_k\}$, the sequences belonging to this particular basic open set can behave arbitrarily.)

Still, even if you prefer to use basis to generating topologies, it might be useful to work (at least sometimes) with subbasis. For example, here in the case of product topology, you can see that the description using subbasis is much simpler.