Understanding the proof that countable product of second countable space is second countable under the product topology This is a very short proof but I dont understand the reason the proof was done this way and why we have a countable basis for the product topology.
"A countable product of second countable spaces is second countable."
Suppose that for each $n$ ,$X_n$ is second countable, say $\mathcal{B}_n$ is a countable basis. Then $$\{\prod_n B_n : B_n = X_n \text{ for all but finitely many } n; \text{ if } B_n \neq X_n, \text{ then } B_n \in \mathcal{B}_n \}$$ is a countable basis for $X = \prod_n X_n$, so  $X$ is second countable.
 A: The proof consists of a few things:
Note that all members of $\mathcal{B}$ are product open sets (of the basic kind: infinite cubes where all but finitely many factors (“sides”) are the whole space in that factor space $X_n$).
The collection $\mathcal{B}$ is countable: to construct an element we make finitely many independent choices from countable sets: first a finite subset of $\Bbb N$: the coordinates that are allowed to be non-trivial (there are countably many finite subsets for a countable set), and next for such a subset $F$ a member $B_n$ from the countable set $\mathcal{B}_n$ for all $n \in F$, and then our member from $\mathcal{B}$ is determined. Underlying this is theorems like the countable union of countable sets is countable and finite products of countable sets is countable. It’s basically set theory.
The collection $\mathcal{B}$ is a base for the product topology. For this it suffices to show that for each product open $O$ and $x \in O$ we can find a member of the base containing $x$ sitting inside $O$. To this end we first find a standard basic open subset $U:=\prod_{n \in \Bbb N} U_n$ such that $x \in U \subseteq O$ and being basic open means there is a finite subset $F$ such that for $n \notin F$, we have $U_n = X_n$ and for $n \in F$ we have that $U_n \subseteq X_n$ is open. So for those $n \in F$: as $x_n \in U_n$ and $\mathcal{B}_n$ is a base for $X_n$ we find $B_n \in \mathcal{B}_n$ such that $x_n \in B_n \subseteq U_n$, and then we define $B = \prod_n B_n$ by also defining $B_n = X_n$ for $n \notin F$. Then $B \in \mathcal{B}$, and $$x \in B =\prod_n B_n \subseteq \prod_n U_n = U \subseteq O$$ as required.
A: Let $\mathcal{B}_{n}:=\left\{N_{i}^{n}: i \in \mathbb{N}\right\}$ be a countable basis for $\tau_{n}$ for each $n \in I$.
Let $\mathrm{\pi}_{n}$ denote the projection of $X$ onto $X_{n}$.
Let $\mathcal{L}_{n}:=\left\{\operatorname{\pi}_{n}^{-1}\left(N_{i}^{n}\right): N_{i}^{n} \in B_{n}\right\}$.
Let $\mathcal{K}_{J}=\left\{\bigcap_{n \in J} L_{n}: \forall n\in J \text{ we have } L_{n} \in \mathcal{L}_{n}\right\}$ for $J \subset I,|J|<\infty$
From the definiton of basis of the product topology,
$\mathcal{B}=\bigcup_{J \subset I,|J|<\infty} \mathcal{K}_{J}$ is a basis of the product space.
Now, each of the $\mathcal{L}_{n}$'s is countable.
Since the $\mathcal{K}_{J}$'s can be identified with a finite product of countable sets, they are countable.
Since countable union of countable sets is countable, $\mathcal{B}$ forms a countable basis of $X$.
