Countable Product of Second-Countable Spaces is Second-Countable

Theorem
Let $I$ be an indexing set with countable cardinality.

Let $\family{\struct{S_\alpha, \tau_\alpha}}_{\alpha \mathop \in I}$ be a family of topological spaces indexed by $I$.

Let $\displaystyle \struct{S, \tau} = \prod_{\alpha \mathop \in I} \struct{S_\alpha, \tau_\alpha}$ be the product space of $\family{\struct{S_\alpha, \tau_\alpha}}_{\alpha \mathop \in I}$.

Let each of $\struct{S_\alpha, \tau_\alpha}$ be second-countable.

Then $\struct{S, \tau}$ is also second-countable.

Proof
Let $X$ and $Y$ be second-countable spaces.

Let $\set{B_{x_i}}_{i \mathop \in \N}$ and $\set{B_{y_i}}_{i \mathop \in \N}$ be bases for their topologies.

By Cartesian Product of Countable Sets is Countable, the basis of $X \times Y$, namely $\set{B_{x_i} \times B_{y_j} }_{i, j \mathop \in \N}$, is also countable.

Let $\displaystyle \prod_{n \mathop \in \N} X_n$ be the product of topological spaces $\struct{X_n, \tau_n}$.

Let $B_n$ be a countable basis for $\tau_n$.

Then let $L_i = \set{\map {\pi_i^{-1}} {N_i}: N_i \in B_i}$.

Let $K_J = \displaystyle \bigcap_{j \mathop \in J} L_j$ for $J \subset \N$, $\size{J} < \infty$.

Then $\displaystyle B = \bigcup_{J \mathop \subset \N} K_J$ forms a basis of the product space.

Now, since the $K_J$'s can be identified with a finite product of countable sets, they are each countable

From Countable Union of Countable Sets is Countable, $B$ forms a countable basis of $\displaystyle \prod_{n \mathop \in \N} X_n$.