Countable Product of Second-Countable Spaces is Second-Countable

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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 $\BB_\alpha = \set{N^\alpha_i : i \in \N}$ be a countable basis for $\tau_\alpha$ for each $\alpha \in I$.

Let $\pr_\alpha$ denote the projection of $S$ onto $S_\alpha$.


Let $\LL_\alpha = \set{\map {\pr_\alpha^{-1}} {N^\alpha_i}: N^\alpha_i \in B_\alpha}$.

Let $\KK_J = \displaystyle \set{\bigcap_{\alpha \mathop \in J} L_\alpha : \forall \alpha \in J: L_\alpha \in \LL_\alpha}$ for $J \subset I$, $\size{J} < \infty$.

From Product Space Basis Induced from Factor Space Bases:

$\displaystyle \BB = \bigcup_{J \mathop \subset I \mathop , \size{J} < \infty} \KK_J$ forms a basis of the product space.


Now, each of the $\LL_\alpha$'s is countable.

Since the $\KK_J$'s can be identified with a finite product of countable sets, they are each countable

From Countable Union of Countable Sets is Countable, $\BB$ forms a countable basis of $S$.

$\blacksquare$


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