[proofplan] The [product topology](/page/Product%20Topology) on $\prod_{k=1}^\infty X_k$ is generated by cylinder sets of the form $\pi_{k}^{-1}(U_k)$, and hence by finite intersections of such sets. Starting from countable bases $\mathcal{B}_k$ for each factor, we form all finite intersections of sets $\pi_k^{-1}(B)$ with $B \in \mathcal{B}_k$. The resulting collection is countable (a countable union of finite products of countable sets) and is a base for the product topology. [/proofplan] [step:Construct the candidate base from finite intersections of cylinder sets] For each $k \in \mathbb{N}$, let $\mathcal{B}_k = \{B_{k,1}, B_{k,2}, \ldots\}$ be a countable base for $(X_k, \tau_k)$. Define the collection \begin{align*} \mathcal{C} := \left\{ \bigcap_{k \in F} \pi_k^{-1}(B_{k, n_k}) : F \subset \mathbb{N} \text{ finite and nonempty},\; n_k \in \mathbb{N} \text{ for each } k \in F \right\}, \end{align*} where $\pi_k: \prod_{j=1}^\infty X_j \to X_k$ denotes the $k$-th projection. Each member of $\mathcal{C}$ has the form \begin{align*} \bigcap_{k \in F} \pi_k^{-1}(B_{k, n_k}) = \prod_{k=1}^\infty V_k, \quad \text{where } V_k = \begin{cases} B_{k, n_k} & \text{if } k \in F, \\ X_k & \text{if } k \notin F. \end{cases} \end{align*} [guided] The [product topology](/page/Product%20Topology) on $X = \prod_{k=1}^\infty X_k$ is defined as the coarsest topology making every projection $\pi_k: X \to X_k$ continuous. A subbasis for this topology consists of all sets $\pi_k^{-1}(U_k)$ with $k \in \mathbb{N}$ and $U_k \in \tau_k$. A base is then formed by taking all finite intersections of subbasis elements: \begin{align*} \pi_{k_1}^{-1}(U_{k_1}) \cap \pi_{k_2}^{-1}(U_{k_2}) \cap \cdots \cap \pi_{k_m}^{-1}(U_{k_m}). \end{align*} To build a *countable* base, we replace each $U_{k_i} \in \tau_{k_i}$ with a basis element $B_{k_i, n_{k_i}} \in \mathcal{B}_{k_i}$. The resulting collection is \begin{align*} \mathcal{C} := \left\{ \bigcap_{k \in F} \pi_k^{-1}(B_{k, n_k}) : F \subset \mathbb{N} \text{ finite and nonempty},\; n_k \in \mathbb{N} \text{ for each } k \in F \right\}. \end{align*} Each member of $\mathcal{C}$ is a "box" that constrains finitely many coordinates to lie in specified basis elements and leaves all other coordinates unconstrained. [/guided] [/step] [step:Verify that $\mathcal{C}$ is countable] We show that $\mathcal{C}$ is countable by expressing it as a countable union of countable sets. For each finite set $F = \{k_1, \ldots, k_m\} \subset \mathbb{N}$, the family of all intersections $\bigcap_{k \in F} \pi_k^{-1}(B_{k, n_k})$ is indexed by $(n_{k_1}, \ldots, n_{k_m}) \in \mathbb{N}^m$, which is countable (a finite product of countable sets is countable). The collection of all finite subsets of $\mathbb{N}$ is itself countable. Therefore \begin{align*} \mathcal{C} = \bigcup_{\substack{F \subset \mathbb{N} \\ F \text{ finite}}} \left\{ \bigcap_{k \in F} \pi_k^{-1}(B_{k, n_k}) : n_k \in \mathbb{N} \text{ for } k \in F \right\} \end{align*} is a countable union of countable sets, hence countable by [Countable Union of Countable Sets](/theorems/755). [guided] We must verify that $\mathcal{C}$ is countable despite being parametrised by a finite set $F$ and a tuple of natural numbers. Fix a finite set $F = \{k_1, \ldots, k_m\} \subset \mathbb{N}$. The intersections indexed by $F$ are parametrised by $(n_{k_1}, \ldots, n_{k_m}) \in \mathbb{N}^m$. Since $\mathbb{N}^m$ is countable (induction: $\mathbb{N}^1 = \mathbb{N}$ is countable, and if $\mathbb{N}^m$ is countable then $\mathbb{N}^{m+1} = \mathbb{N}^m \times \mathbb{N}$ is a product of two countable sets, hence countable), the sub-family associated to each fixed $F$ is countable. The collection of all finite subsets of $\mathbb{N}$ is countable: each finite subset $F \subset \mathbb{N}$ is contained in $\{1, \ldots, N\}$ for some $N$, and $\{1, \ldots, N\}$ has finitely many subsets, so the finite subsets of $\mathbb{N}$ form $\bigcup_{N=1}^\infty \mathcal{P}(\{1, \ldots, N\})$, a countable union of finite sets. Therefore $\mathcal{C}$ is a countable union of countable sets. By [Countable Union of Countable Sets](/theorems/755), the collection $\mathcal{C}$ is countable. [/guided] [/step] [step:Verify that $\mathcal{C}$ is a base for the product topology] We must show that every open set in the product topology is a union of members of $\mathcal{C}$. Since the sets $\pi_k^{-1}(U_k)$ with $U_k \in \tau_k$ form a subbasis, it suffices to show that every finite intersection of subbasis elements is a union of members of $\mathcal{C}$. Let $W = \bigcap_{k \in F} \pi_k^{-1}(U_k)$ be such a finite intersection, where $F \subset \mathbb{N}$ is finite and $U_k \in \tau_k$ for each $k \in F$. Since $\mathcal{B}_k$ is a base for $\tau_k$, each $U_k$ is a union of basis elements: \begin{align*} U_k = \bigcup_{n \in I_k} B_{k, n} \end{align*} for some index set $I_k \subset \mathbb{N}$. Therefore \begin{align*} W = \bigcap_{k \in F} \pi_k^{-1}\!\left(\bigcup_{n \in I_k} B_{k,n}\right) = \bigcap_{k \in F} \bigcup_{n \in I_k} \pi_k^{-1}(B_{k,n}). \end{align*} Distributing the intersection over the unions (since the intersection is finite): \begin{align*} W = \bigcup_{(n_k)_{k \in F} \in \prod_{k \in F} I_k} \; \bigcap_{k \in F} \pi_k^{-1}(B_{k, n_k}). \end{align*} Each set $\bigcap_{k \in F} \pi_k^{-1}(B_{k, n_k})$ belongs to $\mathcal{C}$, so $W$ is a union of members of $\mathcal{C}$. Since every basic open set in the product topology (i.e., every finite intersection of subbasis elements) is a union of members of $\mathcal{C}$, and every open set is a union of basic open sets, it follows that every open set is a union of members of $\mathcal{C}$. Therefore $\mathcal{C}$ is a countable base for the product topology, and $\prod_{k=1}^\infty X_k$ is second-countable. [guided] To confirm that $\mathcal{C}$ is a base, we verify the defining property: every open set in the product topology is a union of members of $\mathcal{C}$. Recall that a subbasis for the product topology consists of all sets $\pi_k^{-1}(U_k)$ with $k \in \mathbb{N}$ and $U_k \in \tau_k$. The standard base for the product topology consists of all finite intersections of such subbasis elements: \begin{align*} W = \pi_{k_1}^{-1}(U_{k_1}) \cap \cdots \cap \pi_{k_m}^{-1}(U_{k_m}). \end{align*} Every open set is a union of these basic open sets, so it suffices to show that each such $W$ is itself a union of members of $\mathcal{C}$. Fix $W = \bigcap_{k \in F} \pi_k^{-1}(U_k)$ with $F = \{k_1, \ldots, k_m\}$ finite and $U_k \in \tau_k$. Since $\mathcal{B}_k$ is a base for $\tau_k$, we write $U_k = \bigcup_{n \in I_k} B_{k,n}$ for some $I_k \subset \mathbb{N}$. Preimages commute with unions: $\pi_k^{-1}(U_k) = \bigcup_{n \in I_k} \pi_k^{-1}(B_{k,n})$. Substituting: \begin{align*} W &= \bigcap_{k \in F} \bigcup_{n \in I_k} \pi_k^{-1}(B_{k,n}). \end{align*} We now distribute the finite intersection over the unions. Since $F$ is finite, the distributive law for intersections over unions applies (this is where finiteness of $F$ is essential --- for an infinite intersection, the distributive law would require the Axiom of Choice and produce an uncountable union). The result is: \begin{align*} W = \bigcup_{(n_k)_{k \in F} \in \prod_{k \in F} I_k} \; \bigcap_{k \in F} \pi_k^{-1}(B_{k, n_k}). \end{align*} Each intersection $\bigcap_{k \in F} \pi_k^{-1}(B_{k, n_k})$ belongs to $\mathcal{C}$ by definition. So $W$ is a union of members of $\mathcal{C}$. Since every basic product-open set is a union of members of $\mathcal{C}$, and every product-open set is a union of basic product-open sets, transitivity of "is a union of" gives that every product-open set is a union of members of $\mathcal{C}$. The collection $\mathcal{C}$ is therefore a countable base for the product topology. [/guided] [/step]