Product Space Basis Induced from Factor Space Bases

Theorem
Let $\family {\struct{S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be an indexed family of topological spaces for $\alpha$ in some indexing set $I$.

Let $\BB_\alpha$ be a basis for the topology $\tau_\alpha$ for each $\alpha \in I$.

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

Let $\BB$ be the set of all cartesian products of the form $\displaystyle \prod_{\alpha \mathop \in I} U_\alpha$ where:
 * for all but finitely many indices $\alpha : U_\alpha = S_\alpha$
 * for all $\alpha \in I : U_\alpha \ne S_\alpha \implies U_\alpha \in \BB_\alpha$

Then $\BB$ is a basis for the topology on the product space $\struct{S, \tau}$.

Proof
Let $x = \family {x_\alpha}_{\alpha \mathop \in I} \in S$.

Let $W \in \tau$ such that $x \in W$.

From Natural Basis of Tychonoff Topology, the set $\BB'$ of cartesian products of the form $\displaystyle \prod_{\alpha \mathop \in I} V_\alpha$ where:
 * for all $\alpha \in I : V_\alpha \in \tau_\alpha$
 * for all but finitely many indices $\alpha : V_\alpha = X_\alpha$

is a basis for $\tau$.

Therefore, there exists $V = \displaystyle \prod_{\alpha \mathop \in I} V_\alpha \in \BB'$:
 * $x \in V \subseteq W$

Let $J = \set{\alpha \in I : V_\alpha \ne S_\alpha}$.

By definition of $\BB'$, $J$ is finite.

Now for all $\alpha \in J$, $x_\alpha \in V_\alpha$ and $V_\alpha \in \tau_\alpha$.

Since $\BB_\alpha$ is a basis for $\tau_\alpha$ then for all $\alpha \in J$ there exists $N_\alpha \in \BB_\alpha$:
 * $x_\alpha \in N_\alpha \subseteq V_\alpha$

For all $\alpha \in I$, let:
 * $U_\alpha = \begin {cases} N_\alpha & : \alpha \in J \\ S_\alpha & : \alpha \notin J \end {cases}$

Let $U = \displaystyle \prod_{\alpha \mathop \in I} U_\alpha$.

For $\alpha \in J$:
 * $U_\alpha = N_\alpha \subseteq V_\alpha$
 * $x_\alpha \in N_\alpha = U_\alpha$
 * $U_\alpha = N_\alpha \in \BB_\alpha$

For $\alpha \notin J$:
 * $U_\alpha = S_\alpha = V_\alpha$
 * $x_\alpha \in S_\alpha = U_\alpha$

Thus:
 * $x \in U$
 * $U \subseteq V \subseteq W$

By the definition of $\BB$ and because $J$ is finite:
 * $U \in \BB$.

The result follows.