Product Space Local Basis Induced from Factor Spaces Local 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 $\displaystyle \struct{S, \tau} = \displaystyle \prod_{\alpha \mathop \in I} \struct{S_\alpha, \tau_\alpha}$ be the product space of $\family{\struct{S_\alpha, \tau_\alpha}}_{\alpha \mathop \in I}$.

Let $x = \family{x_\alpha} \in S$.

Let $\BB_\alpha$ be a local basis for $x_\alpha$ in the topological space $\struct{S_\alpha, \tau_\alpha}$ for each $\alpha \in I$.

Let $\BB_x$ 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 $\alpha \in I : U_\alpha \neq S_\alpha \implies U_\alpha \in \BB_\alpha$

Then $\BB_x$ is a local basis for $x$ in the product space $\struct{S, \tau}$.

Proof
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 $J = \set{\alpha \in I : V_\alpha \neq S_\alpha}$.

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 local 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 $\BB_x$ and since $J$ is finite:
 * $U \in \BB_x$.

The result follows.