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 $\struct {S, \tau} = \ds \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ 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 $\ds \prod_{\alpha \mathop \in I} U_\alpha$ 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 Product Topology, the set $\BB'$ of cartesian products of the form $\ds \prod_{\alpha \mathop \in I} V_\alpha$ 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 $V = \ds \prod_{\alpha \mathop \in I} V_\alpha \in \BB'$:
- $x \in V \subseteq W$
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 $U = \ds \prod_{\alpha \mathop \in I} U_\alpha$.
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.
$\blacksquare$