Natural Basis of Product Topology

Theorem
Let $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be an indexed family of topological spaces where $I$ is an arbitrary index set.

Let $X$ be the cartesian product of $\family {X_i}_{i \mathop \in I}$:
 * $\displaystyle X := \prod_{i \mathop \in I} X_i$

Then the natural basis on $X$ is the set $\BB$ of cartesian products of the form $\displaystyle \prod_{i \mathop \in I} U_i$ where:
 * for all $i \in I : U_i \in \tau_i$
 * for all but finitely many indices $i : U_i = X_i$

Proof
Let $\NN$ denote the natural basis on $X$.

By definition of the natural basis, $\NN$ is the basis generated by:
 * $\SS = \set {\pr_i^{-1} \sqbrk U: i \in I, \, U \in \tau_i}$

where for each $i \in I$, $\pr_i: X \to X_i$ denotes the $i$th projection on $X$:
 * $\forall \family {x_j}_{j \mathop \in I} \in X: \map {\pr_i} {\family {x_j}_{j \mathop \in I} } = x_i$

Lemma 2
From Synthetic Basis formed from Synthetic Sub-Basis:
 * $\NN = \set{\bigcap \mathcal F: \mathcal F \subseteq \mathcal S, \, \mathcal F \text{ is finite}}$

From Lemma 1 and Lemma 2 it follows that:
 * $\NN \subseteq \BB$.

Lemma 3
From Lemma 3 and Synthetic Basis formed from Synthetic Sub-Basis:
 * $\BB \subseteq \NN$

The result follows from the definition of set equality.