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 $\family {X_i}_{i \mathop \in I}$:

$\ds X := \prod_{i \mathop \in I} X_i$

Then the natural basis on $X$ is the set $\BB$ of cartesian products of the form $\ds \prod_{i \mathop \in I} U_i$ where:

for all $i \in I : U_i \in \tau_i$

for all but finitely many indices $i : U_i = X_i$

Finite Product

Let $n \in \N$.

For all $k \in \set {1, \ldots, n}$, let $\struct {X_k, \tau_k}$ be topological spaces.

Let $\ds X = \prod_{k \mathop = 1}^n X_k$ be the (finite) cartesian product of $X_1, \ldots, X_n$.

Then the natural basis on $X$ is:

$\BB = \set {\ds \prod_{k \mathop = 1}^n U_k : \forall k : U_k \in \tau_k}$

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 1

$\SS \subseteq \BB$

$\Box$

Lemma 2

$\forall B_1, B_2 \in \BB : B_1 \cap B_2 \in \BB$

$\Box$

From Synthetic Basis formed from Synthetic Sub-Basis:

$\NN = \set {\bigcap \FF: \FF \subseteq \SS, \, \FF \text{ is finite} }$

From Lemma 1 and Lemma 2 it follows that:

$\NN \subseteq \BB$.

Lemma 3

$\ds \forall B \in \BB : B = \bigcap_{j \mathop \in J} \pr_j^{-1} \sqbrk {U_j}$

where:

$\ds B = \prod_{i \mathop \in I} U_i$

$J = \set{j \in I : U_i \ne X_i}$ is finite.

$\Box$

From Lemma 3 and Synthetic Basis formed from Synthetic Sub-Basis:

$\BB \subseteq \NN$

The result follows from the definition of set equality.

$\blacksquare$