# Natural Basis of Product Topology/Lemma 1

## Lemma for Natural Basis of Product Topology

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

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

For each $i \in I$, let $\pr_i: X \to X_i$ denote the $i$th projection on $X$:

$\forall \family {x_j}_{j \mathop \in I} \in X: \map {\pr_i} {\family {x_j}_{j \mathop \in I} } = x_i$

Let $\SS = \set {\pr_i^{-1} \sqbrk U: i \in I, \, U \in \tau_i}$

Let $\BB$ be the set 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 $i : U_i = X_i$

Then:

$\SS \subseteq \BB$

## Proof

Let $i \in I, \, U \in \tau_i$.

Then:

 $\ds \pr_i^{-1} \sqbrk U$ $=$ $\ds \set {\family{x_j} : x_i \in U, \forall j \ne i : x_j \in X_j}$ $\ds$ $=$ $\ds \prod_{i \mathop \in I} V_i$

where:

$V_i = U$
$\forall j \ne i : V_j = X_j$

By definition of $\BB$:

$\pr_i^{-1} \sqbrk U \in \BB$

Since $i$ and $U$ were arbitrary:

$\SS \subseteq \BB$

$\blacksquare$