# Natural Basis of Product Topology/Lemma 3

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

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

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:

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

where:

$\ds B = \prod_{i \mathop \in I} U_i$
$J = \set{j \in I : U_i \ne X_i}$ is finite.

## Proof

Let $B \in \BB$.

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

where

for all $i \in I : U_i \in \tau_i$
for all but finitely many indices $i : U_i = X_i$

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

Then $J$ is a finite set and:

$\forall i \in I \setminus J : U_i = X_i$

For all $j \in J$, let:

$\pr_j^{-1} \sqbrk {U_j} = \ds \prod_{i \mathop \in I} V^j_i$

where:

$V^j_j = U_j$
$\forall i \ne j : V^j_i = X_i$

Then:

 $\ds \bigcap_{j \mathop \in J} \pr_j^{-1} \sqbrk {U_j}$ $=$ $\ds \bigcap_{j \mathop \in J} \paren {\prod_{i \mathop \in I} V^j_i}$ $\ds$ $=$ $\ds \prod_{i \mathop \in I} \paren {\bigcap_{j \mathop \in J} V^j_i}$ General Case of Cartesian Product of Intersections

To complete the proof it remains to show that:

$\forall i \in I : U_i = \bigcap_{j \mathop \in J} V^j_i$

Let $i \in I$.

### Case: $i \not \in J$

Let $i \not \in J$.

Then:

 $\ds \bigcap_{j \mathop \in J} V^j_i$ $=$ $\ds \bigcap_{j \mathop \in J} X_i$ as $\forall j \in J : i \ne j$ $\ds$ $=$ $\ds X_i$ Set Intersection is Idempotent $\ds$ $=$ $\ds U_i$ Definition of $J$

$\Box$

### Case: $i \in J$

Let $i \in J$.

Then:

 $\ds \bigcap_{j \mathop \in J} V^j_i$ $=$ $\ds V^i_i \cap \bigcap_{j \mathop \in J \mathop \setminus i} V^j_i$ $\ds$ $=$ $\ds U_i \cap \bigcap_{j \mathop \in J \mathop \setminus i} V^j_i$ Definition of $V^i_i$ $\ds$ $=$ $\ds U_i \cap \bigcap_{j \mathop \in J \mathop \setminus i} X_i$ Definition of $V^j_i$ when $j \ne i$ $\ds$ $=$ $\ds U_i \cap X_i$ Set Intersection is Idempotent $\ds$ $=$ $\ds U_i$ Intersection with Subset is Subset

$\Box$

Thus:

$\ds \forall i \in I : U_i = \bigcap_{j \mathop \in J} V^j_i$

The result follows.

$\blacksquare$