Natural Basis of Product Topology/Lemma 3

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 $\family {X_i}_{i \mathop \in I}$:
 * $\displaystyle X := \prod_{i \mathop \in I} X_i$

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

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

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

Proof
Let $B \in \BB$.

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

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

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

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

Then:

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:

Case: $i \in J$
Let $i \in J$.

Then:

Thus:


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

The result follows.