Natural Basis of Product Topology/Lemma 1
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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:
- $\SS \subseteq \BB$
$\blacksquare$