Natural Basis of Product Topology/Lemma 3
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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 $\family {X_i}_{i \mathop \in I}$:
- $\ds X := \prod_{i \mathop \in I} X_i$
Let $\BB$ be the set of cartesian products of the form $\ds \prod_{i \mathop \in I} U_i$ where:
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
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.
$\blacksquare$