Sigma-Algebra Generated by Complements of Generators
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Theorem
Let $\Sigma$ be a $\sigma$-algebra on a set $X$.
Let $\GG$ be a generator for $\Sigma$.
Then:
- $\GG' := \set {X \setminus G: G \in \GG}$
the set of relative complements of $\GG$, is also a generator for $\Sigma$.
Proof
| \(\ds \forall G \in \GG: \, \) | \(\ds G\) | \(\in\) | \(\ds \Sigma\) | Definition of Generator of Sigma-Algebra | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \forall G \in \GG: \, \) | \(\ds X \setminus G\) | \(\in\) | \(\ds \Sigma\) | Definition of Sigma-Algebra: $\text{(SA3)}$ | |||||||||
| \(\ds \leadsto \ \ \) | \(\ds \forall G \in \GG': \, \) | \(\ds G\) | \(\in\) | \(\ds \Sigma\) | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \map \sigma {\GG'}\) | \(\subseteq\) | \(\ds \Sigma\) | Definition of Sigma-Algebra Generated by Collection of Subsets | ||||||||||
| \(\ds \forall G \in \GG': \, \) | \(\ds G\) | \(\in\) | \(\ds \map \sigma {\GG'}\) | Definition of Sigma-Algebra Generated by Collection of Subsets | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \forall G \in \GG: \, \) | \(\ds X \setminus G\) | \(\in\) | \(\ds \map \sigma {\GG'}\) | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \forall G \in \GG: \, \) | \(\ds X \setminus \paren {X \setminus G}\) | \(\in\) | \(\ds \map \sigma {\GG'}\) | Definition of Sigma-Algebra: $\text{(SA3)}$ | |||||||||
| \(\ds \leadsto \ \ \) | \(\ds \forall G \in \GG: \, \) | \(\ds X \cap G\) | \(\in\) | \(\ds \map \sigma {\GG'}\) | Set Difference with Set Difference | |||||||||
| \(\ds \leadsto \ \ \) | \(\ds \forall G \in \GG: \, \) | \(\ds G\) | \(\in\) | \(\ds \map \sigma {\GG'}\) | Intersection with Subset is Subset | |||||||||
| \(\ds \leadsto \ \ \) | \(\ds \Sigma\) | \(\subseteq\) | \(\ds \map \sigma {\GG'}\) | Definition of Generator of Sigma-Algebra | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \Sigma\) | \(=\) | \(\ds \map \sigma {\GG'}\) | Definition 2 of Set Equality |
Hence the result.
$\blacksquare$