Sigma-Algebra Generated by Complements of Generators

Theorem
Let $\Sigma$ be a $\sigma$-algebra on a set $X$.

Let $\mathcal G$ be a generator for $\Sigma$.

Then:


 * $\mathcal{G}' := \left\{{X \setminus G: G \in \mathcal G}\right\}$

the set of relative complements of $\mathcal G$, is also a generator for $\Sigma$.

Proof
By definition of generator:


 * $\forall G \in \mathcal G: G \in \Sigma$

The third axiom for a $\sigma$-algebra thus ensures that:


 * $\forall G \in \mathcal G: X \setminus G \in \Sigma$

Therefore, by definition of generated $\sigma$-algebra, $\sigma \left({\mathcal{G}'}\right) \subseteq \Sigma$.

However, by the same argument (now applied to $\sigma \left({\mathcal{G}'}\right)$), also:


 * $\forall G \in \mathcal G: X \setminus \left({X \setminus G}\right) \in \sigma \left({\mathcal{G}'}\right)$

and by Set Difference with Set Difference and Intersection with Subset is Subset:


 * $X \setminus \left({X \setminus G}\right) = G \cap X = G$

Thus, it follows that $\Sigma = \sigma \left({\mathcal G}\right) \subseteq \sigma \left({\mathcal{G}'}\right)$.

Hence the result, by definition of set equality.