# Show that the set $\mathcal{S}=\{\bigcup_{i \in I} A_i \subseteq X \mid I \subseteq \mathcal{A}\}$ of unions of partitions of $X$ is a sigma-algebra.

Let $$X$$ be a set and $$\{A_\alpha\}_{\alpha \in \mathcal{A}}$$ be a partition of $$X$$. Define $$\mathcal{S}=\{\bigcup_{i \in I} A_i \subseteq X \mid I \subseteq \mathcal{A}\}$$. Show that $$\mathcal{S}$$ is a sigma-algebra

To show this I pick $$F \in \mathcal{S}$$. Now $$F = \bigcup_{i} A_i$$. So it’s union of some elements that partition $$X$$. Now to show that this is a sigma-algebra I need to show that $$\emptyset \in \mathcal{S}$$, $$F^c \in \mathcal{S}$$ and if I take more elements $$F_1, F_2, \dots$$, then $$\bigcup_{k=1}^\infty F_k \in \mathcal{S}$$.

The complement is $$F^c = \left(\bigcup_{i}A_i \right)^c = \bigcap_{i} A_i^c$$, but how do I know that the intersection of the complements is in $$\mathcal{S}$$?

Also if $$F_k \in \mathcal{S}$$, then $$\bigcup_{k} F_k = \bigcup_{k} \left( \bigcup_{i} A_i \right)$$, but this also doesn’t seem to be easy to work with? Is there another way to show that a set is sigma-algebra?

$$\left(\bigcup_{i \in I} A_i\right)^c = \bigcup_{i \notin I} A_i$$
Also note that $$\bigcup_{n=1}^\infty \bigcup_{i \in I_n} A_i = \bigcup_{i \in \bigcup_{n=1}^\infty I_n} A_i$$
• Could the first equality hold since $I \subseteq \mathcal{A}$? I’m trying to figure out what it is saying. Sep 25, 2021 at 14:54
• Yes with $i \notin I$ I mean $i \in I^c = \{x \in \mathcal{A}: x \notin I\}$ Sep 25, 2021 at 14:55