I'm new to this concept of a $\sigma$-algebra generated by a collection of subsets.
Let $\Omega = \{a, b, c, d\}$ and $$\begin{align}
&\mathcal{F}_1 = \{\Omega, \emptyset, \{a\}\} \\
&\mathcal{F}_2 = \{\Omega, \emptyset, \{a\}, \{b, c, d\}\}\text{.}
\end{align}$$
I wish to show that
$$\sigma\langle \mathcal{F}_1 \rangle = \bigcap_{\mathcal{F} \in \mathcal{I}(\mathcal{F}_1)}\mathcal{F} = \mathcal{F}_2$$
where $\mathcal{I}(\mathcal{F}_1) = \{\mathcal{F}: \mathcal{F}_1 \subset \mathcal{F} \text{ and }\mathcal{F} \text{ a }\sigma\text{-algebra on }\Omega \}$.
I know immediately from this that any $\sigma$-algebra $\mathcal{F}$ must, at the very least, contain elements of $\mathcal{F}_1$: $$\{\Omega, \emptyset, \{a\}\}$$
but other than literally listing every possible collection of subsets of $\Omega$ and checking which are $\sigma$-algebras, and then intersecting such sets, I don't see what's an efficient way to do this.
Is my way of thinking correctly or is there a quicker way?
Best Answer
Yes. but one quick shortcut is that the complements of the sets are in the sigma algebra as well, so you can easily see $\{b,c,d\}$ must be in $\sigma\langle\mathcal{F}_1\rangle$.