$\sigma$-algebra generated by a subset 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?
 A: 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 $\sigma\langle\mathcal{F}_1\rangle$.
A: Here are some more general statements that imply the desired equation. We assume the following without proof.  


*

*The intersection of $\sigma$-algebras is again a $\sigma$-algebra.

*A set of the form $\lbrace \emptyset, \Omega, A, A^c \rbrace$ is a $\sigma$-algebra.


Let $\mathscr{G}$ (particular case $\mathcal{F}_1$) be any set of subsets of $\Omega$. $\mathscr{A}:=\bigcap_{\mathcal{F} \in \mathcal{I}(\mathscr{G})}\mathcal{F}$ is nonempty because the powerset $\mathscr{P}(\Omega)$ has $\mathscr{P}(\Omega) \in \mathcal{I}(\mathscr{G})$. It follows that $\mathscr{G} \subset \mathscr{A}$ and that, because of assumption 1, $\mathscr{A}$ is a $\sigma$-algebra. If $\mathscr{A}'$ is another $\sigma$-algebra with $\mathscr{G} \subset \mathscr{A}'$, then, because $\mathscr{A}'$ is one of the sets over which the intersection is taken, we have $\mathscr{A} \subset \mathscr{A}'$. So $\mathscr{A}$ satisfies the definition of the smallest $\sigma$-algebra generated by $\mathscr{G}$, i.e. $\sigma(\mathscr{G}) = \mathscr{A}$.
We conclude $\sigma(\mathcal{F}_1) = \bigcap_{\mathcal{F} \in \mathcal{I}(\mathscr{\mathcal{F}_1})} \mathcal{F}$.
We now show that $\sigma(\mathcal{F}_1) = \mathcal{F}_2$. Note that $A \in \mathcal{F}_1$ implies $A \in \sigma(\mathcal{F}_1)$ and $\lbrace a \rbrace \in \sigma(\mathcal{F}_1)$ implies $ \lbrace b,c,d \rbrace = \lbrace a \rbrace^c \in \sigma(\mathcal{F}_1)$, so that $\mathcal{F}_2 \subset \sigma(\mathcal{F}_1)$. By assumption 2, $\mathcal{F}_2$ is a $\sigma$-algebra. We obtain $\sigma(\mathcal{F}_1) = \mathcal{F}_2$.
Note
It holds in general that $\sigma(\lbrace A \rbrace) = \sigma(\lbrace \emptyset, \Omega, A \rbrace) = \lbrace \emptyset, \Omega, A, A^c \rbrace$, we could have proved this instead of proving this for our particular case in the last step.
