Verifying my proof of the intersection of $\sigma$-algebras is also a $\sigma$-algebra It has been a while since I've done proofs so I would like to verify if my proof is correct. I feel like I cheated or skipped a step but I am not sure how. Here is what I have to prove:
Let $F_1$ and $F_2$ be two $\sigma -$Algebras over a set $\Omega$. Show that $F_1 \cap F_2$ is also a $\sigma -$Algebra over a set $\Omega$
Claim:
$F_1 \subset F_2$ or  $F_2 \subset F_1$ or $F_1 \cap F_2 = \{\emptyset, \Omega \}$
Clearly if $F_1 \cap F_2 = \{\emptyset, \Omega \}$ then it is a $\sigma -$Algebra over a set $\Omega$
Now, for the nontrivial case, WLOG, I claim $F_1 \subset F_2$
Assume by contradiction this is not the case.
Let $A \subset F_1$ non-empty such that $A \cap F_2 = \emptyset$ so that $F_1 \backslash A \subset F_2$ $\quad$ ($F_1$ minus $A$)
$\therefore \quad (F_1 \backslash A)^c \subset F_2$ as $F_2$ is closed under set complements
$\therefore \quad (F_1 \cap A^c)^c \subset F_2$
$\therefore \quad (F_1^c \cup A) \subset F_2$
$\therefore \quad A \subset F_2$ CONTRADICTION
$\therefore \quad F_1 \subset F_2$
$\therefore \quad F_1 \cap F_2 = F_1$ which is $\sigma -$Algebra over a set $\Omega$
Any help would be greatly appreciated, thank you!
 A: What is the definition of $\sigma$-algebra? Given a non-empty set $\Omega$, we say that a collection $\mathcal{F}$ of subsets from $\Omega$ is a $\sigma$-algebra iff the following conditions hold:

*

*$\Omega\in\mathcal{F}$,

*If $X\in\mathcal{F}$, then $X^{c}\in\mathcal{F}$,

*If $(A_{n})_{n\in\mathbb{N}}$ is a sequence of sets such that $A_{n}\in\mathcal{F}$, then $\bigcup_{n\geq 1}A_{n}\in\mathcal{F}$.

Based on such definition, we can prove the desired claim.
Firstly, notice that $\Omega\in\mathcal{F}_{1}\cap\mathcal{F}_{2}$: that is because $\Omega\in\mathcal{F}_{1}$ and $\Omega\in\mathcal{F}_{2}$.
Similarly, if $X\in\mathcal{F}_{1}\cap\mathcal{F}_{2}$, then $X^{c}\in\mathcal{F}_{1}\cap\mathcal{F}_{2}$. Indeed, if $X\in\mathcal{F}_{1}\cap\mathcal{F}_{2}$, then $X\in\mathcal{F}_{1}$ and $X\in\mathcal{F}_{2}$. Hence, by the definition of $\sigma$-algebra, one concludes that $X^{c}\in\mathcal{F}_{1}$ and $X^{c}\in\mathcal{F}_{2}$, that is to say, $X^{c}\in\mathcal{F}_{1}\cap\mathcal{F}_{2}$.
Finally, if $(A_{n})_{n\in\mathbb{N}}$ is a sequence of subsets from $\Omega$ such that $A_{n}\in\mathcal{F}_{1}\cap\mathcal{F}_{2}$, then $A_{n}\in\mathcal{F}_{1}$ and $\mathcal{A}_{n}\in\mathcal{F}_{2}$. Therefore, due to the definition of $\sigma$-algebra, one concludes that $\bigcup_{n\geq 1}A_{n}\in\mathcal{F}_{1}$ and $\bigcup_{n\geq 1}A_{n}\in\mathcal{F}_{2}$, whence it results that $\bigcup_{n\geq 1}A_{n}\in\mathcal{F}_{1}\cap\mathcal{F}_{2}$, and we are done.
Hopefully this helps!
Additional comments
Such property allows us to define the $\sigma$-algebra generated by an arbitrary family of subsets $\mathcal{E}$ from $\Omega$ as the (unique) collection of sets $\sigma(\mathcal{E})$ satisfying the next properties:

*

*the set $\sigma(\mathcal{E})$ is a $\sigma$-algebra,

*the set $\sigma(\mathcal{E})$ contains $\mathcal{E}$,

*if there is another $\sigma$-algebra $\mathcal{G}$ over $\Omega$ such that $\mathcal{G}\supseteq \mathcal{E}$, then $\mathcal{G}\supseteq\sigma(\mathcal{E})$.

A: Your error is in concluding that $(F_1 \backslash A)^c \subset F_2$. The fact that $F_2$ is closed under complements means that $B \in F_2$ implies $B^c \in F_2$, not that $B \subset F_2$ implies $B^c \subset F_2$.
The other answer by Átila shows how the correct proof goes.
