I'm practicing my proof-writing and was hoping you could let me know if this proof looks good. I would like to know if the proof is incorrect, if there are parts that are overly wordy/complicated, or if I'm missing some element of a proof that is helpful to see, if not strictly necessary.
The Prompt:
Let $X$ be a nonempty set and $A$ a countable collection of pairwise disjoint subsets of $X$ with $\emptyset \in A$. Prove that:
$\sigma (A) = \{Z \in 2^X : Z \in \{\cup B, X\setminus\cup B\}$ for some $B \subset A\}$
My Proof:
$\sigma(A)$ is closed under countable union, so $\cup B \in \sigma(A)$ for any $B \subset A$. It is also closed under complementation, so for any set $C \in \{Z \in 2^X | Z = \cup B$ for some $B \subset A\}$, $X \setminus C \in \sigma(A)$. Define the following as $Y$, and notice:
$Y = \{Z \in 2^X | Z = \cup B$ for some $B \subset A\} \cup \{Z \in 2^X | Z = X \setminus \cup B$ for some $B \subset A\} =$
$\{Z \in 2^X | Z \in \{\cup B, X \setminus \cup B\}$ for some $B \subset A\} \subset \sigma(A)$
Since $Y$ is closed under complementation, countable union, has $A$ as a subset, and contains both $\emptyset$ (an element of $A$) and $X$ ($X \setminus \{\emptyset \}$), it is a $\sigma$-algebra and is thus $\sigma$($A$).
You need to conclude that $\sigma(A)\subset Y$. In particular, you have shown that $Y$ is a sigma algebra of $A$ and by definition $\sigma(A)$ is the smallest such sigma algebra, so $\sigma(A)\subset Y$ (the intersection of two sigma algebras is another sigma algebra).