I am testing my hand at some introductory measure theory problems and have come across a tough (relatively speaking) problem. I think having my proof verified would assist with my learning.
I am interested in any critique or lambasting of my proof, be it focusing on inaccuracies or cohesiveness, I do not care.
Problem
Let $\mathcal{C}$ be a countable partition of a set $E$. Show that every element of $\sigma \mathcal{C}$ is a countable union of elements taken from $\mathcal{C}$.
Attempt
$\text{ }$ We begin by constructing a set $\mathcal{E}$ consisting of all possible countable unions of elements taken from $\mathcal{C}$. We want to first show that $\mathcal{E}$ is a $\sigma$-algebra, i.e. $\mathcal{E}$ is closed under countable unions and closed under complements. Since $\mathcal{E}$ consists entirely of countable unions of elements from $\mathcal{C}$, the union of any amount of elements from $\mathcal{E}$ will also be a countable union of elements from $\mathcal{C}$, which means that this union will be in $\mathcal{E}$, so $\mathcal{E}$ is closed under countable unions. Now consider some arbitrary element $D \in \mathcal{E}$. Since $D$ is a countable union of elements from $\mathcal{C}$, i.e. $D \subseteq \bigcup_{B \in \mathcal{C}} B$, and since $\bigcup_{B \in \mathcal{C}} B = E$ (given $\mathcal{C}$ is a countable partition of $E$), we have that $E \setminus D \subseteq \bigcup_{B \in \mathcal{C}} B$. This means that $E \setminus D$ is a countable union of elements from $\mathcal{C}$, so $E \setminus D \in \mathcal{E}$. Since $D$ was arbitrary, we have that $\mathcal{E}$ is closed under complements.
$\text{ }$ Finally, we need to show that $\mathcal{E} = \sigma \mathcal{C}$, which we can do by proving that $\sigma \mathcal{C} \subseteq \mathcal{E}$ and $\mathcal{E} \subseteq \sigma \mathcal{C}$. For the first case, we know that the countable union of all elements of $\mathcal{C}$, i.e. $\bigcup_{B \in \mathcal{C}} B = \mathcal{C} = E$, is an element of $\mathcal{E}$. Let $A_1 \cap A_2 \cap \dotsm$ be the intersection of all $\sigma$-algebras that contain $\mathcal{C}$. By the definition of $\sigma \mathcal{C}$ as the smallest $\sigma$-algebra to contain $\mathcal{C}$, $A_1 \cap A_2 \cap \dotsm = \sigma \mathcal{C}$, and since $\mathcal{E}$ is a $\sigma$-algebra that contains $\mathcal{C}$, we have that $\sigma \mathcal{C} \subseteq \mathcal{E}$. For the second and final case, $\mathcal{C} \subseteq \sigma \mathcal{C}$ and the fact that $\sigma \mathcal{C}$ is a $\sigma$-algebra implies that all countable unions of elements taken from $\mathcal{C}$ are also contained in $\sigma \mathcal{C}$. Since we constructed $\mathcal{E}$ to consist exclusively of all countable unions of elements taken from $\mathcal{C}$, we have that $\mathcal{E} \subseteq \sigma \mathcal{C}$. Since $\sigma \mathcal{C} \subseteq \mathcal{E}$ and $\mathcal{E} \subseteq \sigma \mathcal{C}$, we have that $\sigma \mathcal{C} = \mathcal{E}$, as desired.
Ideas
Only part that might seem problematic is "$\mathcal{C} \subseteq \sigma \mathcal{C}$ and the fact that $\sigma \mathcal{C}$ is a $\sigma$-algebra implies that all countable unions of elements taken from $\mathcal{C}$ are also contained in $\sigma \mathcal{C}$."
Also, please point out if I switched $\in$ and $\subseteq$ anywhere. This is an annoying issue I need to prune out of my behaviour.
Best Answer
Your attempt is right, but it can be improved.
Let $\mathcal{E}$ be the set consisting of all possible countable unions of elements taken from $\mathcal{C}$. You already proved that it is closed by countable union and by complement. It is trivial that $\emptyset \in \mathcal{E}$. So $\mathcal{E}$ is a $\sigma$-algebra.
Clearly, for each $A \in \mathcal{C}$, $A \in \mathcal{E}$. So, $\mathcal{C} \subset \mathcal{E}$. So $$\sigma(\mathcal{C}) \subseteq \mathcal{E}$$
which means that every element of $\sigma( \mathcal{C})$ is a countable union of elements taken from $\mathcal{C}$ (and answer the Problem as stated). But we can go a little further.
On the other hand, any countable unions of elements taken from $\mathcal{C}$ is an element of $\sigma(\mathcal{C})$. So $$\mathcal{E} \subseteq \sigma(\mathcal{C})$$ So $$\sigma(\mathcal{C}) = \mathcal{E}$$
which means that the elements of $\sigma( \mathcal{C})$ are precisely the countable union of elements taken from $\mathcal{C}$.