Generated $\sigma$ - algebra example Let $\Omega = \mathbb{Z}$.
Consider $E_1 := \big\{\{2n|n \in \mathbb{Z}\}\big\}$ and $E_2 := \big\{\{2n\} | n \in \mathbb{Z}\big\}$. 
Find the generated $\sigma$-algebra $\sigma$($E_1$) and $\sigma$($E_2$).
So this solution is from my teacher:
$$\sigma(E_1) = \{ \emptyset, E_1, E_1^c, \Omega \} = \big\{ \emptyset, \{2n|n \in \mathbb{Z}\}, \{2n+1|n \in \mathbb{Z}\}, \mathbb{Z} \big\}.$$
$$\sigma(E_2) = \big\{ \mathcal{P}(E_2),  \{2n+1|n \in \mathbb{Z}\}, ( \{2n+1|n \in \mathbb{Z}\} \cup C ) | C \in \mathcal{P}(E_2) \}$$
Remark: $\mathcal{P}$ is the power set.
So is this really correct?
To me $\sigma(E_1)$ is only correct if we would have $E_1 = \{2n|n \in \mathbb{Z}\}$ not $\big\{\{2n|n \in \mathbb{Z}\}\big\}$. And for $\sigma(E_2)$ I would say that we have $\big\{ C, \{2n+1|n \in \mathbb{Z}\}, ( \{2n+1|n \in \mathbb{Z}\} \cup C ) | C \in \mathcal{P}(E_2)\}$.
 A: Your objection to the second one is correct. The correct answer looks something like
$$ \sigma(E_2) = \mathcal{P}(\{2n \mid n \in \mathbb Z\}) \cup \{ \{2n + 1 \mid n \in \mathbb Z \} \cup C \mid C \in \mathcal{P}(\{2n \mid n \in \mathbb Z\}) \}. $$
There's no need to include $\{2n + 1 \mid n \in \mathbb Z \}$ seperately because we can simply take $C = \varnothing$. And, unless I'm missing something, we require $\mathcal{P}(\{2n \mid n \in \mathbb Z\})$ instead of  $\mathcal{P}(E_2)$ because the latter is neither a subset of $\mathbb{Z}$ nor a collection of subsets of $\mathbb Z$.
The first one is wrong but I wouldn't say it's for the reason you mention.  Note that the set $E_1$ has to be a collection of subsets of $\mathbb Z$. I.e. a subset of $\mathcal P(\mathbb Z)$. In this case the subset is $E_1 = \{A\}$ where $A$ is the set of even integers. On the other hand, $A$ is not a collection of subsets of $\mathbb Z$ so if we set $E_1 = A$ then $\sigma(E_1)$ isn't defined (unless you define it to mean $\sigma(\{A\})$). We always have
$$ \sigma(\{A\}) = \{\varnothing, A, A^c, \Omega\}.$$
The mistake here is that
$$ \sigma(E_1) \ne \{\varnothing, E_1, E_1^c, \Omega\} $$
because again $E_1 \not\subseteq \mathbb Z$.
