The algebra generated by the family and the sigma algebra generated by the family. Properties. For any family $B \subset 2^{X} $ there exists the smallest algebra A containing B, we call this algebra the algebra generated by the family B and denote it $\sum (B)$.
For any family $B \subset 2^{X} $ there exists the smallest sigma algebra A containing B, we call this sigma algebra the sigma algebra generated by the family B and denote it $\sigma(B)$.
Prove (for any family of sets $B$):
1) $\sum(\sum(B)) = \sum(B)$
2) $ \sigma(\sigma(B))=\sigma(B)$
3) $\sum(\sigma(B))=\sigma(\sum(B))=\sigma(B)$
Attempt of proof:
To be honest I can see these properties only in examples, but I don't know how to prove it in generally.
Example:
$X=\{1,2,3 \}$, and considering $\sigma(\{1\})$ and $\sum(\{1\})$
 A: 1) 
$\Sigma(B)$ is evidently the smallest algebra that contains $\Sigma(B)$ so $\Sigma(\Sigma(B))=\Sigma(B)$
2) 
same reasoning as under 1) but now for $\sigma$-algebras.
3) 
$\sigma(B)$ is evidently the smallest algebra  that contains $\sigma(B)$ so $\Sigma(\sigma(B))=\sigma(B)$.
$\sigma(B)$ is an algebra that contains $B$ so that $B\subseteq\Sigma(B)\subseteq\sigma(B)$. 
This implies that $\sigma(B)\subseteq\sigma(\Sigma(B))\subseteq\sigma(\sigma(B))=\sigma(B)$ hence  $\sigma(\Sigma(B))=\sigma(B)$.
A: Proof of 1).
$\sum (\sum (B))$ is an algebra by definition. 
$\sum (\sum (B)) \supseteq \sum(B) \supseteq B$ again by definition.
It remains to prove that $\sum (\sum (B))$ is the smallest algebra containing $B$ (hence we will have the equality $\sum (\sum (B))= \sum (B)$).
Now, let $A$ be any algebra containing $B$, we need to show that $\sum (\sum (B)) \subseteq A$. Since $\sum(B)$ is the smallest algebra containing $B$, we have $\sum(B) \subseteq A$. Since $\sum (\sum (B))$ is the smallest algebra containing $\sum(B)$, we have $\sum (\sum (B)) \subseteq A$. This completes the proof of 1).
Proof of 2): the same argument as 1).
Proof of 3): 
One equality is proved by monotonicity (in the sense that $X \subseteq Y \Rightarrow \sigma (X) \subseteq \sigma (Y)$)
$$\sigma ( \sigma (B)) = \sigma(B) \subseteq \sigma ( \Sigma (B)) \subseteq \sigma ( \sigma (B))$$
hence all these are equal.
Finally, recalling that all sigma-algebras are algebras (hence $\Sigma (X) \subseteq \sigma (X)$),
$$\sigma (B) \subseteq \Sigma ( \sigma (B)) \subseteq \sigma ( \sigma (B))= \sigma (B)$$ hence all these are equal.
