Let $\mathcal{A}$ be an algebra for $X \neq \emptyset$. Show $\mathcal{A}$ is $\sigma$-algebra for $X$ if and only if the following statement holds:
$E_{i} \in \mathcal{A}$ and $E_{i} \subset E_{i+1}$ with $i \in \mathbb{N}$, then $\bigcup_{i=1}^{\infty} E_{i} \in \mathcal{A}$.
I was trying to prove this as follows:
$\Rightarrow$ If $\mathcal{A}$ is $\sigma$ algebra, then by definition of $\sigma$ algebra we have that for any sequence $\lbrace E_{i}\rbrace_{i \in \mathbb{N}}$ where $E_{i} \in \mathcal{A}$ for every $i \in \mathbb{N}$, then $\bigcup_{i=1}^{\infty} E_{i} \in \mathcal{A}$ particulary if the sequence is increasing as the statement holds. So we are done.
$\Leftarrow$ Having the statement as hypothesis I want to prove that $\mathcal{A}$ is a $\sigma$ algebra. As $\mathcal{A}$ is an algebra for $X$ we already have that $X \in \mathcal{A}$ and $\mathcal{A}$ is closed under compliments. So, I only need to prove $\lbrace E_{i}\rbrace_{i \in \mathbb{N}}$ where $E_{i} \in \mathcal{A}$ is closed under enumerable union.
Case 1- If the sequence is increasing then $\lbrace E_{i}\rbrace_{i \in \mathbb{N}}$ where $E_{i} \in \mathcal{A}$ is closed under enumerable union as we are supposing the statement.
Case 2.- If the sequence is increasing then $\lbrace E_{i}\rbrace_{i \in \mathbb{N}}$ where $E_{i} \in \mathcal{A}$ is decreasing.
Case 3.- If the sequence is is such $\lbrace E_{i}\rbrace_{i \in \mathbb{N}}$ where $E_{i} \in \mathcal{A}$ is such is neither increasing or decreasing.
If my idea of proving this by cases how can I end up this proof as Im pretty up out of ideas to prove case 2 and case 3.
Other ways to prove this are welcome. Thanks
If $E_n, n \in \omega$ is any sequence in the algebra $\mathcal{A}$, define $$E'_n = \bigcup_{i \le n} E_i \in \mathcal{A}$$
as algebras are closed under finite unions. Also the sequence $E'_n$ is increasing in $n$ by definition, and it's easy to check that
$$\bigcup_n E_n = \bigcup_n E'_n \in \mathcal{A}$$
the last holding by the assumption on the increasing sequences of sets. No need for case distinctions.