Determine the smallest sigma-algebra A on $\Omega$ We consider the sample space $\Omega =\mathbb{N}$. Determine the smallest $\sigma$-algebra $\mathfrak A$ on $\Omega$ to which applies: $\\$
If $n \in\mathbb{N}$ is an even number, then $\left\{{n}\right\} \in \mathfrak A.$ 
I don't quite know. Can anybody please help?

So far I have come to the following:
First, we claim that any subset $A \subseteq \mathbb N$ containing only even numbers must be in $\mathfrak A$. If A is finite, $A=\left\{{a_1,..., a_n}\right\}$, then we set
$$A_i = \left\{a_i\right\} \text{  for  } i =1,...n$$
According to the precondition $A_i \in \mathfrak A$, so according to the third axiom for $\sigma$-algebras also applies
$$ {\bigcup_{i=1}^n A_i\in \mathfrak A}. $$
If A is infinite, but A is in any case still countable (since whole $\mathbb N$ is countable), we can write $A=\left\{{a_1,..., a_n,...}\right\}$. Let's put
$$A_i = \left\{a_i\right\} \text{    for   }  i \in N $$
$A \in \mathfrak A$ are after precondition, so after the third axiom for sigma-algebras also applies here
$$ {\bigcup_{i=1}^\infty A_i\in \mathfrak A}. $$
Thus necessarily all subsets of $\mathbb N$, which contain only even numbers, must be in $\mathfrak A$.

I don't know how to go on from here.
Thank you!
 A: What you did is right. Let $E,O\subset\mathbb N$ be the subsets of even and odd nubmers respectively. You showed that $E$ and all its subsets are in $\mathfrak A$. Since $\mathfrak A$ needs to have complements, you also have $O\in\mathfrak A$. And then 
$$
\mathfrak A=\{\varnothing, \mathbb N, O, E, \text{subsets of $E$}, \text{ sets } O\cup B\ \text{ where }B\subset E\}.
$$
This is closed under taking complements and unions. 
A: The following answer is a summary of the answers from @Martin Argerami and me.

Let $E \subset \mathbb N$ be the subset of even nubmers and $O \subset \mathbb N$ be the subset of odd nubmers.
So for $n\in \mathbb N$ let's define then $E_n=\{n\ \mathrm{even}  \,|\,  n\in \mathbb N\}$ and $O_n=\{n\ \mathrm {odd}  \,|\,  n\in \mathbb N\}$
The union of these $E_n$ and $O_n$ over all subsets of $\mathbb N$ does actually give all the subsets of even and odd numbers.
We claim now that $ E$ must be in $\mathfrak A$. According to the precondition $E_n \in \mathfrak A$.
The only condition that could possibly cause difficulties here is the third. Here we distinguish two cases:
$1. \,$ If $E$ is finite: 
according to the third axiom for $\sigma$-algebras, it must apply here:
$$E_n=\{n\ \mathrm{even}  \,|\,  n\in \mathbb N\} \Rightarrow  E= {\bigcup_{n \in \mathbb N} E_n\in \mathfrak A}. $$
Note that: $E=\left\{{e_1,..., e_n}\right\}$
$2.\,$ If $E$ is infinite:
Let's put
$$E_i=\{i\ \mathrm{even}  \,|\,  i\in \mathbb N\} \Rightarrow E = {\bigcup_{i=1}^\infty E_i\in \mathfrak A}. $$
Note that: $E=\left\{{e_1,..., e_n,...}\right\}$
Thus necessarily $E$ and $E_n$ must be in $\mathfrak A$.
Since $\mathfrak A$ needs to have complements, we also have $O\in\mathfrak A$. And then 
$$
\mathfrak A=\{\varnothing, \mathbb N, O, E,E \cup O\}.
$$
"This is closed under taking complements and unions". 
