Problem : We consider on $\mathbb{N}$, for all $n\geq 0$ the sigma algebra $ \mathcal{F}_n = \sigma(\{0\},\{1\},\{2\}, \dots , \{n\})$ show that the sequence of sigmas $(\mathcal{F}_n, n\geq 0)$ is increasing but $\cup_{n\geq0} \mathcal{F}_n $ is not a sigma algebra.
Solution : Let $ \mathcal{F} = \cup_{n\geq0} \mathcal{F}_n $ suppose $\mathcal{F}$ is a sigma algebra, we have :
$\{2n\} \in \mathcal{F}_{2n} \subset \mathcal{F} $ and $2\mathbb{N}=\cup_{n\geq0} \{2n\}$
This way, there exists $n_0$ such that $2\mathbb{N} \in \mathcal{F}_{2n}$. But the only elements of infinite cardinal of $\mathcal{F}_{n_0}$ are of the form $\mathbb{N} \setminus A$ , where $A$ is a subset of $\{0,1,\dots,n_0\}$ we obtain a contradiction.
End of solution
What I thought of doing first was trying to verify the three axioms of a sigma algebra, using a result of the lecture that a union sigma algebras is not a sigma algebra,
But I don't understand the way this solution is constructed,
why are we supposing $\mathcal{F}_{2n}$ a subsequence of a sigma algebra is not necessarily a sigma algebra, Is there some result I have missed?
there exists $n_0$ such that $2\mathbb{N} \in \mathcal{F}_{2n}$ , why there exist? if $\mathcal{F}_{2n}$ is a sigma algebra, $\Omega = 2\mathbb{N}$ is in $\mathcal{F}_{2n}$
"But the only elements of infinite cardinal of $\mathcal{F}_{n_0}$ are of the form $\mathbb{N} \setminus A$" , I spent an hour looking for what this means I don't get it, did I miss another result here?
How did we use this fact : "$(\mathcal{F}_n, n\geq 0)$ is increasing but ..." I mean how does it contradict anything? I am guessing it is either trying to use the monotone class theorem "the Dynkin system" which I don't see how it relates to this problem, or I have missed some result ...