The set $\mathcal{A}=\{X \in \mathcal{P}(\mathbb{N}): |X| \textrm{ or } |X^c| \textrm{ is finite}\}$ is no $\sigma$-algebra because, for example, the set of even numbers can be written as the countable infinite union of the sets $\{2n\} \in \mathcal{P}, n \in \mathbb{N}$, but of course its complement, the odd numbers, is not finite.
In the following task, I am a bit confused by the notation used for the set $\mathcal{A}$:
Let $M_n=\{X \in \mathcal{P}(\mathbb{N}): X \subset \{1,…,n\} \textrm{ or } X^c \subset \{1,…,n\}\}$
Show that $\bigcup_{n=1}^{\infty}M_n$ is no $\sigma$-algebra on the natural numbers.
My naive opinion is that $\bigcup_{n=1}^{\infty}M_n$ means something different than $\mathcal{A}$.
For example,
$\bigcup_{n=1}^{\infty}[-n,n] = \mathbb{R}$ (I), similarly
$\bigcup_{n=1}^{\infty}\{2n\} = \mathbb{2N}$ (II), or
$\bigcup_{n=1}^{\infty}\{\textrm{even numbers between 2 and }n\} = \mathbb{2N}$
Now, by uniting the $M_n$ we are also including $\bigcup_{n=1}^{\infty}\{\textrm{even numbers between 2 and }n\} = \mathbb{2N}$ in the union because $\{\textrm{even numbers between 2 and }n\} \in M_{n}$ which would imply that $\mathbb{2N} \subset \bigcup_{n=1}^{\infty}M_n$. So the argument stated above does not work, in fact, I think $\bigcup_{n=1}^{\infty}M_n$ is a $\sigma$-algebra in that case.
To me, the infinite union removes the "finite" restriction of the sets $M_n$, similar to (I) and (II).
Am I just getting something wrong, or should the set in question be written like the description of $\mathcal{A}$ ?
Best Answer
By definition, $S\in\bigcup_{n=1}^\infty M_n$ iff there exists some $n$ with $S\in M_n$.
There is no $n$ with $2\Bbb N\in M_n$.