Let $\Omega$ be a set.
If $\Omega$ is finite, then any $\sigma$-algebra on $\Omega$ is finite.
If $\Omega$ is infinite and countable, a $\sigma$-algebra on $\Omega$ cannot be infinite and countable.
What about if $\Omega$ is not countable ? Is it possible to find an uncountable $\Omega$ with a $\sigma$-algebra that is infinite and countable ?
Best Answer
Suppose that $\lvert \Omega\rvert\ge\aleph_0$, and $\mathscr M\subset\mathscr P(\Omega)$ is a $\sigma-$algebra. We shall show that: $$ \textit{Either}\,\,\,\, \lvert\mathscr M\rvert<\aleph_0\quad or\quad \lvert\mathscr M\rvert\ge 2^{\aleph_0}. $$ Define in $\Omega$ the following relation: $$ a\sim b\qquad\text{iff}\qquad \forall E\in\mathscr M\, (\,a\in E\Longleftrightarrow b\in E\,). $$ Clearly, "$\sim$" is an equivalence relation in $\Omega$, and every $E\in\mathscr M$ is a union of equivalence classes. Also, for every two different classes $[a]$ and $[b]$, there are $E,F\in\mathscr M$, with $E\cap F=\varnothing$, such that $[a]\subset E$ and $[b]\subset F$.
Case I. If there are finitely many classes, say $n$, then each class belongs to $\mathscr M$, and clearly $\lvert \mathscr M\rvert=2^n$.
Case II. Assume there are $\aleph_0$ classes. Fix a class $[a]$, and let $\{[a_n]:n\in\mathbb N\}$ be the remaining classes. For every $n\in\mathbb N$, there exist $E_n,F_n\mathscr\in M$, such that $[a]\subset E_n$, $[a_n]\subset F_n$ and $E_n\cap F_n=\varnothing$. Clearly, $[a]=\bigcap_{n\in\mathbb N} E_n\in\mathscr M$, and thus $\lvert \mathscr M\rvert=2^{\aleph_0}$.
Case III. If there are uncountably many classes, we can pick infinite countable of them $[a_n]$, $n\in\mathbb N$, and disjoint sets $E_n\in\mathscr M$, with $[a_n]\subset E_n$, (using the Axiom of Choice), and then realise that the $\sigma-$algebra generated by the $E_n$'s has the cardinality of the continuum and is a subalgebra of $\mathscr M$.