There is the exercise:
Does there exist an infinite $\sigma$-algebra which has only countably many members?
Intuitively I think that the answer is no but I've really trouble to proof it.
This is one proof But I have few questions about it.
I don't understand this part in this proof: " If not, then every other measurable set of $X$ must intersect both $E$ and $E^c$ nontrivially."
I also don't understand why is $F_n$ a set of size $n$ measurable sets? for example $F_2$ has three members . by this logic $F_n$ should have $n+1$ members. why do we need to make $n$ sufficiently large? Any help would be appreciated.
Since $\varnothing\ne E\ne X$, this is the exact negation of "$E$ or $E^c$ contains a nonempty measurable $E'$ properly contained in it", i.e. of "there exists a nonempty $E'\in\mathfrak M$ such that $E'\subsetneq E$ or $E'\subsetneq E^c$".
Now, may If $S$ is an infinite $\sigma$ algebra on $X$ then $S$ is not countable help?