Proof that $sigma$-algebra is not countable (proof revision)
0
$begingroup$
This problem has already several answers here Prove that an infinite sigma algebra contains an infinite sequence of disjoint sets and is uncountable and it has been asked a lot times before. But I'd like a revision of my proof. Given the complexity of the solutions I've seen I believe my proof is wrong , but I cannot determine what is my mistake. This is the problem: Let $mathcal{M}$ be an infinite $sigma$ algebra on a nonempty set $X$ . Show that (a) $mathcal{M}$ contains an infinite sequence of disjoint sets (b) $mathcal{M}$ is not countable So for $(a)$ , let ${E_j}_{j=1}^infty subseteq mathcal{M}$ be a sequence of elements in $mathcal{M}$ . Define $F_i$ as $$F_k = E_kbackslash left( bigcup_{i=1}^{k-1} E_i right) = E_k cap left( bigcup_{i=1}^{k-1} E_i right)^c$$...