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
$$F_k = E_k\backslash \left( \bigcup_{i=1}^{k-1} E_i \right) = E_k
\cap \left( \bigcup_{i=1}^{k-1} E_i \right)^c$$
so clearly $\{F_j\}_{j=1}^\infty \subseteq \mathcal{M}$ and they are disjoint.
For $(b)$, suppose there exists and injection $f$ from $\mathcal{M}$ to $\omega$. Then $\bar{f}= f\restriction_{ \{F_i\}_{i=1}^\infty}$ is inyective and $Im(\bar{f}) = \omega$. But for any $k$ we have $F_k^c \in \mathcal{M}$ and $F_k^c \notin \{F_i\}_{i=1}^\infty$ (because $X = F_k \cup F_k^c$ and the $F_i$ are disjoint). So $\bar{f}(F_k^c) \notin \omega$ (since $\bar{f}$ is an injection), but this contradicts $Im(\bar{f}) = \omega$.
Thanks in advance!
There is no guarantee that the $F_k$ will not be empty from some $k$ onwards. You will need to do more work to ensure we get non-empty sets.
The proof of uncountability you gave makes no sense. Why is the image of $\bar{f}$ equal to $\omega$? You just claim that.