# Proof that $\sigma$-algebra is not countable (proof revision)

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 $$Im(\bar{f}) = \omega$$.

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.
• Thanks a lot for your answer. Since $f$ is an injection and I assumed the $F_i$ are all distinct, I thought the restriction of $f$ to $F_i$ would be a bijection to $\omega$ (for any $k \in \omega$ there is a corresponding $F_k$), thus $Im(\bar{f}) = \omega$. Sadly, I can't see why is that wrong. Would you explain me please? thanks! – mate89 Dec 19 '18 at 4:23
• @mate89 the image can be any infinite subset of $\omega$ – Henno Brandsma Dec 19 '18 at 5:11
• Hmmm I see. One last question: if I have a sequence $\{x_i\}_{i=0}^\infty$ where they are all disctinct ($x_i \ne x_j$ for $j \ne i$), and we set $A := \{x_i\}_{i=0}^\infty$ can we say something about $card(A)$? is it possible to define an injection such that we can show that $card(A) = \aleph_0$. Thanks again! – mate89 Dec 19 '18 at 5:22
• @mate89 by definition $i \to x_i$ is an injection so $A$ is trivially countably infinite. – Henno Brandsma Dec 19 '18 at 5:42