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.

  • $\begingroup$ 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! $\endgroup$ – mate89 Dec 19 '18 at 4:23
  • $\begingroup$ @mate89 the image can be any infinite subset of $\omega$ $\endgroup$ – Henno Brandsma Dec 19 '18 at 5:11
  • $\begingroup$ 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! $\endgroup$ – mate89 Dec 19 '18 at 5:22
  • $\begingroup$ @mate89 by definition $i \to x_i$ is an injection so $A$ is trivially countably infinite. $\endgroup$ – Henno Brandsma Dec 19 '18 at 5:42

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