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I am trying to work out the following question. Any hints are very much appreciated.

Let $\mathcal{F}$ consist of all $A \subseteq \Omega $ such that either $A$ is a countable set or $A^{\complement}$ is a countable set

  1. Verify that $\mathcal{F}$ is a $\sigma$-algebra.
  2. Show that $\mathcal{F}$ is countably generated if and only if $\Omega$ is a countable set.

My attempt is as follows (please point out if there are any inconsistencies in arguments).

Proof: $\mathcal{F}$ is a $\sigma$-algebra

  1. $\varnothing$ is a countable set. Hence $\varnothing\in \mathcal{F}$. Since $\Omega^{\complement} = \varnothing$, hence $\Omega \in \mathcal{F}$

  2. For any set $A \in \mathcal{F}$, $A^{\complement} \in \mathcal{F}$ (this is by definition of $\mathcal{F}$, since either $A$ or $A^{\complement}$ is countable)

  3. Consider any countable collection of sets $A_1, A_2, A_3, \cdots \in \mathcal{F}$. If $A_i$ 's are all countable then $\bigcup_{i=1}^{\infty} A_i \in \mathcal{F}$ (since countable union of countable sets is countable). If for any $j$, $A_j$ is such that $A_j^{\complement}$ is countable then consider the set $B = \bigcap_{i=1}^{\infty} A_i^{\complement}$. Since $A_j^{\complement}$ is countable hence $B$ is atmost countable. Therefore $B \in \mathcal{F}$ and so $B^{\complement} \in \mathcal{F}$. Which implies $\bigcup_{i=1}^{\infty} A_i \in \mathcal{F}$. Hence $\mathcal{F}$ is closed under countable unions.

Proof: $\mathcal{F}$ is countably generated if and only if $\Omega$ is a countable set

If $\Omega$ is a countable set i.e. $\Omega = \{\omega_1, \omega_2, \omega_3, \cdots \}$, then as suggested by the above construction $\mathcal{F} = 2^{\Omega}$. And since $2^{\Omega}$ is the $\sigma$-algebra generated by the collection of singletons $\{\omega_i\}$, which is a countable collection, hence $\mathcal{F}$ is countably generated.

However I am unable to prove the only-if statement. Thanks for your suggestions.

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Let it be that $\mathcal A$ is a countable collection of subsets of $\Omega$ with $\sigma(\mathcal A)=\mathcal F$.

Then $\mathcal A\subseteq\mathcal F$ and because every $A\in\mathcal A$ may be replaced by its complement without loosing the property $\sigma(\mathcal A)=\mathcal F$ we can take every $A\in\mathcal A$ to be countable.

Consequently $\cup\mathcal A$ is countable.

Now define: $\mathcal F_0=\{A\in\wp(\Omega)\mid A\subseteq\cup\mathcal A\text{ or } A^{\complement}\subseteq\cup\mathcal A\}$.

It is straightforward to prove that $\mathcal F_0$ is a $\sigma$-algebra.

This with $\mathcal A\subseteq \mathcal F_0$ and consequently $\mathcal F\subseteq\mathcal F_0$.

Now assume that $\Omega$ is not countable.

Then some $\omega\in\Omega$ must exist with $\omega\notin\cup A$.

Next to that $\{\omega\}^{\complement}$ is uncountable hence cannot be a subset of $\cup A$.

Then $\{\omega\}\notin\mathcal F_0$ contradicting that $\{\omega\}\in\mathcal F\subseteq\mathcal F_0$.

This contradiction tells us that the assumption that $\Omega$ is uncountable is wrong.

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