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Let $E=\{\{\omega \} : \omega \in \Omega \}$.

$\sigma ( E) = \{ A \subseteq \Omega : A \ is \ countable \ or \ A^c \ is \ countable \} $ is a $\sigma$-algebra generated by the set $E$ i.e. the smallest $\sigma$-algebra containing $E$ (already proved).

Prove: $\sigma (E)$ is equal to partitive set of $\Omega$ if and only if $\Omega$ is a countable set.

One side: Suppose $\Omega$ is countable. $\sigma (E)$ contains all the sets that are countable or their complement is countable. That is true for every subset of a countable set, so it leads $\sigma (E)$ is a partitive set of $\Omega$.

Other side: please help

Petra
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    What do you mean with "partitive set"? It seems to be "powerset", but if so then I would no speak of "a" partitive set, but rather of "the" partitive set. – drhab Nov 29 '18 at 11:38
  • By partitive set I mean the set of all subsets of $\Omega$. – Petra Nov 29 '18 at 12:47
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    Usually that is called the powerset of $\Omega$. The term "partitive set" is new to me. – drhab Nov 29 '18 at 13:07

1 Answers1

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If $\Omega$ is not countable then it can be written as disjoint union of two uncountable subsets $\Omega_1,\Omega_2$.

See here for a proof of that.

So then $\Omega_1,\Omega_2\notin \sigma(E)$.

drhab
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  • Why do you think that they are not in $\sigma(E)$. If their complements are countable than they are in $\sigma(E)$. – Petra Nov 29 '18 at 12:49
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    $\Omega_1^{\complement}=\Omega_2$ hence is not countable and $\Omega_2^{\complement}=\Omega_1$ hence is not countable. – drhab Nov 29 '18 at 13:05