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I have shown that the $\sigma$-algebra generated by the set of all countable and cocountable sets in $\mathbb{R}$, i.e. $\sigma \left( \{A\subset \mathbb{R} : A \text{ or } A^\complement \text{ is countable } \}\right)$ is equal to the $\sigma$-algebra generated by the set of singletons in $\mathbb{R}$, which must be uncountably generated.

Is this equivalence sufficient to show, that $\sigma \left( \{A\subset \mathbb{R} : A \text{ or } A^\complement \text{ is countable } \}\right)$ is also uncountable? I am having trouble seeing whether or not an uncountably generated $\sigma$-algebra can be equivalent to a countably generated $\sigma$-algebra in some cases.

Also, how can this $\textit{uncountably}$ generated $\sigma$-algebra be a subset of the Borel algebra on $\mathbb{R}$, which is $\textit{countably}$ generated?

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    $\begingroup$ Since your two $\sigma$-algebras are equal, the former (the countable-cocountable) is uncountably generated because they are the same set. As for the intuition behind your second question, I'd say it is something similar to why an open subset of a compact space needn't be compact - think of $\mathbb{Q} \cap [0;1]$. $\endgroup$ – Simone Ramello Jul 15 '19 at 17:32
  • $\begingroup$ What does uncountably generated mean? Does it mean that the sigma algebra cannot be generated by a countable family? $\endgroup$ – mechanodroid Jul 16 '19 at 9:38
  • $\begingroup$ Better then "uncountably generated" would be "not countably generated". So far, we have not seen a proof of it in this thread. $\endgroup$ – GEdgar Jul 16 '19 at 10:19
  • $\begingroup$ @GEdgar. If $A$ is a countable subset of the $\sigma$-algebra on $\Bbb R$ generated by the countable subsets of $\Bbb R$ then the $\sigma$-algebra $B$ on $\Bbb R$ generated by $A$ is also generated by the set $A''$ of countable members of $A\cup \{\Bbb R \setminus a: a\in A\}$. Hence every countable member of $B$ is a subset of the countable set $\cup A''.$ So if $C$ is a non-empty countable subset of $\Bbb R \setminus \cup A''$ then $C\not \in B.$ $\endgroup$ – DanielWainfleet Jul 16 '19 at 13:20
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As to the second question: The Borel algebra can be generated from open intervals with rational endpoints. However, these generators do not belong to the countable-cocountable sigma algebra. There is no contradiction.

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