# Showing that the $\sigma$-algebra of countable and cocountable sets on $\mathbb{R}$ is uncountably generated, and also a subset of the Borel algebra

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?

• 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 $\mathbb{Q} \cap [0;1]$. – Simone Ramello Jul 15 '19 at 17:32
• What does uncountably generated mean? Does it mean that the sigma algebra cannot be generated by a countable family? – mechanodroid Jul 16 '19 at 9:38
• Better then "uncountably generated" would be "not countably generated". So far, we have not seen a proof of it in this thread. – GEdgar Jul 16 '19 at 10:19
• @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 $C\not \in B.$ – DanielWainfleet Jul 16 '19 at 13:20