Let $\Omega$ any uncountable set and $\mathcal{O}$ is the collection of all subsets of $\Omega$ which are countable or have countable complements be the collection. We want to show that $\mathcal{O}$ forms a $\sigma$-algebra. Note that the $\emptyset \subset \Omega$. Hence $\emptyset \in \mathcal{O}$. We see that the first condition is satisfied since both elements are already in $\mathcal{O}$.

Similarly, the complement of $\emptyset$ is $\Omega$ and the complement of $\Omega$ is $\emptyset$, which are both in $\mathcal{O}.$

How would I finish the statement of complements and Unions?


The complement of countable set is in $\mathcal{O}$ by definition. Since countable Union of countable sets is countable, countable Union is closed in $\mathcal{O}$.

  • $\begingroup$ Do you mean complement instead of component? $\endgroup$ – Username Unknown Feb 2 '16 at 5:39

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