If R is $\sigma -$ring then {$E\in X|E\in R \text { or} \ E^c \in R$ } is a $\sigma$ -algebra
$\sigma -$ring : A family of sets $R\in P(X)$ is called a $\sigma $-ring if it is closed under countable unions and differences of 2 sets.
$\sigma-$ algebra: A family of sets $R\in P(X)$ is called a $\sigma- $algebra if it is closed under countable unions and every element has a complement in it.
I had proved that it is enough to show that if $X\in R$ then $R$ is a $\sigma-$ algebra
Case 1) If $E,E^c\in R$ then by union we are done.
Case 2) If only either $E$ or $E^c \in R,$
then I am not able to proceed.
I wanted to solve this. Please give me only a hint.
Thanks in advance.
Best Answer
Let $\mathcal A:=\{E\in\wp(X)\mid E\in\mathcal R\text{ or }E^{\complement}\in\mathcal R\}$ where $\mathcal R$ denotes a $\sigma$-ring on set $X$.
Then it is immediate that $\mathcal A$ is closed under complements.
Now observe that a $\sigma$-ring is closed under countable intersections.
This because we can write: $$\bigcap_{n=1}^{\infty}R_n=R_1\setminus\bigcup_{n=2}^{\infty}(R_1\setminus R_n)$$
Now let $A_n\in\mathcal A$ for $n=1,2,\dots$ and let it be that $A_n\in\mathcal R$ for $n\in I$ and $A_n^{\complement}\in\mathcal R$ for $n\in J$ where we take $I$ and $J$ to be disjoint and covering subsets of $\mathbb N$.
Then $P:=\bigcap_{n\in I}A_n\in\mathcal R$ and $Q:=\bigcup_{n\in J}A_n^{\complement}\in\mathcal R$.
Consequently $\bigcap_{n=1}^{\infty}A_n=P\setminus Q\in\mathcal R\subseteq\mathcal A$.
Proved is now that $\mathcal A$ is closed under countable intersections, and combined with the fact that it also closed under complements it can be proved now that $\mathcal A$ is closed under countable unions: $\bigcup_{n=1}^{\infty}A_n=\left(\bigcap_{n=1}^{\infty}A_n^{\complement}\right)^{\complement}$.
Proved is now that $\mathcal A$ is a $\sigma$-algebra.