$\mathscr{C}=\{E\subset\Omega:E\in\mathscr{R}\:\text{or}\:\Omega-E\in\mathscr{R}\}$ is an algebra

Question

Let $\mathscr{R}$ be a ring of subsets of $\Omega$ and $\mathscr{C}=\{E\subset\Omega:E\in\mathscr{R}\:\text{or}\:\Omega-E\in\mathscr{R}\}$.

Show that $\mathscr{C}$ is an algebra.

To show $\mathscr{C}$ is an algebra:

i) $\emptyset\in\mathscr{C}$

ii) $A\in\mathscr{C}$ and $B\in\mathscr{C}$ so $A\cup B\in\mathscr{C}$

iii)$A\in\mathscr{C}$ and $B\in\mathscr{C}$ so $A\setminus B\in\mathscr{C}$

iv)$A\in\mathscr{C}$ so $A^c=\Omega\setminus A\in\mathscr{C}$

For example if I want to check $\mathscr{C}$ is closed for unions then:

If $A\in\mathscr{R}$ and $B\in\mathscr{R}$then $A\cup B\in\mathscr{R}$ because $\mathscr{R}$ is a ring hence closed for finite unions or $\Omega\setminus(A\cup B)=(A\cup B)^c=A^c\cap B^c\in\mathscr{R}$ once $\mathscr{R}$ is closed for finite intersections. This concludes $A\cup B$ is in $\mathscr{C}$.

Question:

It was pointed out that this argument I wrote is wrong once I does not prove $A\cup B$ is in $\mathscr{C}$. But I fail to see why? Can someone tell me what is failing?

Thanks in advance!

Answer 1

You have to take sets $A$ and $ B$ from $\mathscr{C}$ and not from $\mathscr{R}$. Now do some case analysis

Case 1: $A\in\mathscr{R}$ and $B\in\mathscr{R}$ then $A\cup B\in\mathscr{R}\implies A\cup B\in\mathscr{C}$

Case 2: $A\in\mathscr{R}$ and $B^c\in\mathscr{R}$ then $A-B\in\mathscr{R}\stackrel{?}{\implies} A\cup B\in\mathscr{C}$

Case 3: $A^c\in\mathscr{R}$ and $B^c\in\mathscr{R}$ then $(A\cup B)^c= A^c\cap B^c\in\mathscr{R}\stackrel{}{\implies} A\cup B\in\mathscr{C}$

However, I'm not sure how to deal with case 2.

Answer 2

After greedoid answer I am able to post an answer to the question. So we have 4 cases:

1: $A\in\mathscr{R}$ and $B\in\mathscr{R}$ then $A\cup B\in\mathscr{R}\implies A\cup B\in\mathscr{C}$

2: $A\in\mathscr{R}$ and $B^c\in\mathscr{R}$ then $(A\cup B)^c=A^c\cap B^c=B^c\cap A^c=B^c\setminus A\in\mathscr{R}{\implies} A\cup B\in\mathscr{C}$

3: $A^c\in\mathscr{R}$ and $B^c\in\mathscr{R}$ then $(A\cup B)^c= A^c\cap B^c\in\mathscr{R}\stackrel{}{\implies} A\cup B\in\mathscr{C}$

It was intended to show that either $A\cup B\in\mathscr{R}$ or $(A\cup B)^c\in\mathscr{R}$,which are the conditions defined in $\mathscr{C}$ for $A\cup B$ to belong to $\mathscr{C}$. The remaining sections of the exercise are about the same reasoning.