I have been stalled on the following question for a while now:
"Suppose $A$ and $B$ are algebras on the sample space $\Omega$. Prove that if $C = A \cap B$ then $C$ is an algebra on $\Omega.$"
I think I understand that this is true because the largest possible algebra is the set of all subsets and if we let $A$ be the largest possible algebra on $\Omega$ and let $B$ be an arbitrary algebra on $\Omega$ then $ A \cap B$ will always equal $B$ which is an algebra.
Since this topic is still new to me, any opinions improving or correcting my assumption would be very much appreciated.
As Kavi Rama Murthy points out, you have only considered the special case of $A=\mathcal{P}(\Omega)$. As you point out, this implies $A\cap B=B$ for any $B\subset\mathcal{P}(\Omega)$. However, you still have to prove the case of $A\neq\mathcal{P}(\Omega)$!
Hint: Let me get you started. Suppose $A$ and $B$ are both algebras. Let $C=A\cap B$. To prove that $C$ is an algebra we have to check three things:
I'll prove (2) and let you prove the rest. Let $c\in C$. Then, $c\in A$ and $c\in B$. Since $A$ and $B$ are algebras (and hence closed under complement), it follows that $\Omega\setminus c\in A$ and $\Omega\setminus c\in B$. Therefore, $\Omega\setminus c\in C$. That is, $C$ is closed under complements.
Comment: are you sure you did not mean $\sigma$-algebra? If you did, replace "finite unions" above by "countable unions". Otherwise, ignore this comment.