Taking intersection in $\Omega := \{0, 1\}^{2}$ I am trying to prove the importance of closure for intersection in the Monotone Class Theorem.
The hint says let $\Omega:=\{0,1\}^2 $ and $\mathcal{C}:=\{\{(0,0),(0,1)\},\{(1,0),(1,1)\},\{(0,0),(1,0)\}, \{(0,1),(1,1)\},\Omega\}$ and let $P_{0}$ be the uniform probabilty on $\mathcal{C}$.
The problem is that I do not understand the structure of these sets. How do I take intersections of two elements in $\mathcal{C}$ i.e. what is  $\{(0,0),(0,1)\}\cap\{(1,0),(1,1)\}$.
Can someone clear this up for me? I think that I can work further from that.
Thanks in advance.
EDIT: I do know understand it a bit. Following the Monotone Class Theorem I let $\mathcal{B}$ be the smallest class containing $\mathcal{C}$ which is closed under increasing limits and by differences. So making B closed under differences I add $\emptyset, {(0,0)},{(0,1)},{(1,0)},{(1,1)}$. Now I need to make $\mathcal{B}$ closed under increasing limits. I think this implies that $\mathcal{B}=2^{\Omega}$, but I do not see how to construct a sequence of events in $\mathcal{B}$ to create for example $\{\{(0,0)\},\{(0,1)\},\{(1,0)\}\}$
 A: You take intersections of the pairs of coordinates, not each component separately.  Since $(0,0) \not \in \{(1,0),(1,1)\}$ and $(0,1) \not \in \{(1,0),(1,1)\}$, we have $\{(0,0),(0,1)\} \cap \{(1,0),(1,1)\} = \emptyset.$  For another example, $\{(1,0),(1,1)\} \cap \{(0,0),(1,0)\} = \{(1,0)\}$.
If we let $\mathcal B$ be the smallest class of sets containing $\mathcal C$ that is closed under increasing unions and differences, we have
\begin{align*}
\mathcal B &= \{\emptyset, \{(0,0)\}, \{(0,1)\}, \{(1,0)\}, \{(1,1)\}, \{(0,0),(0,1)\}, \{(1,0),(1,1)\}, \\
& \quad \quad \quad \{(0,0),(1,0)\}, \{(0,1),(1,1)\}, \Omega\}.
\end{align*}
You have already verified that $\mathcal B$ is closed under set differences.  To see that $\mathcal B$ is closed under increasing unions, note that the only increasing sequences of sets things like $\emptyset \subseteq \{(0,0)\} \subseteq \{(0,0),(0,1)\} \subseteq \Omega$, and it can be easily verified that these are all in $\mathcal B$.  Essentially, since $\mathcal B$ contains only finitely many sets, all increasing sequences of sets are finite and hence their union is trivially in $\mathcal B$ (if $A_1 \subseteq A_2 \subseteq \cdots \subseteq A_n$ with $A_1,...,A_n \in \mathcal B$ then $\bigcup_{k=1}^n A_k = A_n \in \mathcal B$).
