A set with non-$\sigma$-algebra monotone class Working on this very basic analysis problem:

Find an example of set $X$ with its monotone class $\mathscr M$ such that $\emptyset, X \in \mathscr M$, but it is not a $\sigma$-algebra.

My solution is $X = \mathbb R$, since $\mathscr P(X)$ is monotone class of $X$ but $\mathscr P(X)$ is not $\sigma$-algebra since the 4th. condition of $\sigma$-algebra says that $\bigcap_{i=1}^{\infty} X_i$ and $\bigcup_{i=1}^{\infty} X_i$ are for countable intersections and unions only, whereas $X$ is uncountable.
Here are what I would like to know:
(1) Is my solution correct? 
(2) Regardless of whether it is right or wrong, could you please give me another example?
Thank you for your time and help.
 A: I don't think your solution is correct. In particular If $X_i \in \mathcal M$, then $\bigcup_{i = 1}^n X_i, \bigcap_{i = 1}^n X_i \in \mathcal M$ is going to be true (assuming $M = \mathcal P(X)$ is the power set). Yes the unions are countable, but all you need to show to get that the set is a $\sigma$-algebra is that countable intersections, unions are in the set, which is obviously true, given the definition of the power set..
Consider the following example: Let $X = \{a, b\}$ and let $\mathcal M = \big\{ \emptyset, \{a\}, \{a, b\} \big\}$. Observe that $\emptyset, X \in \mathcal M$. 
If $A_k \nearrow A$, it is either an eventually constant sequence, in which case we're done, or it would have to be a subsequence of $\emptyset \subseteq \{a\} \subseteq X$ each of whose limit is in $\mathcal M$. 
If $A_k \searrow A$, it is either an eventually constant sequence, in which case we're done, or it would have to be a subsequence of $X \supseteq \{a\} \supseteq \emptyset$, each of whose limits is in $\mathcal M$. 
Conclude that $\mathcal M$ is a monotone class that is not an algebra and hence not a $\sigma$-algebra. Notice it's not an algebra because it is not closed under complements: $X \setminus \{a\} = \{b\}$ is not in the set.
