Give an example of a monotone class $\mathcal{G}$ on $\mathbb{R}$ that satisfies: (a) \mathbb{R} belongs to $\mathcal{G},$ and (b)if $A \in \mathcal{G} $ then its complement $A^c$ is in $\mathcal{G},$ but it is not a $\sigma-$algebra.
Here is my trial:
consider the following set $\{(-\infty, a), (-\infty, a], \emptyset, [a, \infty), (a, \infty)\}$, is this a correct example?if so, where is the increasing and decreasing sequences in $\mathcal{G},$? if not, could anyone provide me with an example please?
Monotone Class Definition:
We say that $\mathcal{G}$ is a monotone class if whenever $\{A_{k}\}$ is an increasing and $\{B_{k}\}$ is a decreasing sequence in $\mathcal{G},$ then $\cup A_{k}$ and $\cap B_{k}$ are in $\mathcal{G},$ as well.
Note that I found this question here A monotone class in $\mathbb{R}$ which is closed under complement but is not a sigma-algebra but it does not fully answer my questions.
Your example is correct except that you forgot to include $\mathbb R$. The only way you get a decreasing sequence in this class is by considering $(-\infty, a_n)$ with $(a_n)$ decreasing, $(-\infty, a_n]$ with $(a_n)$ decreasing or $(a_n,\infty) $ with $(a_n)$ increasing, or $[ a_n, \infty)$ with $(a_n)$ increasing. Similarly we can write down increasing sequences.
It is not a sigma algebra because $\{0\}=(-\infty,0]\cap [0,\infty)$ does not belong to it.