I have two sets:
$$ G_{1} = \{(-\infty,a]:a\in \mathbb{R}\}, G_{2} = \{(a-1,a]:a\in \mathbb{R}\} $$
and I want to show two things
(a) that $G_{1}$ is stable under intersection while $G_{2}$ is not stable under intersection.
I am arguing that $(-\infty,a]$ generates $\mathscr{B}(\mathbb{R})$ so $G_{1}$ kan be considered a Borel $\sigma$-algebra and all $\sigma$-algebra are stable under intersection. Is that argument sound? How would you argue that $G_{2}$ is not stable under intersection?
(b) that $\delta(G_{1}) = \sigma(G_{1})$ and $\sigma(G_{1})=\mathscr{B}(\mathbb{R})$.
I know that if a Dynkin system is stable under finite intersection then it is also a $\sigma$-algebra and I have a Theorem saying that if a $\sigma$-algebra generator $\mathscr{G} \subset \mathscr{P}(X)$ is stable under finite intersection then $\delta(G_{1}) = \sigma(G_{1})$, but how do I actually show this with $G_{1}$?
About $\sigma(G_{1})=\mathscr{B}(\mathbb{R})$, has it something to do with the family of closed sets in the $\sigma$-algebra?
Any inputs?
Question (a)
Take $(-\infty, a] = A \in G_1$ and $(-\infty, b] = b \in G_1$, with $a < b$. Then:
$$C = A \cap B = A \in G_1.$$
As a consequence, $G_1$ is stable under intersection. Note that in case that $a > b$, then
$$C = A \cap B = B \in G_1.$$
Take $(a-1, a] = A \in G_2$ and $(b-1, b] = b \in G_2$, with $a < b-1$. Then:
$$C = A \cap B = \emptyset \not\in G_2.$$
As a consequence, $G_2$ is not stable under intersection. Note that in case that $b < a -1$, then
$$C = A \cap B = \emptyset \not\in G_2.$$