Sigma Algebra property

Question

Let $\mathcal{C}=\{(-\infty,x]:x\in\mathbb{R}\}$ and $\mathcal{C}_0=\{[a,b]:-\infty<a<b<\infty\}$. I want to show that $\sigma(\mathcal{C})=\sigma(\mathcal{C}_0)$. I think the general idea of the proof is to show that $A\subseteq \mathcal{C}\implies A\subseteq\sigma(\mathcal{C}_0)\implies\sigma(\mathcal{C}_0)\subseteq\sigma(\mathcal{C})$ and vice versa. My difficulties lie in understanding why if $A\subseteq \mathcal{C}\implies A\subseteq\sigma(\mathcal{C}_0)$ is true, then, $\sigma(\mathcal{C}_0)\subseteq\sigma(\mathcal{C})$. Can someone help me with this step?

Answer 1

Note that $ (-\infty,x]= \cup_{n=1}^{\infty}[-n,x]$. As $ [-n,x]\in C_0 \implies \cup_{n=1}^{\infty}[-n,x] \in \sigma(C_0)\implies \sigma(C)\subseteq \sigma(C_0).\\$

$[a,b]= [a,\infty)\cap(-\infty,b] $ and since $(-\infty,a] \in \sigma(C) \implies(-\infty,a]^c= (a,\infty) \in \sigma(C). \\$ $[a,\infty)= \cap_{n=1}^{\infty}(a-\frac{1}{n}, \infty) \in \sigma(C)\\$. $\implies [a,\infty)\cap(-\infty,b] = [a,b] \in \sigma(C)\implies \sigma(C_0) \subseteq \sigma(C).\\$