# Why this is not a sigma algebra

Let be $\Omega$ the interval $(0,1]$ and let $\mathcal{F}$ be the set of all sets of the form $(a_0,a_1]\cup(a_2,a_3]\cup\cdots\cup(a_{n-1},a_n]$, where $0\le a_0\le a_1\le\cdots\le a_n\le 1$. Show that $\mathcal{F}$ is not a $\sigma$-algebra.

I have attempted a similar question:
Let be $\Omega = [0,1]$ and $G = \{[0,\frac{1}{3}],[\frac{1}{2},1]\}$. What is $\sigma(G)$?
My answer is the following, not sure whether this is correct:

$$\sigma(G) = \left\{\emptyset, \left[0,1\right], \left[0,\frac{1}{3}\right], \left[\frac{1}{2},1\right], \left[\frac{1}{3},\frac{1}{2}\right], \left[0,\frac{1}{3}\right]\cup\left[\frac{1}{2},1\right], \left[\frac{1}{3},1\right], \left[0,\frac{1}{2}\right]\right\}.$$

But using a similar reasoning, I wasn't able to to prove the first question is not a $\sigma$-algebra. I think if we apply all properties of $\sigma$-algebra, the resulting sets are all of the form $(a_0,a_1]\cup(a_2,a_3]\cup\cdots\cup(a_{n-1},a_n]$. My intuition is to try to come up with a set of the form $[a_{i-1},a_i)$, but I found no clue.

Side questions: We know that the usual subsets of $\mathbb{R}$ are in $\mathcal{B}$($\mathbb{R}$). And any one point set, $\mathbb{N}$ and $\mathbb{Q}$ are in $\mathcal{B}$($\mathbb{R}$). But does $\mathcal{B}$($\mathbb{R}$) contains the set of all irrational numbers? Why?

Any helps will be greatly appreciated. Thanks!

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Your "similar question" isn't similar enough. If $a_0,\ldots,a_n$ is a fixed finite set of points, and you find the sigma-algebra generated by the set of intervals bounded by them, then if I'm not mistaken, you'll get a finite sigma-algebra.
But "the set of all sets of the form $(a_0,a_1]\cup(a_2,a_3]\cup\cdots\cup(a_{n-1},a_n]$, where $0\le a_0\le a_1\le\cdots\le a_n\le 1$" would normally mean something else: there are infinitely---indeed uncountably---many sequences $0\le a_0\le a_1\le\cdots\le a_n\le 1$, and all of them would be used in your generating set. This won't give you a finite sigma-algebra.
Consider $\cap_n (a \ - \frac{1}{n}, a]$. This is equal to $\{a \}$, which isn't of the form required to be in $\mathcal{F}$. So $\mathcal{F}$ isn't closed under countable intersections.