Let $S = \{(a,b] \cup [−b,−a) : a < b\text{ are real numbers}\}$. Show that $\sigma(S)$ is smaller than the Borel $\sigma$-algebra of $\mathbb R$. 
Let $S = \{(a,b] \cup [−b,−a) : a < b\text{ are real numbers}\}$. Show that $\sigma(S)$ is smaller than the Borel $\sigma$-algebra of $\mathbb R$.

How can I find a Borel set which is not in the sigma algebra generated by $S$?
If not that, can I use any cardinality argument?
Kindly help.
 A: Hint: Any (nonempty) union or intersection of elements of this $S$ contains both positive and negative reals. This is true of the generating "double intervals" which are symmetrically chosen unions of two half intervals. So when these are combined to fill out the generated sigma algebra, one only gets sets having both positive and negative reals in them (or the empty set). So it would seem one could not generate say the interval $(1,2).$
Added: In a comment to Asaf's answer the OP has asked for clarification. Call any set $Y$ of reals "symmetric"
 provided $y \in Y$ iff $ -y \in Y.$ Note now that each generator of your sigma algebra $S$ is symmetric in this sense. Let $T$ denote the subset of $S$ consisting of its symmetric sets. Show that $T$ is closed under complement, under countable union, and under countable intersection. [These are fairly straightforward to show.] Then $T$ is a sigma algebra, and contains all the generating sets of $S$, and you can conclude that $T=S,$ so that in fact each subset in $S$ is symmetric. Then since $(1,2)$ is not symmetric it does not lie in the sigma algebra $S$.
A: Right off the bat, you can't use cardinality arguments. There are only $2^{\aleph_0}$ Borel sets, and there are clearly $2^{\aleph_0}$ sets in $S$ as well.
For a more explicit set, note that if you could have generated all the open intervals, then you could have generated all the Borel sets. So it suffices to try and find an open interval that cannot be generated by these sets; or alternatively note that you can generate every open interval and disprove that statement.
Note that the elements in $S$ are all symmetric around $0$. What happened when you intersect things which are symmetric around $0$, or take unions of them?
Now find a counterexample.
A: Let $\Sigma = \{ A \subset \mathbb{R}| x \in A \Rightarrow -x \in A \}$. It is easy to check that $\Sigma$ is a $\sigma$-algebra.
It is clear that $S \subset \Sigma$, hence $\sigma(S) \subset \Sigma$.
It is also clear that $\sigma(S) \subset {\cal B}$.
We have $\{1\} \in {\cal B}$ and $\{1\} \notin \Sigma$, so we see that
$\sigma(S) \neq {\cal B}$.
