Base for standard topology on $\mathbb{R}$ Given a subbase $S = \{(a,\infty)\ \mid a \in \mathbb R \}\,\, \cup \{(-\infty, b)\ \mid b \in \mathbb R \}$ for the standard topology on $\mathbb R,$ we can get a base by taking finite intersections. Thus, we have $B = \bigg\{(a, b) := \{x ∈ \mathbb{R} \mid a < x < b\} \bigg| a, b ∈ \mathbb{R}\bigg\}.$ I know subbase is subset of base but intuitively, how are we getting a bigger set by taking finite intersections?
I have read that $B' = \{(a, b) | a, b \in \mathbb R, a < b\}  \cup \{(−\infty, b) | b \in \mathbb{R}\} \cup \{(a,\infty) | a \in \mathbb{R}\}$ is also a base for standard topology on $\mathbb{R}.$ This is making much more intuitive sense to me because $S \subset B'.$
My question is are $B$ and $B'$ the same base? Because it looks like $B'$ is bigger than $B.$
 A: Yes, you’re right, the base $\mathcal S$ generated by the subbase $\mathcal S$ also contains $\mathcal S$, because a single element $S \in \mathcal{S}$ is the intersection of the finite subset $\{S\}$ of $\mathcal{S}$. Intersections of two or more subbase elements (if the finite set contains a “left” and a “right” set) will be intervals (or $\emptyset$, depending on the sets) so the open intervals $(a,b)$ are certainly also in $\mathcal{B}$. People even include $\Bbb R = \bigcap \emptyset$ in the base, as $\emptyset$ is also a finite subset of $\mathcal{S}$ and this intersection $\bigcap \emptyset = X$ holds in the context of families of subsets of a set $X$), so that any collection $\mathcal{S}$ can serve as a subbase (and not only those that cover the whole set $X$, as Munkres and other texts demand).
So they’re not the same base, but luckily in the case of $\Bbb R$ we don’t really need $(\leftarrow, a)= (-\infty,a)$ to be in the base as it’s easily already a union of intervals too, so open by that virtue. They generate the same topology.
