Subbases of a topology Let  $\mathscr { S } $    the class of all closed intervals $[a,b]$, where $a$ and $b$ are rational , i.e. $a,b\in \mathbb{Q}$  and $a<b$ . Show that  $\mathscr{S}\cup \{\{p\}:p\in\mathbb{ Q }\}$  is a basis of the topology $\tau$ of the real line $\mathbb{R}$ generated by $\mathscr {S}$ .
Having problems show that the intersection of two base elements can be expressed as a union of elements of the base.
 A: Let us define $\mathscr{T}:=\mathscr{S}\cup \left\{ \{ p\} :p\in \mathbb{Q}\right\}$.  You first should check that each element of $\mathscr{T}$ is actually open in this topology.  Of course, every element in $\mathscr{S}$ is, and $\{ p\} =[a,p]\cap [p,b]$ for $a<p<b$, and so $\{ p\}$ is also open in this topology.  It is perhaps just easier to think of $\mathscr{T}$ has $\mathscr{T}=\left\{ [a,b]:a\leq b,a,b\in \mathbb{Q}\right\}$, that is, it is the same as $\mathscr{S}$ except we allow that $a=b$.
It is obvious that $\mathscr{T}$ covers $\mathbb{R}$.
Let $x\in [a_1,b_1]\cap [a_2,b_2]$.  Without loss of generality, $a_1\leq a_2$.  Then, in order for this intersection to be non-empty, we must have $a_2\leq b_1$.  But then $[a_1,b_1]\cap [a_2,b_2]=[a_2,b_1]$ (and of course $x\in [a_2,b_1]$).
A: The class $\mathscr{S}'=\mathscr{S}\cup\{\{p\}:p\in\mathbb{Q}\}\cup\{\emptyset\}$ is closed under binary intersections, because the intersection of two closed proper intervals is either a closed proper interval or a singleton, unless it is empty.
Every element of $\mathscr{S}'$ is an intersection of elements of $\mathscr{S}$.
What is $\bigcup\mathscr{S}$?
