$\mathscr{B} = \{ [a, b) | a< b \in \mathbb{R} \}$ is a basis for a Topology in $\mathbb{R}$

Question

I just want to ask if my proof for this problem is correct. $$\mathscr{B} = \{ [a, b) | a< b \in \mathbb{R} \}$$ is a basis for a Topology in $\mathbb{R}$ . Here is my proof:

1. Let $x \in \mathbb{R}$. Choose $B \in \mathscr{B}$ st $a \leq x <b$. Then $x \in B.$

2. Let $B_1= [a_1, b_1), B_2= [a_2, b_2) \in \mathscr{B}$.

   If $B_1 \cap B_2 = \varnothing$ then we are done.

   If otherwise, then there exists $x$ in the intersection. Choose $B_3= [a_3, b_3) \in \mathscr{B}$ st $a_3= \mbox{max}\{a_1, a_2 \}, b_3 =\mbox{min}\{b_1, b_2 \}$. Then $x \in B_3$ and $B_3 \subseteq B_1 \cap B_2$.

Therefore $\mathscr{B}$ is a basis.

Answer 1

Simpler and direct is
$[a,b) \cap [r,s) = [\max(a,r), \min(b,s))$.

In addition, it is necessary to note that every $r \in \mathbb{R}$ is in some base set. $[r, r+1)$ for example.