Show that B = {[a, b) ⊂ ℝ | a < b} is a basis for a topology on ℝ

Answer:Remember that a set [tex]\mathcal{B}[/tex] is a base for some topology on [tex]\mathbb{R}[/tex] if satisfy the following properties:1. [tex]\mathbb{R}=\cup\{B: B\in\mathcal{B}\}[/tex]2. For [tex]B, B^* \in \mathcal{B}[/tex], If [tex]p\in B\cap B^*[/tex] then exist [tex]B_p\in\mathcal{B}[/tex] such that [tex]p\in B_p\subset B\cap B^*[/tex].Now, for [tex]B=\{[a,b)\subset\mathbb{R}| a < b\}[/tex] we verify the above properties:1. It's clear that [tex]\mathbb{R}=\cup_{a,b \text{ with }a<b}[a,b)[/tex]2. Let [tex]B=[a,b), B^*=[c,d) \in \mathcal{B}, p\in B\cap B^*[/tex]. Without loss of generality suppose that [tex]a<c, b <d[/tex]. Then [tex]c\leq p < b[/tex], this implies that [tex]p\in[c,b)[/tex] and [tex]B_p=[c,d) \in \mathcal{B}[/tex] and [tex]B_p\subset B\cap B^*[/tex].Then, B satisfy the two properties. This show that B is a basis for a topology in [tex]\mathbb{R}[/tex]