Question

Show that B = a < b is a basis for a topology on ℝ

Answer 1

Remember that a set
`\mathcal{B}` is a base for some topology on
`\mathbb{R}` if satisfy the following properties:

1.
`\mathbb{R}=\cup\{B: B\in\mathcal{B}\}`

2. For
`B, B^* \in \mathcal{B}`, If
`p\in B\cap B^*` then exist
`B_p\in\mathcal{B}` such that
`p\in B_p\subset B\cap B^*`.

Now, for
`B=\{[a,b)\subset\mathbb{R}| a < b\}` we verify the above properties:

1. It's clear that
`\mathbb{R}=\cup_{a,b \text{ with }a<b}[a,b)`

2. Let
`B=[a,b), B^*=[c,d) \in \mathcal{B}, p\in B\cap B^*`. Without loss of generality suppose that
`a<c, b <d`. Then
`c\leq p < b`, this implies that
`p\in[c,b)` and
`B_p=[c,d) \in \mathcal{B}` and
`B_p\subset B\cap B^*`.

Then, B satisfy the two properties. This show that B is a basis for a topology in
`\mathbb{R}`