I need to prove:
Show that the basis $B' = \{(a,b); a<b, a, b \in \mathbb{Q}\}$, of subsets of $\mathbb{R}$, spans the usual topology of $\mathbb{R}$. First of all, the usual topology of $\mathbb{R}$ is the topology generated by the open intervals of the form $(a,b)$. What do I need to show? I need to show that this is a topology, but how to show that it's the usual one?
For example, I know that $B'$ is a basis because for every point $x$ there is an interval such that $x\in (a,b)$ and if $x\in B_1\cap B_2$, that is, another open interval, then there is another open interval that contains these two.
How do I show that this is the usual topology? And what changes when we work with intervals defined by rationals?
Best Answer
To show that the two basis $\mathscr B=\{(a,b):a,b\in \Bbb R\}$ and $\mathscr B^{'}=\{(a,b):a,b\in \mathbb Q\} $generate the same topology we need to show that :
For the first one use the fact there exists a rational between any two reals(since $\overline {\Bbb Q}=\Bbb R)$
The second one is obvious