Topologies on $\mathbb{R}$

Question

I have found this in the internet:

Suppose$$\mathcal B=\{[a,b)\mid a\in\Bbb R\wedge b\in\Bbb Q\wedge a<b\},$$ $$\mathcal B_1=\{[a,b)\mid a\in\Bbb R\wedge b\in\Bbb R\wedge a<b\},$$and$$\mathcal B_2=\{[a,b)\mid a\in\Bbb Q\wedge b\in\Bbb Q\wedge a<b\}.$$Show that $\mathcal B$ and $\mathcal B_1$ are two different bases for the same topology on $\Bbb R$. Is $\mathcal B_2$ another basis for the same topology?

I already know that $\mathcal{B}_1$ is the Sorgenfrey line. What I did not know is that $b$ doesn't need to be real in order to get the line topology. How can i show this?

I have found that $\mathcal{B}_2$ is also a topology (on $\mathbb{R}$?) but it is not equivalent with the Sorgenfrey line. I cannot find anywhere proofs for this and I don't have an idea how to show this.

Has anyone a proof for this two statements?

In general: Is there somewhere an overview for all the topologies on $\mathbb{R}$ generated by all the intervals?

Answer 1

It is clear that $\mathcal B\subset\mathcal B_1$. On the other hand, if $a,b\in\Bbb R$ and $a<b$, if you take e sequence $(b_n)_{n\in\Bbb N}$ of rational numbers from $[a,b)$ such that $\lim_{n\to\infty}b_n=b$, then $[a,b)=\bigcup_{n\in\Bbb N}[a,b_n)$. So, $\mathcal B$ and $\mathcal B_1$ are bases of the same topology.

You will find an answer to your other question here.

Answer 2

$\mathcal{B}$ is a base for the Soregnfrey line, indeed.

$\mathcal{B}_1$ is too: for any basic set $[a,b)$ with $a<b$ both in $\Bbb R$ we can always find a rational $q$ such that $a < q < b$ and then $[a,q) \in \mathcal{B}_1$ is a subset of $[a,b)$. So these are bases for the same topology.

$\mathcal{B}_2$ is also a base but by definition for some topology $\tau_2$ that is second countable (as $\mathcal{B}_2$ is a countable family of sets!). And the Sorgenfrey line does not have a countable base. So $\tau_2$ is not the Sorgenfrey topology. You can also show that $[\sqrt{2}, 2)$ is open in the Sorgenfrey line but not open in $\tau_2$.