Is true that these two topologies are finer than the standard topology on line and are not comparable I was asked to exibe two topologies non comparable and finer than the standard topology of line. I stated:


*

*The Sorgenfrey line $\mathbb{R}_l$ and

*The topology whose base is defined as:
$$\{(a,b) : a < b, a,b \in \mathbb{R}\}\cup \{(a,b)\cap \mathbb{Q} : a < b, a, b\in \mathbb{R}\}$$
Is this right?
 A: You are right. For the Sorgenfrey line, note that its base sets are the half open intervals $[a,b)$ with $a,b \in \mathbb R$, and we have
$$
 (a,b) = \bigcup_n \left[a+\frac{1}{n}, b\right)
$$
hence it contains every set open in the standard topology. For your second example it refines the standard topology by construction, and as every set open in the standard topology contains some irrational number, the refinement is proper.
But one more obvious topology might be $(\mathbb R, \mathcal P(\mathbb R))$, i.e. the discrete topology where every subset is open. This refines each topology, in particualr the standard one.
A: Call the second topology $R_Q.$ There are many ways to show that $R_l$ and $R_Q$ are incomparable. 
(1).  No member of the given base $B$ for $R_Q$ has a $<$-least member. And $[0,1)$ is $R_l$-open. If $[0,1)$ were $R_Q$-open then $0\in b\subset [0,1)$ for some $b\in B.$ But then  $\min b=0,$ a contradiction. Therefore $[0,1)\in R_l$  \  $R_Q.$
(2).  Every member of the standard base for $R_l$ is an infinite $<$-convex set. And $Q\cap (0,1)$ is $R_Q$-open. So if $Q\cap (0,1)$ were $R_l$-open it would have an infinite $<$-convex subset. Therefore $Q\cap (0,1)\in R_Q$ \ $R_l.$
