$\mathbb{R}$ equiped with topology generated by $(a,b)$ and $(a,b)\cap \mathbb{Q}$ $\mathbb{R}$ equiped with topology generated by $(a,b)$ and $(a,b)\cap \mathbb{Q}$
then which of the following statements is correct?


*

*It is normal

*It is regular

*$\mathbb{R}\setminus \mathbb{Q}$ is dense 

*$\mathbb{Q}$ is dense.
I am not able to find out what is the role here of the open sets of the form $\mathbb{Q}\cap (a,b)$. Thank you for help.
 A: I’m going to assume that you mean the topology $\tau$ that has as a subbase the family $\mathscr{S}$ of all sets of the forms $(a,b)$ and $(a,b)\cap\Bbb Q$ for $a,b\in\Bbb R$ with $a<b$.
Notice that if $a<b$ and $c<d$, then
$$(a,b)\cap\Big((c,d)\cap\Bbb Q\Big)=\Big((a,b)\cap(c,d)\Big)\cap\Bbb Q\;.$$
If $(a,b)\cap(c,d)=\varnothing$, this is empty. Otherwise let $u=\max\{a,c\}$ and $v=\min\{b,d\}$; then $(a,b)\cap(c,d)=(u,v)$, and
$$(a,b)\cap\Big((c,d)\cap\Bbb Q\Big)=(u,v)\cap\Bbb Q\;.$$
This is the key step in showing that $\mathscr{S}$ is closed under finite intersections and is therefore actually a base for $\tau$.
Let $\mathscr{E}$ be the usual topology on $\Bbb R$; $\mathscr{E}$ is generated by the open intervals $(a,b)$, so $\mathscr{E}\subseteq\tau$. That is, every set that’s open in the usual topology is also open in $\tau$. However, $\tau$ has lots of open sets that aren’t in $\mathscr{E}$: every set of the form $(a,b)\cap\Bbb Q$ is open in $\tau$ but not in $\mathscr{E}$. Take the set $(0,1)\cap\Bbb Q$, for instance: it consists of the rational numbers lying strictly between $0$ and $1$, without any of the irrationals in $(0,1)$. It’s not open in the usual topology: for instance, $\frac12$ is in it, but every $\epsilon$-interval $\left(\frac12-\epsilon,\frac12+\epsilon\right)$ around $\frac12$ contains irrational numbers and therefore cannot be a subset of $(0,1)\cap\Bbb Q$.
Now let $U\in\tau$; then $U$ is the union of members of $\mathscr{S}$, so there are families $\mathscr{I}_0$ and $\mathscr{I}_1$ of open intervals in $\Bbb R$ such that
$$U=\bigcup_{I\in\mathscr{I}_0}I\cup\bigcup_{I\in\mathscr{I}_1}(I\cap\Bbb Q)\;.$$
Let $U_0=\bigcup\mathscr{I}_0$ and $U_1=\bigcup\mathscr{I}_1$; then $U_0$ and $U_1$ are ordinary Euclidean open sets, and $U=U_0\cup(U_1\cap\Bbb Q)$. Thus, every open set in the topology $\tau$ has the form $U_0\cup(U_1\cap\Bbb Q)$ for some Euclidean open sets $U_0,U_1\in\mathscr{E}$.

Every $\tau$-open set is the union of an ordinary open set and a relatively open subset of $\Bbb Q$ in its usual topology.

For example, $\Bbb Q$ is open in $\tau$: it’s the union of the ordinary open set $\varnothing$ and the relatively open subset $\Bbb Q$ of $\Bbb Q$. The set $(0,1)\cup(2,3)\cup\{q\in\Bbb Q:q>0\}$ is open in $\tau$ because it’s the union of the ordinary open set $(0,1)\cup(2,3)$ and the set of positive rationals, which is a relatively open subset of $\Bbb Q$ in the usual topology on $\Bbb Q$.

*

*One of the examples in the preceding paragraph gives you the answer to (3).


*(4) isn’t too hard either. Suppose that $\alpha$ is irrational and that $U\in\tau$ is a $\tau$-nbhd of $\alpha$. We know that $U=U_0\cup(U_1\cap\Bbb Q)$ for some ordinary Euclidean open sets $U_0$ and $U_1$. Is it possible for this set not to contain any rational numbers?


*For (2) show that $\operatorname{cl}_\tau\Big((a,b)\cap\Bbb Q\Big)=[a,b]$ whenever $(a,b)\cap\Bbb Q\in\mathscr{S}$, and use this to show that $\langle\Bbb R,\tau\rangle$ is not regular.


*Once you have (2), (1) is trivial, since $\tau$ is a $T_1$ topology.
