I am trying to prove that the topological space $(\mathbb R, \tau)$ is regular, where $\tau$ is all the subsets $U \subseteq \mathbb R$ open in $\mathbb R$ with the usual topology plus all the sets of the form for $U \cap \mathbb Q$. Note that $B_{r} (x) = (x -r, x +r)$
My attempt: Let $p \in \mathbb Q$, $U$ open in $(\mathbb R,\tau)$ with $p \in U$ and let $V \subseteq \mathbb R $
open in $\mathbb R$ with the usual topology such that $\mathbb R\setminus\mathbb Q \subseteq V$. Then for every $x \in \mathbb R\setminus \mathbb Q$, there exists $\varepsilon_x > 0$ such that
$$ \bigcup_{x\in \mathbb R\setminus\mathbb Q} B_{\varepsilon_x}(x) \subseteq V.$$and between any to real numbers, there is a rational number, so
$B_{\varepsilon_x}(x) \cap \mathbb Q \neq \emptyset$ and hence $V \cap
\mathbb Q \neq \emptyset$.Now, if $U \subset \mathbb R$ then $U $ is either $U$ is open in $\mathbb R$ with usual
topology or $U = G \cap \mathbb Q$ for some $G$ open in $\mathbb R$ with the
usual topologyCase 1: If $U$ is open in $\mathbb R$ with usual topology, then
for every $u \in U$, there exists $\varepsilon_u > 0$ such that $$ \bigcup_{u\in U}B_{\varepsilon_u}(u)\subseteq U,$$and between every two rational numbers, there is an irrational
number, that is, there exists $y \in V \cap B_{\varepsilon_u}(u) \neq
\emptyset$ for some $u \in U$Case 2: If $U = G \cap \mathbb Q$ for any $u_1, u_2 \in G\cap Q$ with $u_1 < u_2$, then there
exists $x \in V$ such that $u_1 \leq x \leq u_2$. Moreover, if $x -\varepsilon_x \leq u_1 $ and $u_2 \leq x +\varepsilon_x $, then $U
\cap V \neq \emptyset$. Otherwise, there exists $u_1^\prime, u_2^\prime
\in B_{\varepsilon_x} (x)$ and again $U \cap V \neq \emptyset$. So
$(\mathbb R,\tau)$ is not regular.
I am on the right track or is some clear misunderstanding? Also, the last part is very clunky, so I welcome suggestions on how to streamline it.
Best Answer
Indeed $\mathbb{P}$ (the irrationals) is closed in $\tau$. Suppose $0 \in U$ and $\mathbb{P} \subseteq V$ and both are open in $\tau$, and disjoint.
As $V$ contains the irrationals it is in particular Euclidean open (it's not of the new type).
Also, $0 \in (-\delta, \delta) \cap \mathbb{Q} \subseteq U$ for some $\delta>0$. Let $p$ be an irrational in $(-\delta, \delta)$. Then for some small open interval $I$ around $p$ inside $(-\delta, \delta)$ we have that $I \subseteq V$ (by openness of $(-\delta, \delta) \cap V$ in the Euclidean topology). But if $q$ is a rational inside $I$ (which exists) then this $q \in U \cap V$ showing that $U$ and $V$ cannot be disjoint after all, contradiction.