Find the interior and closure of these sets(Intersection between two topological spaces) This question is in my lecture's note.
$Q)$ There are three topological spaces that $(\mathbb{R},T_1)$, $(\mathbb{R},T_2)$ and  $(\mathbb{R},T_3)$ 
Here the $T_1$ is a countable complement topology(cocountable topology) and $T_2$ is a usual Topology on $\mathbb{R}$ respectively.
Plus the $(\mathbb{R},T_3) $ is a topology having the basis as "$T_1 \cap T_2$"
Find the $int(A)$, $\bar A$, $int(B)$ and  $\bar B$ on $T_3$ for $A= \mathbb{R} - \{ {1 \over n} \vert n \in \mathbb{N}  \}, B = \mathbb{Q}$ 

There are three-type open sets $G$(or closed set $F$) on $T_3$


*

*$G_1$(open set in $T_1$) considering $G_1 \cap \mathbb{R}$  for $\mathbb{R}$ in $T_2$. Hence the $G_1$ is a open on $T_3$. Likewise $F_1$(closed set in $T_1$) is a closed set in $T_3$

*$G_2$(open set in $T_2$) considering $G_2 \cap \mathbb{R}$  for $\mathbb{R}$ in $T_1$. Hence the $G_2$ is a open on $T_3$. Likewise $F_2$(closed set in $T_2$) is a closed set in $T_3$

*Surely $G_1 \cap G_2$ is a open and $F_1 \cap F_2$ is a closed on $T_3$ respetively.
So my answer is $int(A) = A $, $\bar A =\mathbb{R} $, $int(B)=\phi$ and  $\bar B = \mathbb{Q}$
(e.g. When the case of the $A$, $A$ is a open on $T_1$ Hence it is a open on $T_3$ )
But the lecture claimed $int(A) = A-\{0\} $, $\bar A =\mathbb{R} $, $int(B)=\phi$ and  $\bar B = \mathbb{R}$.
What did I had a mistake?(I believe his answer is not correct.) Any help would be appreciated.
 A: $T_1\cap T_2$ is the collection of sets that are open in both the co-countable and the usual topology, so it’s the family of all $U\in T_2$ such that $\Bbb R\setminus U$ is countable. Another way to say this is that $T_1\cap T_2$ is the family of all $\Bbb R\setminus F$ such that $F$ is a countable closed set in the usual topology on $\Bbb R$.
This family is closed under taking finite intersections, because the union of finitely many countable closed sets in the usual topology is a countable closed set in the usual topology. Thus, $T_1\cap T_2$ is a base for the topology $T_3$, and $T_3$ is simply the set of all unions of members of $T_1\cap T_2$. Note that this implies that every non-empty member of $T_3$ is co-countable and open in the usual topology. That is, $T_1\cap T_2$ is actually closed under taking arbitrary unions, so in fact $T_3=T_1\cap T_2$.
Let’s find $\operatorname{cl}_{T_3}A$. Certainly it includes all of $A$, so the only question is whether it includes any of the points $\frac1n$ as well. If $U$ is an open nbhd of $\frac1n$ in the topology $T_3$, then $\Bbb R\setminus U$ is countable, so $U$ is uncountable. And $\Bbb R\setminus A$ is only countable (why?), so $U\nsubseteq\Bbb R\setminus A$, and hence $U\cap A\ne\varnothing$. Thus, every open nbhd of $\frac1n$ intersects $A$. What does this tell you about $\operatorname{cl}_{T_3}A$?
I’ll leave it at this for now; see if you can finish off the other two parts as well.
