# Interior and closure $\mathbb{Q}$ in $\mathbb{R}$, but not with the euclidean topology

I have to compute the interior and closure of $\mathbb{Q}$ in $\mathbb{R}$, but I know that in the Euclidean topology, the interior is the empty set and the closure is whole $\mathbb{R}$. I have to compute these when $\mathbb{R}$ is endowed with the following topologies:

1. The discrete topology: the topology whose opens sets are all subsets of $X$.
2. The co-finite topology: the topology whose open sets are the empty set and complements of finite subsets of $X$
3. The co-countable topology: the topology whose open sets are the empty set and complements of subsets of $X$ which are at most countable.

I think this lemma would be very useful: $x$ is in the interior of $A$ if and only if there exists $U\in \mathcal{B}_x$ (a basis of neighborhoods around $X$) such that $U \subset A$. And $x$ in the closure of $A$ if and only if, for all $U \in\mathcal{B}_x, U \cap A \neq \emptyset$

Who can help me? Thank you!

The following is also useful: the interior of $A$ is the largest open subset of $A$, and the closure of $A$ is the smallest closed subset that contains $A$.
In the discrete topology, all subsets are open and closed. By the above the interior of $\mathbb{Q}$ is $\mathbb{Q}$ and so is its closure.
In the cofinite topology, the only closed sets are the finite ones, and $\mathbb{R}$ itself. So the closure of any infinite set is $\mathbb{R}$. The only open sets are the empty set and all sets with finite complements, none of which fit inside $\mathbb{Q}$. So the interior is empty here.
In the co-countable topology $\mathbb{Q}$ is closed, so equal to its own closure. No non-empty open subset fits inside the rationals, as in the cofinite case, so again the interior is empty.