The interior of $\mathbb{R} \times \mathbb{Q}$ A question says, find the closure and interior of the sets $\mathbb{R} \times \mathbb{R}$ and  $\mathbb{R} \times \mathbb{Q}$. The answers say $\mathbb{R}^2$ and $\emptyset$ respectively for both. Why isn't the interior of  $\mathbb{R} \times \mathbb{Q}$, $\mathbb{R} \times \emptyset$ because the interior of $\mathbb{R}$ is $\mathbb{R}$? Does $\mathbb{R} \times \emptyset$ even make sense as a set?
Edit: For clarity I mean in $\mathbb{R}^2$ with the usual topology. I was told one should always assume the usual topology when it is not specified. 
 A: Either you’ve phrased the given answers unclearly, or at least one of them is wrong: the closures of $\Bbb R\times\Bbb R$ and $\Bbb R\times\Bbb Q$ are both $\Bbb R\times\Bbb R$, since every non-empty open subset of $\Bbb R\times\Bbb R$ intersects both $\Bbb R\times\Bbb R$ and $\Bbb R\times\Bbb Q$. The interior of $\Bbb R\times\Bbb R$ is also $\Bbb R\times\Bbb R$: $\Bbb R\times\Bbb R$ is itself an open set in $\Bbb R\times\Bbb R$. The only one of the four that is $\varnothing$ is the interior of $\Bbb R\times\Bbb Q$.
$\Bbb R\times\varnothing$ does indeed make sense as a set: by definition it’s $\{\langle x,y\rangle:x\in\Bbb R\text{ and }y\in\varnothing\}$, which is clearly just $\varnothing$, since there are no $y\in\varnothing$. Thus, $\Bbb R\times\varnothing$ actually does turn out to be the right answer after simplification. And it is true in general that $$\operatorname{int}(A\times B)=(\operatorname{int}A)\times(\operatorname{int}B)\;,$$ though this is something that does require proof.
