Boundary and Interior Points of the set: Rational Numbers While I was studying metric spaces, I saw this question on the book:

Let $Q$ be the set of all rational numbers. What are $\operatorname{int}\Bbb Q$ and $∂\Bbb Q$?

I think the answer should be: $\operatorname{int}\Bbb Q$ is all the rational numbers, and the $∂Q$ is all the real numbers. But I do not have the answer and do not feel confident with my answer. Is my answer true?
edit: To make it clear why I though $\operatorname{int}\Bbb Q$ is all the rational numbers: if I take $X_0 = \frac 3 2$ and $r=1$, $(X_0-r,X_0+r)$ is a subset of $\Bbb Q$ and as long as I choose $X_0$ a rational number, for all values of $r$, I can get a rational number.
 A: You have to specify what metric space you are working in. If you say that $\mathbb{Q}$ is a subset of the metric space $(\mathbb{R},d)$ where $d$ is the usual metric given by the usual absolute value then we get that $\operatorname{int}{\mathbb{Q}} = \emptyset$ since if we take any $x \in \mathbb{Q}$ and $r \in \mathbb{R}$, $r>0$, then any open ball (i.e. open interval) centered at $x$ will contain irrationals, and hence it will not lie entirely in $\mathbb{Q}$.
Also we have $\partial \mathbb{Q} = \mathbb{R}$ since the boundary is defined to be the closure minus the interior, and the closure if simply $\mathbb{R}$ since if $x \in \mathbb{R}$ then $x$ "adheres" to $\mathbb{Q}$, i.e. there will be some sequence of rational numbers tending to $x$ (for instance, a decimal expansion).
A: It depends on the topology we adopt. In the standard topology or $\mathbb{R}$ it is $\operatorname{int}\mathbb{Q}=\varnothing$ because there is no basic open set (open interval of the form $(a,b)$) inside $\mathbb{Q}$ and $\mathrm{cl}\mathbb{Q}=\mathbb{R}$ because every real number can be written as the limit of a sequence of rational numbers. It also follows that 
$$\partial\mathbb{Q}=\mathrm{cl}\mathbb{Q}\setminus \mathrm{int}\mathbb{Q}=\mathbb{R}.$$
Comment on your edit: The definition of the interior of a set $C$ is the largest (wrt $\subseteq$) open set inside $C$, i.e., it is the union of all open sets in $C$. If (and only if) you can't find a basic open set inside $C$ whatsoever, then the interior of $C$ is empty.
A: Ok, here is a explanation why Int($\mathbb{Q}$) is empty.
We know that a point $p$ is an interior point of any set $S$ if there is a neighborhood of $p$ contained entirely in the set $S$. Now suppose that any real number $r$ is interior point of set $\mathbb{Q}$, then there should be a neighborhood of $r$ contained entirely in $\mathbb{Q}$, but we can't find any such neighborhood, because the set of irrationals is dense in $\mathbb{R}$.
Hence int($\mathbb{Q}$) is empty.
