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 $intQ$ and $∂Q$?


I think the answer should be: $intQ$ 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 $intQ$ is all the rational numbers: if I take $X_0$ = 3/2 and r=1, $(X_0-r,X_0+r)$ is a subset of $Q$ and as long as I choose $X_0$ a rational number, for all values of r, I can get a rational number.
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 $r \in \mathbb{R}$, $r&gt;0$, 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).
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.