TITLE: To understand some terminology of metric spaces
QUESTION [9 upvotes]: In the course of my study of metric spaces I've come across some terminology which I can't seem to understand completely.
So, assuming $X=\mathbb{R}$, and $\mathbb{Q}\subset X$ is the set of the rational numbers, what exactly are:

$\mathbb{Q}^\circ$
int $\mathbb{Q}$
$\partial\mathbb{Q}$
$\bar{\mathbb{Q}}$

Everywhere I looked, there is always a quite confusing explanation.

REPLY [3 votes]: $\Bbb{Q}° = \operatorname{int}(\Bbb{Q})$

In general, given a metric space $X$ we have that $x$ is an interior point of $X$ if there exist an $\epsilon > 0$ such that $B(x,\epsilon) \subset X$. The set $\operatorname{int}(X)$ consist of all of the interior points of $X$.
In our case we want to find $\operatorname{int}(\mathbb{Q})$. Let's see that $\operatorname{int}(\mathbb{Q}) = \emptyset $. Suppose its not, so there exist $x \in \operatorname{int}(\mathbb{Q})$ but what does this mean? Well, that there exists $\epsilon >0$ such that $B(x,\epsilon) \subset \Bbb{Q}$ but, we know that between any two real numbers there exist an irrational number between them, (since the irrationals are dense in $\Bbb{R}$) so, we know that there must be an irrational $y \in (x,x+\epsilon)$ which contradicts the fact that $B(x,\epsilon)=(x-\epsilon,x+\epsilon) \subset \Bbb{Q}$ So $\operatorname{int}(\mathbb{Q})=\emptyset$

$\operatorname{cl}(\mathbb{Q})=\overline{\Bbb{Q}}$

The closure of a set $X$ is the union between $X$ and its accumulation points, that is $\operatorname{cl}(X)=X \cup X'$ where $X'$ is the set of accumulation points. So, what is an accumulation point? Well, $x$ is said to be an acummulation point of $X$ if for every $\epsilon >0$, $B(x,\epsilon) \cap X \neq \emptyset$. And therefore $X'$ consist of all the acumulation points of $X$
Now in our case we have $\Bbb{Q}$ and we want to find out $\Bbb{Q}'$. Lets see that $\Bbb{Q}'= \mathbb{R}$. Take $x \in \Bbb{R}$, we want to see that $x \in \Bbb{Q}'$ (the other inclusion is trivial). Ok since $\Bbb{Q}$ is dense in $\Bbb{R}$ we have that for any $\epsilon >0$, $B(x,\epsilon)\cap \Bbb{Q} \neq \emptyset$ and therefore $x \in \Bbb{Q}'$, concluding that $\operatorname{cl}(\Bbb{Q})=\Bbb{Q} \cup \Bbb{Q}'=\Bbb{R}$.

Finally, $\partial(X)=\operatorname{cl}(X)-\operatorname{int}(X)$. So in our case, $\partial(\Bbb{Q})=\operatorname{cl}(\Bbb{Q})-\operatorname{int}(\Bbb{Q})=\Bbb{R}-\emptyset = \Bbb{R}$