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:
Everywhere I looked, there is always a quite confusing explanation.
I believe the first two denote interior, the third boundary, and the fourth closure.
The interior of a set is the largest open set included in it. Equivalently, it is the union of all open sets included in it: $$\operatorname{int} S\equiv\bigcup\{U\,|\,U\text{ is open and }U\subseteq S\}.$$
Analogously, the closure of a set is the smallest closed set that includes it. Equivalently, it is the intersection of all closed sets including it: $$\overline S\equiv\bigcap\{C\,|\,C\text{ is closed and }S\subseteq C\}.$$
The boundary of a set is simply the set of points in the closure but not in the interior: $$\partial S\equiv \overline S\setminus\operatorname{int} S.$$
In the case of the rationals, one has that \begin{align*} \operatorname{int}\mathbb Q=&\,\varnothing,\\ \overline{\mathbb Q}=&\,\mathbb R,\\ \partial{\mathbb Q}=&\,\mathbb R. \end{align*}
Why? Note that every non-empty open subset of $\mathbb R$ contains a non-empty open interval and every non-empty interval, in turn, contains irrational numbers. Hence, $\mathbb Q$ cannot have any non-empty open set as its subset, so the largest open set included in it is the empty set.
On the other hand, if $C\subseteq\mathbb R$ is a closed set containing $\mathbb Q$, then $C$ must contain all irrationals, too, because every irrational is a limit of some sequence of rational numbers and the set $C$ is closed. Hence, it must be the case that $C=\mathbb R$. It follows that the smallest closed set containing $\mathbb Q$ is, in fact, $\mathbb R$.
Int $\mathbb{Q} = \emptyset$, since, if you have an arbitrary $x \in \mathbb{Q}$, no matter how you choose an $r>0$, the x-centered ball($B(x,r)$), with $r$ radius will always contain an irrational number, therefore it is not an internal point. Since our $x$ was arbitrary, the set of interal points must be empty. $\rightarrow Int \mathbb{Q} = \emptyset$
Let us discuss $\bar{\mathbb{Q}}$ next. $\bar{\mathbb{Q}} = \mathbb{Q} \cup \mathbb{Q'}$. Let us have an arbitrary $x \in \mathbb{R}$, no matter how you choose an $r>0$, the x-centered ball($B(x,r)$), with $r$ radius will always contain a rational number, therefore, every $x$ will be element of $\mathbb{Q'}$, so $\bar{\mathbb{Q}}= \mathbb{Q} \cup \mathbb{Q'}= \mathbb{Q} \cup \mathbb{R}=\mathbb{R}$. For that, we tend to say, that rational numbers are dense.
$\partial\mathbb{Q}=\bar{\mathbb{Q}}$ \ $Int(\mathbb{Q})$. Since we discussed, that $Int(\mathbb{Q}) = \emptyset$, that will be $\mathbb{R}$ too.
By the way, $\mathbb{Q}^\circ$ and int $\mathbb{Q}$ are equivalent.
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 $\operatorname{int}(\mathbb{Q})=\emptyset$
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 $\operatorname{cl}(\Bbb{Q})=\Bbb{Q} \cup \Bbb{Q}'=\Bbb{R}$.
