What are the boundaries for subsets of $\mathbb{Q}$ in $\mathbb{R}$? Empty sets? Let $E := \{ p \in \mathbb{Q} : 0 < p < \sqrt 2 \}$. 
(1) $\sqrt 2$ is not an inner point of $E$ because $\forall \delta \in \mathbb{R} : \sqrt 2 \not \in \bigcup_{x \in I} B(x, \delta) \subset E \subset \mathbb{Q} \subset \mathbb{R}$ and $\sqrt 2$ is not an outer point because $\forall \delta \in \mathbb{R} : \sqrt 2 \not \in \bigcup_{x\in I_{2}} B(x, \delta) \subset E^{c} \subset \mathbb{Q} \subset \mathbb{R}$. So $\sqrt 2$ is a boundary point of $\mathbb{Q}$ in $\mathbb{R}$.
(2) Notice that $0 \in \mathbb{R}$ and $0 \in \mathbb{Q}$. Since $\forall \delta > 0 : 0 \in \bigcup_{x\in I_{3}} B(x, \delta) \subset E \subset \mathbb{Q} \subset \mathbb{R}$ so $0$ is an inner point. $0$ is also an outer point because $\forall \delta > 0 : 0 \in \bigcup_{x \in I_{4}} B(x, \delta) \subset E^{c} \subset \mathbb{Q} \subset \mathbb{R}$. No boundary point.
[conlclusion, investigating]
Can you declare the borders in some other numbers such as complex numbers or some other way? I think one cannot do it but is there some result about it?
 A: Associated with a subset $E$ in a topological space $X$ there are three subsets of $X$ which are pairwise disjoint and cover $X$. They are


*

*the set of interior points of $E$ or $\operatorname{int}E$, which consists of the points having a neighborhood which is contained in $E$,

*the set of exterior points of $E$ or $\operatorname{ext}E$, which consists of the points having a neighborhood which is contained in $X\setminus E$, and

*the set of boundary points of $E$ or $\partial E$, which consists of the points whose every neighborhood intersects $E$ and $X\setminus E$.


Let's see what these sets are in the case of $E=\{p\in\mathbb{Q}:0<p<\sqrt{2}\}\subseteq\mathbb{R}$.
First take any point $x<0$. Then $(x-|x|,x+|x|)\subseteq(-\infty,0)\subseteq\mathbb{R}\setminus E$, which means $x$ is an exterior point of $E$.
Then take any point $x>\sqrt{2}$ and denote $r=x-\sqrt{2}>0$. Then $(x-r,x+r)\subseteq(\sqrt{2},\infty)\subseteq\mathbb{R}\setminus E$. Again $x$ is an exterior point of $E$.
Finally take any point $x\in[0,\sqrt{2}]$ and a neighborhood $U$ of $x$. The neighborhood $U$ contains an interval $(x-r,x+r)$ for some (small) $r>0$. Because $x$ is in $[0,\sqrt{2}]$, this interval $(x-r,x+r)$ intersects the interval $(0,\sqrt{2})$, and we know that if the intersection of two open intervals is nonempty, then it (being an open interval) contains both a rational and an irrational number. Furthermore the rational numbers in $(x-r,x+r)\cap(0,\sqrt{2})$ are in $E$ and the irrational numbers are in $\mathbb{R}\setminus E$. We have shown that every neighborhood of $x$ intersects both $E$ and $\mathbb{R}\setminus E$, which means $x$ is a boundary point of $E$.
The conclusion is that $\operatorname{int}E=\emptyset$, $\operatorname{ext}E=\mathbb{R}\setminus[0,\sqrt{2}]$ and $\partial E=[0,\sqrt{2}]$.
A: The closure of $E$ in $\mathbb{R}$ is $[0,\sqrt 2]$ and the interior of $E$ in $\mathbb{R}$ is empty. So the boundary of $E$ in $\mathbb{R}$ is the closed interval $[0,\sqrt{2}]$.
A: I think the word you're looking for is boundary. The boundary of a set $A \subset X$ (in $X$) is defined as $\overline{A} \setminus A^\circ$ and it's sometimes written $\partial A$. In your example the boundary of $E$ in $\mathbb{Q}$ is $\{0\}$ as $(0,\sqrt{2})\cap \mathbb{Q}$ is open in $\mathbb{Q}$ and $E$ has $0$ as a limit point.
