Take $X = \mathbb{R}$, and find the interior, boundary and closure of the following subsets:
a) $E_1 = \{1,\frac{1}{2},\frac{1}{3},…\}$
b) $E_2 = [0,1)$
c) $ E_3 = \mathbb{R} \setminus \mathbb{Q}$
For a) the interior must be empty since since every open ball around a point in $E_1$ will contain points not in $E_1$. It also seems like every point in $E_1$ is a boundary point since for every $x \in E_1$ an open ball around $x$ will contain points both in $E_1$ and not in $E_1$. Lastly it seems like the closure of $E_1$ is $E_1 \setminus \{1\} $ since I don't see how you could make a sequence in $E_1$ converging to $1$.
b) The interior I believe is $(0,1)$, boundary $\{0,1\}$ since all open balls at $0$ and $1$ contain elements both in and out of $E_2$. If Im not wrong the closure of a set is the union of the boundary and interior so the closure must be $[0,1]$?
c)
Empty interior because no open ball around an irrational number contains only irrationals. Boundary should equal $E_3$ because all open balls around an irrational number will contain both irrational and rational numbers. And lastly the closure is empty? I believe it's not possible to make a sequence of irrationals converge to any element in $\mathbb{R}$
Would appreciate if someone could help verify/correct my answers.
Best Answer
First of all I discuss all the case here on $\mathbb{R}$ with usual topology.
(a) Interior of $E_1=\{\frac{1}{n}: n \in \mathbb{N} \} $ is empty as every neighborhood of each point contains irrationals, but they are not belonged in $E_1$
Closure of $E_1$, $\bar{E_1}=E_1∪\{0\}$ , as $0$ is the only limit point of $E_1$.
Boundary of $E_1=E_1∪\{0\}$ , as each interval containing the point of $E_1$ also contains the points of exterior of $E_1$.
(b) Interior of $E_2=[0,1)$ is $(0,1)$ as each point in $E_2$ except $0$ has a neighborhood which is entirely contained in $E_2$.
Closure of $E_2$ , $\bar{E_2}=E_2∪[0,1]=[0,1]$ , as each point of $E_2$ together with $\{1\}$ is a limit of $E_2$ , as every neighborhood of these points contains infinitely many points of $E_2$.
Boundary of $E_2=\{0,1\}$, as every neighborhood of these points contains the interior as well as the exterior of $E_2$.
(c) Interior of $E_3=\mathbb{R}\setminus\mathbb{Q}$ is empty as every neighborhood of each point contains rationals, but they are not belonged in $E_3$
Closure of $E_3$,$\bar{E_3}=E_{3}∪\mathbb{R}=\mathbb{R}$ ,as $E_3$ is dense in $\mathbb{R}$.
Boundary of $E_3=\mathbb{R}$ , each neighborhood of any reals contains infinitely many points of $E_3$ as well as of $\mathbb{Q}$.