general topology: closure, int, derivative inf, sup, of a sequence I'm working on some exercises on topology on a book I found. What I'm trying to find is the $\operatorname{Int}{X}$, $\partial{X}$, $\bar{X}$, ${X}'$, $\inf X$ and $\sup X$. And if the terms of the sequence form an open or closed set.
Consider the following set $X=\{ x_n \mid n \in \Bbb N \}$ where
\begin{align*}
x_{n} =
\begin{cases}
\dfrac{1}{n} + 2   & \text{if n is even}\\\\ 
\dfrac{n}{n+1} - 1 & \text{if n is odd}&
\end{cases}.
\end{align*}
The topology that we are working is the usual topology in $\mathbb{R}.$
 A: You haven't written the topological space in which you are workin. For different spaces you may have different results.
However, let suppose that we work in $\mathbb{R}$ with the usual topology.
Then, notice that your set is defined using a sequence, hence all terms of the sequence belong to the set.
Since you are using a book, you can refer to the definitions of the notions you want to find there.
By the definition of the interior of the set, yes $Int X= \emptyset .$
Before answering the other questions, notice that the given sequence is constructed with the help of two monotone bounded subsequences $a_k =\frac{1}{2k}+2=\frac{4k+1}{2k}$ and $b_k=\frac{2k-1}{2k}-1=-\frac{1}{2k}$. The limits of these two subsequences are $\lim_{k\to\infty}a_k=2$ and $\lim_{k\to\infty}b_k=0$, hence $\overline{X}=X\cup\{2,0\}, \; \partial{X}=X\cup\{2,0\}, \; X'=\{2,0\}, \; \sup(X)=\max(X)=2+\frac{1}{2}$ and $\inf(X)=\min(X)=-\frac{1}{2}$.
If you need more detailed answer for any of these please write down in the comment.
