Find the set $B$ such that its derived set $B´=A$ help with this excercise, 
Find the set $B$ such that its derived set $B´=A$ where,
$$A=\{1,\frac{1}{2},\frac{1}{3},...\}$$
How??
I try but none... :(
 A: Such a set does not exist for the usual topology on $\mathbb{R}$, because a derived set containing each $\frac{1}{n}$ must also contain 0:

*

*Suppose every $1/n$ $(n\in \mathbb{N})$ is a limit point of $B$.


*Then $B$ contains arbitrarily close approximations to each $1/n$.


*Then 0 must also be a limit point of $B$.


*Indeed, if $1/n$ gets arbitrarily close to 0, and members of $B$ get arbitrarily close to each $1/n$, then members of $B$ get arbitrarily close to 0.
(To see this formally, pick $\epsilon > 0$. The ball around 0 of radius $\frac{\epsilon}{2}$ contains some $1/n$. The ball around $1/n$ of radius $\frac{\epsilon}{2}$ contains some $y\in B$. The distance between $y$ and 0 is at most $\epsilon = \frac{\epsilon}{2} + \frac{\epsilon}{2}$, by the triangle inequality. Hence for each ε, we can find $y\in B$ within $\epsilon$ of 0, so 0 is a limit point of $B$.)


*Hence if the set of limit points of $B$ contains $\{\frac{1}{n} : n\in \mathbb{N}\}$, it must also contain 0.


*There is no set $A$ whose set of limit points includes $\{\frac{1}{n} : n \in \mathbb{N}\}$ and nothing else.

If our goal is to include $\{\frac{1}{n} : n \in \mathbb{N}\}$ and very little else, we can define $H = \{ \frac{1}{n} : n\in \mathbb{N}\}$, so that $H\cup \{0\}$ has zero as its sole limit point. Then define your set to be:
$$B = \{ x + y \mid x, y \in H\} \cup H.$$
Then if we fix $x$ and vary $y\in H$ in the definition of $B$, we can see that each $x\in H$ is a limit point of $B$, and so is 0. It should be straightforward to see whether any other point $x\notin H$ is a limit point.
(Following answer to distantly-related question: Example of a countable compact set in the real numbers)
A: HINT: You have not provided enough information to give a unique answer, but one answer could come from studying the integral $\int_0^1{x^m}dx=\left[\frac{x^{m+1}}{m+1} \right]_0^1=\frac{1}{m+1}$ where $m$ is a nonnegative integer, with x being a real number between $0$ and $1$.
If you integrate the function again (being a repeated or successive integral depending on which book on calculus you read), what is the result?
$$\frac{1}{m+1}\int_0^1{x^{m+1}}dx=\frac{1}{m+1}\left[\frac{x^{???}}{???} \right]_0^1=\frac{1}{(m+1)(???)}$$
Therefore if the sets B and A are related by the rules of calculus, then you get an answer. Of course any number of other plausible associations and answers may exist.
