help with this doubts about analysis. (usual metric) help with this doubts about analysis. (usual metric) (Please answer)
Find the set $B$ such that its derived set $B´=A$ where $A=\{\frac{1}{n},n\in \mathbb{Z^+}\}$. 
is there any theorem to say that any derived set is closed? If this is so, then the problem would not have a solution.
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I told my teacher about something that he had read that the derived set should be closed, but he told me that he was confusing the derived set with  closure set, (Could you explain this to me?)
I was thinking of the possible proof that the derived set is closed.
Theorem: $A$ is closed if and only if $A^{\prime} \subseteq A $
if $(A^{\prime})^{\prime}\subseteq A^{\prime}$ then $A^\prime$ is closed, propuse it a my teacher but, he said that it is not always true..
how proof that $(A^{\prime})^{\prime}\subseteq A^{\prime}$ for every $A$
I've already asked this exercise here, but every time I have class, I'm confused. and im sorry my English is bad.
 A: A derived set is the set of all limit points of the set. If $S$ be any set then $S'=\{x: x \ \text{is a limit point of } S\}.$
Derived set is closed.
Let $S'$ be the derived set of $S$. Let $x$ is a limit point of $S'$. I will show that $x\in S'$ i.e. $x$ is a limit point of $S $. Since $x$ is a limit point of $S'\implies \forall r>0,\ B(x,r)\cap S'\setminus \{x\}\neq \emptyset.$

To show: $x\in S'$ i.e. $x$ is a limit point of $S$, i.e. for every $r>0,$ $B(x,r)\cap S\setminus\{x\}\neq \emptyset$

Let $y\in B(x,r)\cap S'\setminus \{x\}$. So $d(y,x)<r$. If $y\in S$ then we are done. If $y\notin S\implies y\in S'.$ Let $s=\min\{ d(x,y),r-d(x,y) \}$, so that $s>0$. Now since, $y\in S' $ implies for the obtained $s$ we will get $B(x,s)\cap S'\setminus \{x\}$. Now check that $z\in B(x,s)\cap S\setminus\{x\}.$  Hence, $x\in S'$.
A: First we consider the original problem. Let's look at the elements of $B=\{n^{-1}\,|\,n\in\mathbb Z\}$ and suppose $A'=B$ for some set $A$. Let $x=\frac1n\in B.$ Now we consider the fact that for any set $A,$ we get $A\subseteq A'$ and as $x$ is a limit point, say it is the limit of the sequence $a_k$ (with each $a_k$ in $A$). Wlog every element of $a_n$ must be at most a distance  $\frac12\left(\frac1n-\frac1{n+1}\right)$ from $x$ (we can do this by the definition of a limit and by ignoring the initial part of the sequence not within this distance). Now every element $a_n$ is in $A\subseteq A'=B$ and so we find that $a_n=x$ is constant and so $B\subseteq A$. Now observe that $0=\lim_{n\to\infty}\frac1n$ is a limit point of $A$ but is not in $B$. Therefore contradiction.
Now for the other parts of the question:


*

*In a metric space every derived set is closed. A set is closed if and only if it contains all of it's limit points (that is $A$ is closed if $A=A'$). This is straightforward to prove if your definition of a closed set is that it is the complement of an open set. The next part is that for any set $A,$ we have $(A')'=A',$ that is, you need to prove that a limit of limit points is itself a limit point.

*On closure. In general the closure of a set is the smallest closed superset of it (precisely it is the intersection of all closed supersets). In a metric space this is the same as the derived set. In some general topological spaces these are not always the same. In topological spaces limits are not anywhere near as nice as they are in metric spaces and so all definitions involving limits are less nice.

