What is the derived set of $A=\{1,\frac12,\ldots,\frac1n,\ldots\}$ in $(\mathbb{R},|\cdot|)$ Please how to find $A'$ where $A=\{1,\frac12,\ldots,\frac1n,\ldots\}$
i know that $x\in A' \Longleftrightarrow \forall V\in \mathcal{V}_x, V\setminus\{x\}\cap A \neq \emptyset$ but i dont know how to apply this to the given $A$ .
Thank you 
 A: $A'=\{0\}$.
For any $\epsilon>0$, there is a $\dfrac1{n}\in A$ and $\dfrac1{n}<\epsilon$, i.e. $\dfrac1{n}\in(0,\epsilon)=V_0$ ($V_0$ is open). All other points in $A$ are isolation point, i.e. for any $n$, there is a $\epsilon>0$ 
$$
\left(\dfrac1{n}-\epsilon,\dfrac1{n}+\epsilon\right)\cap \left(A-\left\{\dfrac1{n}\right\}\right)=\varnothing\tag1
$$ 
Edit: 
Consider any point $\dfrac1{n}\in A$. Since $\{\dfrac1{n}\}$ is monotonic decreasing, the closest point to its left is $\dfrac1{n+1}$ and the closest point to its right is $\dfrac1{n-1}$. Take 
$$
\epsilon<\min{(\frac1{n}-\frac1{n+1},\frac1{n-1}-\frac1{n})}=\frac1{(n+1)n}
$$ Then for any $\dfrac1{k}\in A, \:k\ne n$ 
$$
\left|\dfrac1{k}-\dfrac1{n}\right|>\epsilon\quad\text{and so }\quad \dfrac1{k}\notin \left(\dfrac1{n}-\epsilon,\dfrac1{n}+\epsilon\right) 
$$
So $(1)$ holds and $\dfrac1{n}$ is isolation point.
A: Observe that $\lim_{n\rightarrow\infty}\frac{1}{n}=0$.
So let $\epsilon>0$ be arbitrary.
Thus there is $N\in\mathbb N$ such that for each $n>N$ , $|\frac{1}{n}-0|<\epsilon$ .
So we have that $0\in A'$ .
Now observe that $\frac{1}{n}-\frac{1}{n+1}=\frac{1}{n(n+1)}\geq\frac{1}{n(n+n)}\geq\frac{1}{2n^2}$
So whenever you pick a point you can find an open set such that it does not have non empty intersection with $A$.
So $A'$ contains only $0$.
A: $0$ is the only limit point, which exactly this sequence goest to.
