Limit points of a Set and Limit of Sequence Consider a set $A$= {$\frac{1}{n}:n \in N $ }, we want to find all the limit points of the set.
Now there is a theorem which states that 
"A number $c\in R$ is a limit point of a subset $A$ of $R$ if and only if there exists a sequence $(a_n)$ in $A$ such that $lim(a_n)$=$c$ and $a_n\neq c$ for all $n\in N$"
By that theorem it is obvious that $0$ is one of the limit points because it is the limit of the sequence $<\frac{1}{n}>$. But is that the only limit point ? Can't we  obtain another convergent sequence using the elements of $A$. Now I do understand that every sub-sequence of the the sequence $<\frac{1}{n}>$ has $0$ as its limit. But it is not necessary that the elments of the set {$\frac{1}{n}$} needs to be arranged in the same order as the elments of the sequence $<\frac{1}{n}>$. If $0$ is the only limit point can somebody prove it ?
 A: If $x>0$ then $(\frac x 2, 2x)$ is an  open set containing $x$ and it does not contain any number of the form $\frac 1  n$ with $n>\frac 2 x$. But  a limit point must contain infinitely many points of the sequence. Hence $x$ is not a limit point. I will leave it to you write out  similar proof for $x<0$.
A: To determine the limit points of $A$, you can use the theorem about sequences to find a limit point (this works in all metric spaces, not just $\Bbb R$, or more generally in first countable $T_1$ spaces) but it's not the most convenient way to show that you've foudn all of them. For that, it's easy to consider $A \cup A'$, when you have a candidate set from sequence limits $A'$ and show it is closed, which in this case is simple, as $\{\frac1n: n \in \Bbb N\} \cup \{0\}$ is compact (so closed), or directly show (as Kabo does) that all points not in $A'$ are not limit points of the set by finding a neighbourhood of it that only contains at most finitely many points of that set. 
A: For $c$ to be a limit point of $A$ it is necessary that there is a sequence in $A\setminus \{c\}$ converging to $c.$
If there is an open interval $U_c$ with $c\in U_c$ and $U_c\cap (A\setminus  \{c\})=\emptyset$ then no sequence in $A\setminus \{c\}$ can converge to $c.$ Because there is an $r>0$ such that $(c-r,c+r)\subset U_c,...$ so any $a\in A\setminus \{c\}$ satisfies $[a\ge c+r$ or $a\le c-r]$, so $|a-c|\ge r.$
If $c> 1$ let $U_c=(1,1+c).$
If $c<0$ let $U_c=(c-1,0).$
If $c=1$ let $U_c=(\frac {1}{2},2).$
If $0<c<1$ and $c\not \in A,$ there is a (unique) $n_c\in \Bbb N$ with $\frac {1}{1+n_c}< c<\frac {1}{n_c},$ so let $U_c=(\frac {1}{1+n_c},\frac {1}{n_c}).$
If $1\ne c=\frac {1}{n}\in A,$ let $U_c=(\frac {1}{n+1},\frac {1}{n-1}).$ 
