Limit Point Question Question: Given a bounded sequence of real numbers, let $L$ denote the set of limit points of the sequence. Show that L is closed subset of$\mathbb R.$
Attempt: I was thinking of showing $L$ contains all its limit points to show its closed, but I wasn't sure how to proceed with this. Should I use contradiction? 
 A: *

*If you can use the result "A set in $\Bbb R^n$ is closed $\iff $ it contains al its limit points", then all you have to do is to show that $L$ contains all its limit points as you said. Just simply assume $l$ is a limit point of $L$. Then there is a sequence $(l_n) $ of elements in $L$ which converge to $l$. Now for each $n \in \Bbb N, \;$ $l_n$ is a limit point of the original set and hence you can construct a sequence of elements in the original set which converges to $l$ (Think about how to do this rigorously). Then $l$ is a limit point of the original set and hence $l \in L$. 

*To use the base definition: 
Consider the complement of the set $L$ denoted by $L^C$. Let $l \in L^C$ be arbitrary. If there is no neighbourhood around $l$ which is not entirely contained in $L^C$ i.e. if it is impossible to construct a neighbourhood around $l$ which does not have any elements of $L$ then $l$ is a limit point of $L$ and due to arguments in the previous bullet it is also a limit point of the original set contradicting the assumption. Then there is a neighbourhood around $l$ entirely contained in $L^C$ which would mean it is open and hence $L$ is closed. 

A: If $x$ is a limit point of $L$, then every neighbourhood $N$ of $x$ contains a limit point $y$ of the sequence, hence infinite many points of it (since $N$ neighbourhood of $y$ too). This shows that $x$ is a limit point of the sequence too and that's why contained in $L$.
A: An easy proof will be as follows:
Lets denote the set of all limit points of $ L $ by $L^d$.Let $A$ denote range of the sequence.To show $L$ is closed sufficient to show $L^d\subseteq L$.(check that).
Let $x\in L^d$.then $(x-r,x+r)\cap L\setminus \{x\}\neq \phi $.Let $y\in (x-r,x+r)\cap L\setminus \{x\}$.Now $(x-r,x+r)$ is an open neighbourhood of $y$ and also $y\in L\setminus \{x\}$.So $(x-r,x+r) \cap A $ will contain infinitely many points.So $(x-r,x+r) \cap A\setminus\{x\}\neq \phi $.thus $x\in L$
