Real Analysis - Limit points and Open set. The statement is If $A \subset \mathbb{R} $ is any set, $I$ is an open set that contains a limit point of $L_{A}$ - the set of all limit points of $A$, then $I$ also contains a point of $A$. I approach this by letting $x$ in A with $x$ is a limit point of A. So for every $\epsilon >0$ the deleted neighborhood $V^{*}(a) \cap A$ is nonempty. From this how do I show that $x$ is also in $I$.  
 A: If $I$ is an open set containing a limit point of $A$, call it $x$ as you have. Then since $I$ is open it contains a ball around $x$, say $(x-\epsilon,x+\epsilon)$. Then since $x$ is a limit point of $A$ we know that as you've already pointed out $((x-\epsilon,x+\epsilon))\cap A\not=\emptyset$, so since $((x-\epsilon,x+\epsilon))\subseteq I$ it follows that $((x-\epsilon,x+\epsilon))\cap A\subseteq I$, hence $I$ contains an element of $A$. 
A: Let $X$ be any topological space and $A\subset X.$ As per  the comments by the proposer, a point $x\in X$ is a limit point of $A$ iff every nbhd of $x$ contains a member of $A.$
Suppose  $y$ is a limit point of the set $L_A$ of limit points of $A$.  If $U$ is any nbhd of  $y$ then there is an open $V$ with $y\in V\subset U.$
Now $V$ is a nbhd of $y$ so there exists $x\in V\cap L_A.$ But  $V$ is also a nbhd of $x$ (because $V$ is open and $x\in V$ ), and $x\in L_A$,  so there exists $a\in A\cap V.$ 
So $a\in A\cap U$ because $V\subset U.$
So any nbhd $U$ of $y$ contains  a member of $A.$ So $y\in L_A.$
This holds in every topological space.
