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A point $x \in X$ is a limit point of a subset S of X, if every ball $B(x;\varepsilon)$ contains infinitely many points of S. Show that x is a limit point of S iff there is a sequence {$x_{j}$} in S that converges to x, where $x_{j}\neq x$ for all j. Show that the set of limit points of S is closed

I have managed to prove the "iff-part" and tried to use that result for proving the closure, so here is my try;

Let $E$ be the set of limit points of $S$ and $x\in E$. Then there exists a sequence {$x_{n}$}$_{n=1}^{\infty}$ in $S$ with $x_{n}\neq x$ $\forall j\geq 1$ such that $x_{n}\rightarrow x$.

By the definition om limit, $\forall \varepsilon >0$ there $\exists N>0$ such that $d(x_{n},x)< \varepsilon$ whenever $n\geq N$

clearly $x_{n} \in B(x;\varepsilon)$ and $x_{n} \in E$ , $\forall n\geq N$, hence $x_{n} \in B(x;\varepsilon)\hat E$ , $\forall n\geq N$.

My first thougt is that i've only proved that $E \subset \bar{E}$, which is trivial. I'm i right? Please give me some feedback on this one!

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What's your definition of closed? You could try showing that $X \setminus E$ is open. What can you say about a point $x \in X \setminus E$? –  SpamIAm Jan 12 at 22:59
    
See here :math.stackexchange.com/questions/15766/… –  user10444 Jan 12 at 23:03
    
E is closed if all points of E are adherent points. –  TheOscillator Jan 12 at 23:25
    
So supposing that $x \in X$ \ $E$ we have that x is not a limit point of $S$. So by definition there $\exists \varepsilon >0$ such that $|B(x;\varepsilon)\cap S| < \infty $. And clearly this holds $\forall y \in B(x;\varepsilon)$ which implies that $y\in X$\ $E$, hence $B(x;\varepsilon)\subset X$\ $E$. From this we conclude that $X$\ $E$ is open, so consequently $E$ is closed. Is this argument valid? –  TheOscillator Jan 12 at 23:42
    
Only finitely many points $s_1,...,s_k$ of $S$ might belong to your $B(x;\epsilon)$ and so if there are any, let $r=\text{min}\left\{d(x,s_i)\right\}$. Then $B(x;r)\cap S = \emptyset$. –  David Peterson Jan 13 at 0:05
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