$\sup$ of set of limit points is in the set Let $b_n$ a bounded sequence in $\mathbb R$. Let $A\neq \emptyset$ be the set of all its limit points. I want to prove that $\sup A\in A$.
I am really unconfident with my idea:
Proof:
Assume $\sup A\not\in A$.
Then $a<\sup A$ for all $a\in A$ because $\sup\not\in A$.
I kow that $A$ is bounded. So it follows that there is a $\overline a\in A$ such that $a\leq\overline a$ for all $a\in A$. But then it's $a\leq\overline a<\sup A$ and so $\overline a$ has to be supremum of $A$, a contradiction. q.e.d.
So is it working? 
or any better idea?
My problem is if you are moving an $\varepsilon$ away from $\sup A$ you are in the set, right? and therefore there is no $\overline a$ such that $a\leq\overline a<\sup A$.
 A: First of all we notice $A\neq\emptyset$ is a consequence of Bolzano-Weierstrass.
For $\varepsilon>0$ let $(\sup(A)-\varepsilon;\sup(A)+\varepsilon)$ be a neighborhood of $\sup(A)$. So $\exists$ $a\in A$ so that $\sup(A)-\varepsilon<a\leq\sup(A)<\sup(A)+\varepsilon$ and so it follows $a\in(\sup(A)-\varepsilon;\sup(A)+\varepsilon)$.
$(\sup(A)-\varepsilon;\sup(A)+\varepsilon)$ is a neighborhood of $a$.
Since $a\in A$ is a limit point of $b_n$ therefore every neighborhood of $a$ contains infinitely many terms of $b_n$. So $a\in(\sup(A)-\varepsilon;\sup(A)+\varepsilon)$ contains infintely many terms of $b_n$.
Remember that this is true for all $\varepsilon>0$. So every neighborhood of $\sup(A)$ contains infinitely many terms of $b_n$ and so $\sup A$ is a limit point of $b_n$ and so $\sup(A)\in A$. $\Box$
A: Using the fact that the set of limit points of a set is closed we find that $A$ is closed. Using the definition of a supremum select a sequence $\{x_n\}_{n\in Z^+}$ in $A$ such that for all $n\in Z^+$ we have:
$$sup(A)-1/n<x_n\leq sup(A)$$
Using the squeeze theorem, we know that $\lim_{n\rightarrow \infty}x_n=sup(A)$. Since $A$ is closed, thus $sup(A)\in A$
A: Here is an elementary proof:
Let $\text{sup}(A) = x$, then 
For every $k\in \mathbb{N}$, there exists $x_k \in A$ such that $x - \frac1{2k} < x_k \leq x$. 
But since $x_k$ is a limit point for the sequence $\{b_n\}$, there exists $l_k \in \mathbb{N}$ such that $ x_k - \frac1{2k} < b_{l_k} < x_k + \frac1{2k}$
Using the above inequalities, $x - \frac1{4k} < b_{l_k} < x + \frac1{4k}$
This means $\displaystyle \lim_{k \to \infty} b_{l_k} = x$. But then you found a subsequence $\{b_{l_k}\}$ that converges to $x$, so $x$ is a limit point.
P.S: By elementary, I am assuming you have no knowledge of a closed set. My solution is basically a detailed version of Amr's solution. 
A: Let $S=\sup A$ and $(S- \varepsilon, S + \varepsilon)$ a neighbourhood of $S$. By definition of $\sup$ there exist $x_n$ such that $S- \varepsilon \le x_n \le S$, Thus $x_n \in(S- \varepsilon, a + \varepsilon) $  and $S \in A$ by definition of limit point.
