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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 $a\leq\overline a<\sup A$.

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"Let $\emptyset\neq A$ be the set of all its limit points" is conventional notation for "Let $\emptyset$ be the set of all its limits point, and by the way it is different from $A$", which I'm pretty sure is not what you mean. – Henning Makholm Dec 15 '12 at 19:34
@rodriguez888 are you familiar with the concept of limit superior? – T. Eskin Dec 15 '12 at 19:44
"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 $a \in A$" The ' a $\overline{a} \in A$ ' part is not true for any set $A$. To see that, try to find $\overline{a}\in A$ for $A = \{ 1 - \frac1{n} \}$ such that $a \leq \overline{a}$. You have to use the fact that $A$ is the set of limit points somewhere in the proof. – Isomorphism Dec 15 '12 at 20:19
I don't know why you are saying that. An epsilon neighborhood of the supremum of $A$ will contain a limit point of the sequence and a small enough interval embedded in that epsilon neighborhood will (in most cases) contain infinitely many points of the sequence (by the definition of limit point)! – Isomorphism Dec 15 '12 at 22:26
@Isomorphism thanks. I see my mistake. But $A$ is bounded, so there is a $\overline a$ with the properties above. why is this not clear for you? – user45389 Dec 15 '12 at 22:36
up vote 3 down vote accepted

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 $\sup(A)\in A$. $\Box$

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@Isomorphism @ rodriguez888 added another proof. and the problem with $\overline a$ is the existence. there are bounded sequences with accumulation points e.g. $[0;1]$ and it's not obvious why an $\overline a$ in $[0;1]$ exists saying $\sup\{[0;1]\}\not\in [0;1]$. – Euler Dec 16 '12 at 1:47

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 $sup(A)\in A$

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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.

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thanks. but what about my proof above? – user45389 Dec 15 '12 at 19:53
You will have to clarify what you mean by $\overline{a}$. Even if you mean $\overline{a} = \text{sup}(\{b_n\})$, there is no reason for $\overline{a} \in A$. If you write your proof clearly, then we will be able to comment on it. – Isomorphism Dec 15 '12 at 20:03

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.

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