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