When is the supremum\infimum an accumulation point?

Trying to show that if a sequence converges, it either has a maximum, a minimum or both, I reached a dead-end. Assuming it is not constant, it is still bounded and its supremum and infimum aren't equal. Then I assumed that the supremum and infimum are not in the sequence. I want to show that there are two subsequences that converge to each of them but for that to happen I have to show they are accumulation points. I tried to use definition but failed. I know logically that following my assumption they have to be accumulation points but I can't derive it from the definitions. Any help?

• This is a Grammar disaster...I don't know how it turned out like that... Commented Feb 22, 2015 at 19:50

I assume we are working in $\Bbb R$. Let $A$ be a non-empty, bounded set. $\sup A$ is uniquely defined as the number $\alpha$ such that for all $a\in A$, $a\le\alpha$, and if $\beta<\alpha$ there is an $a\in A$ such that $\beta<a<\alpha$. You should be able to use the second statement to show that $\sup A$ is an accumulation point of $A$.
• Can I say that $a$ is an element of the sequence in a punctured vicinity? Commented Feb 22, 2015 at 20:04
• Yes, let $\epsilon>0$ and pick $\beta =\alpha-\epsilon$. Then $a\in B_\epsilon(\alpha)$, and $a\ne\alpha$. Commented Feb 22, 2015 at 20:06
Consider the sequence $\{\mathbf{x}_n\}_{n\in\mathbb{N}} \subseteq A$, where $A \subseteq \mathbb{R}^n$. If this sequence is convergent, then it is bounded. Also, the sequence must converge to $\mathbf{x} \in \overline{A}$. Show that for all $\epsilon > 0$, there are finitely many $\mathbf{x}_n \notin N(\mathbf{x}, \epsilon) \cap A$ where $N(\mathbf{x},\epsilon)$ is some neighborhood of $\mathbf{x}$. Then, use the fact that a finite set is bounded and that $N(\mathbf{x}, \epsilon_0) \cap A$ is bounded for some $\epsilon_0$ to prove your claim.