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?
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asked Feb 22, 2015 at 19:47
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$\begingroup$ This is a Grammar disaster...I don't know how it turned out like that... $\endgroup$– Meitar AbarbanelFeb 22, 2015 at 19:50
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2 Answers
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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 $A$.
answered Feb 22, 2015 at 19:55
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$\begingroup$ How can I turn it into some definition of accumulation point? $\endgroup$ Feb 22, 2015 at 20:04
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$\begingroup$ Can I say that $a$ is an element of the sequence in a punctured vicinity? $\endgroup$ Feb 22, 2015 at 20:04
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1$\begingroup$ Yes, let $\epsilon>0$ and pick $\beta =\alpha-\epsilon$. Then $a\in B_\epsilon(\alpha)$, and $a\ne\alpha$. $\endgroup$ Feb 22, 2015 at 20:06
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4$\begingroup$ @TimRaczkowski but what if A is a finite set. Finite sets don't have accumulation points right? Does that imply that a supremum is either an accumulation point or a maximum for the set? $\endgroup$ Oct 24, 2017 at 12:09
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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.
