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Let $\emptyset = A \subset \mathbb R $ where A is bounded above, and A has a supremum that is not contained in A. Prove the existence of an $(x_n)_{n\in\mathbb N}$ in A such that $\lim_{n->\infty}$ $x_n = supA$.

I'm guessing we can construct an example with the information given. I have been playing around with the idea the that we can order A such that $a_i \leq a_j$ for all $i \leq j$ and define $(x_n)_{n\in\mathbb N}$ such that $x_l = a_l$ for all $l \in \mathbb N$. Hence $x_i\leq x_j$.

I also believe, the actual definition of the limit will have to come in, but here I do not know how to continue.

Any help is appreciated.

Florian Suess
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    Yes, you can order $A$ in a way "such that $a_i\le a_j$ for all $i\le j$" (for instance, since $A$ is already totally ordered, by putting $i(\alpha)=\alpha$ and $a_{i(\alpha)}=i(\alpha)$). However, you have no hope of doing anything like that if you're forced to have the index set be $\Bbb N$ (as the term "sequence" implies). –  Aug 05 '18 at 22:29
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    See also this question. – José Carlos Santos Aug 05 '18 at 22:33
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    Hey I'm having difficultly understanding the similarities between this question (duplicate?) and the original question which is linked. I can't reason how it helps. Here I don't include the $lim sup x_n$, or is that simply $supA$ treated as a constant function or something? – Florian Suess Aug 05 '18 at 22:35
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    No it definitely is a duplicate. lim of supA as n gets big is supA, you can use that. Clever! – Florian Suess Aug 05 '18 at 22:48

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