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Previous Question; Does upper limit and lower limit exist for any sequence in $\mathbb{R}$?

For every sequence $\{s_n\}$ in $\mathbb{R}$, $\{x\in \overline{\mathbb{R}}| s_{n_k} →x\}$ is nonempty.

However, how do i prove the existence of $\sup \{x\in \overline{\mathbb{R}}| s_{n_k} →x\}$ and $\inf \{x\in \overline{\mathbb{R}}| s_{n_k} →x\}$?

Example; $a_{2n}=-n^2$ and $a_{2n+1}=(1/2)^n$

$b_{3n}=-n^2$ and $b_{3n+1}=(1/2)^n$ and $b_{3n+2}=n^2$.

What kind of mathematical property do they share? I don't know where to start my argument.

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    Is choice allowed? :)2012-08-18
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    @David That is a really hard question for me.. I'll try to convert your argument into ZF if it's tedious, but easy.. Yes.2012-08-18
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    @DavidMitra Look at my comment for question http://math.stackexchange.com/questions/183767/does-upper-limit-and-lower-limit-exist-for-any-sequence-in-mathbbr2012-08-18
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    @William I know insisting on ZF is maybe silly, but if you know how to do this, would you help me?2012-08-18
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    If $(s_n)$ is not bounded above, then $\infty$ is an element of your set and thus the supremum of it. Otherwise, your set is bounded above and nonempty. Such sets have supremums.2012-08-18

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Edit: I see I misread your question. My apologies.

Let us take any $E\subseteq\overline{\Bbb R}$, and show that there exists $y\in\overline{\Bbb R}$ such that $y=\sup E$. (The arguments will be similar to show that an infimum exists.)

If $E$ is empty, we say that $\sup E=-\infty$. Suppose not. If $E$ has an upper bound in $\Bbb R$, then by completeness of $\Bbb R$, $E$ has a supremum in $\Bbb R$. Otherwise, we say by definition that $\sup E=+\infty$.