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Suppose $s_n$ is a sequence defined in $\mathbb{R}$. Define $E$ as the set of all subsequential limits of $\{s_n \}$.

Then by the definition of limit superior, $$\limsup\limits_{x\rightarrow \infty} = \sup(E)$$

Suppose $\limsup\limits_{x\rightarrow \infty} = -\infty$, then $E$ can be $\emptyset$.

By Rudion's theorem 3.17:

Let $\{s_n \}$ be a sequence of real numbers. Let $E$ be the set of all subsequential limits of $\{s_n \}$ and $s^* = \sup E = \limsup\limits_{n\to\infty}s_n$. Then $s^*$ has the following two properties:

(a) $s^* \in E$.

(b) If $x> s^*$, there is an integer $N$ such that $n \geq N$ implies $s_n < x$.

Moreover, $s^*$ is the only number with the properties (a) and (b).

We have

$$-\infty \in \emptyset$$

It doesn't really make sense to me. What I have missed above?

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Jay Wang
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    There must be something wrong: $E$ cant' be defined by the $x$ such that ALL sub-sequences converge to $x$, because that would mean that if $E$ is non-empty it would be a single element. I think you meant the set of $x$ for which there THERE EXISTS a sub-sequence that converges to $x$. – Andrei.B Nov 16 '17 at 23:26
  • @Andrei.B Sorry, I meant $E$ is the set of all subsequential limits of ${s_n}$. Then $E$ can be empty? – Jay Wang Nov 16 '17 at 23:28
  • Well it could if you don't consider $+\infty$ as a limit: just take $s_n=n$ for example. But you could also consider limits in $\overline{\mathbb{R}}=\mathbb{R}\cup{+\infty,-\infty}$ and that would not be a problem anymore since this space with its natural topology is compact. – Andrei.B Nov 16 '17 at 23:31
  • @Andrei.B Sorry, I am still confused. If $s_n = n$, then $E$ is empty. $\sup(E) = -\infty$ gives $-\infty \in \emptyset$ again? How does considering $\overline{\mathbb{R}}$ help? – Jay Wang Nov 16 '17 at 23:36
  • It doesn't help, it's just changing frames of work to think out of the box. But going back to just $\mathbb{R}$, saying that $\mbox{sup}(E)=-\infty$ will just mean that every sub-sequence of $(s_n)_n$ diverges to $-\infty$. – Andrei.B Nov 16 '17 at 23:41
  • Actually there's a better definition of $\mbox{lim sup}$ that I prefer: $\mbox{lim sup}\ s_n=\mbox{lim}{N\rightarrow +\infty}\mbox{sup}{k>N}s_n$ I believe that this equivalent definition would make your thinking clearer – Andrei.B Nov 16 '17 at 23:43
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    Check this related post if you need to (https://math.stackexchange.com/questions/1100742/how-to-prove-that-the-limsup-of-a-sequence-is-equal-to-its-greatest-subsequentia?rq=1) – Andrei.B Nov 16 '17 at 23:47

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