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If $(a_n)$ is a real sequence, in lecture we had:
$$\begin{align}\limsup_{n\to\infty} a_n=a \iff &(i)\forall \epsilon >0 \,\exists n_0\in \mathbb{N} :a_n<a+\epsilon \forall n\ge n_0\\\text{ and }&(ii) \forall \epsilon >0 \, \forall m\in \mathbb{N} \, \exists n\ge m :a_n>a-\epsilon\end{align}$$ and such a analogue characterization for lim inf.

I know the definition of $ \limsup\limits_{n\to\infty} a_n=\sup H(a_n)$ with $H(a_n)$ set of all the limitpoints of $(a_n)$. I don't understand the epsilon-characterization of lim sup an, maybe you can draw a picture, explain it in words why it is equivalent (or prove it directly, but I only want to understand it). Later I want to do an example if I have understand this. Maybe you want to help? Regards

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  • $\begingroup$ There is also a third commonly used definition: $\limsup a_n= \lim_{n\to\infty} \sup_{k\ge n} a_k = \inf_{n\in\mathbb N} \sup_{k\ge n} a_k$. $\endgroup$ – Martin Sleziak May 2 '14 at 22:03
  • $\begingroup$ thank you. I will try to understand it $\endgroup$ – RedRose May 2 '14 at 22:55
  • $\begingroup$ This is (to some extent) related: math.stackexchange.com/questions/281807/… $\endgroup$ – Martin Sleziak May 2 '14 at 23:06
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I'm taking your definition of limsup to be the "largest subsequential limit".

i) says that no subsequential limit can be bigger than the limsup.

ii) says that a subsequential limit can be at least as big as the limsup.

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  • $\begingroup$ thank you!it is a helpful answere for me to interprete (i) and (ii). But the equivalence is still a bit uncertain for me $\endgroup$ – RedRose May 2 '14 at 22:53

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