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

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.