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I am struggling to understand what is the limit supremum/infimum. I've been told that it is not the same thing as "the limit of a supremum of a set" (which makes sense since the supremum/infimum is usually a number).

I've consulted with two Analysis books, but none of them seem to be able to convey it what they are trying to say.

I got an example in my notebook that may clarify my confusion

Ex. Consider $\left \{-200,100,1,2,-1,2,-1,1,2,-1 \right \}$

Then let $v_k = \sup \left \{a_n : n \geq k \right \}$ and $\limsup_{n\to\infty} a_n= \lim_{k\to\infty} v_k=2$ and $\liminf_{n\to\infty} a_n=-200$

Can someone explain to me the reasoning (without omitting any details) for the answers? I think I got a feeling for the liminf, but not limsup

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What do you mean? They are numbers – Hawk Oct 1 '12 at 3:52
No, that is the sequence – Hawk Oct 1 '12 at 3:59
up vote 9 down vote accepted

Let's first just recall the definitions of the limit superior and limit inferior. For a sequence $\{a_n\}$, they are $$\limsup_{n \rightarrow \infty} a_n = \lim_{n \rightarrow \infty} \sup_{k\geq n} a_k, \quad \liminf_{n \rightarrow \infty} a_n = \lim_{n \rightarrow \infty} \inf_{k \geq n} a_k.$$ Recall that the supremum is the least upper bound and the infimum is the greatest lower bound. So then the expressions $\sup_{k \geq n} a_k$ and $\inf_{k \geq n} a_k$ are the upper and lower bounds for the tails of the sequence, looking only at terms $k \geq n$.

So the limit superior is asking, how large can the tails of the sequence eventually be? Similarly, the limit inferior is asking, how small can the tails of the sequence eventually be?

Example: Let $a_n = \{100, -100, -1, 1, -1, 1, -1, 1, \ldots\}$. Then $\sup_n a_n = 100$, $\inf_n a_n = -100$, but the $\limsup$ ignores all the large terms that begin in the finite portion of the sequence, so we have $\limsup a_n = 1$. Similarly, the $\liminf$ ignores all the small terms in the beginning, so $\liminf a_n = -1$.

To address the example you gave in your question, the sequence you gave is not an infinite sequence so $\limsup$ and $\liminf$ aren't defined.

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I'd like to add to Christopher A. Wong's answer that the $\liminf$ is the smallest accumulation point while the $\limsup$ is the largest one. Moreover, if you have an understanding of the $\liminf$ already, then consider $\limsup a_n = -\liminf (-a_n)$.

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Oh there is the symmetry of Sup(S) = -inf(-S) – Hawk Oct 4 '12 at 5:10
What do you mean by accumulation point here? – Mathematics Oct 17 '12 at 12:29
An accumulation point of $(a_n)$ is a point such that in each neighborhood of it there infinitely many $a_n$'s. Put differently: A point is an accumulation point of a sequence if it is a limit of a subsequence. – Dirk Oct 17 '12 at 15:46

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