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


3 Answers 3


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.


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 $\limsup a_n = -\liminf (-a_n)$.

  • $\begingroup$ Oh there is the symmetry of Sup(S) = -inf(-S) $\endgroup$
    – Lemon
    Oct 4, 2012 at 5:10
  • $\begingroup$ What do you mean by accumulation point here? $\endgroup$ Oct 17, 2012 at 12:29
  • 1
    $\begingroup$ 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. $\endgroup$
    – Dirk
    Oct 17, 2012 at 15:46

Limit supremum (of a sequence of numbers)

I will try to give intuition only using words, without using any mathematical symbol.

  • What is the smallest number which is greater than or equal to infinitely many members of a sequence? --> limit supremum

Story: To be a candidate for the limit supremum, a number has to be greater than or equal to infinitely many members of the sequence. For this, we need to consider tails of the sequence. Why tails? Why not heads? Because given any member, all the members to the left of it (head) are finite in number as the sequence starts on the left and the starting point is known ([avoiding bi-infinite sequences for now] a sequence is a mapping from the set of natural numbers which starts at 1, so the starting side is known. You can tell the first natural number, the first 2 natural numbers, etc. but cannot tell the last 2 natural numbers), only the right side is infinite. So, where there is talk of infinitely many members, you have to consider a tail. Any tail of the sequence is infinitely long. Hence, the supremum of any tail is greater than or equal to infinitely many members of the sequence. Thus, the supremum of any tail is a candidate for limit supremum. Further, as successive tails are subsets of the previous ones, the corresponding supremums form a decreasing sequence of candidates for the limit supremum. However, the infimum of this sequence of candidates is the smallest number which already bounds infinitely many members of the original sequence from above--which is precisely the purpose. So, we reject all larger candidates, and this infimum, as the best/most parsimonious candidate, is elected as the limit supremum of the original sequence.

Limit infimum (of a sequence of numbers)

  • What is the largest number which is smaller than or equal to infinitely many elements of the sequence? --> limit infimum

You can write the story now.


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