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After reading several alternative definitions of $\lim\sup$ and $\lim\inf$, such as $\lim\sup$ being the supremum of the set of all subsequential limits, I'm still having trouble building the intuition for $\lim\sup$.

One thing that I feel is true, but not sure, is that $\lim\sup$ represents the greatest real number that infinitely many $a_n$ gets close to, and $\lim\inf$ represents the smallest value that infinitely many $a_n$ gets close to. Are these correct statements? If so, how would one go about showing it? Thanks

Tianlalu
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gws
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  • Well both lim sup and lim inf can be infinite, so perhaps you should say "extended real number." – saulspatz Nov 07 '18 at 05:10
  • yes, that's what I meant. Thanks for pointing out. – gws Nov 07 '18 at 05:14
  • Which is the definition of $\limsup$ that you are using? – André Porto Nov 07 '18 at 05:14
  • limsup as the infimum of supremum – gws Nov 07 '18 at 05:15
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    Yes, $\limsup$ is the largest possible limiting value achievable over a subsequence of times that go to infinity, while $\liminf$ is the smallest possible. Intuition comes by considering $\limsup_{t\rightarrow\infty} \cos(t) = 1$ and $\liminf_{t\rightarrow\infty} \cos(t) = -1$. We have $\limsup = \liminf = \lim$ if and only if the regular limit $\lim$ exists. – Michael Nov 07 '18 at 05:27
  • The observations I made in this answer might be of use, especially in seeing the relationship between "infinitely many" and "all but finitely many", two concepts that show up all the time in discussions of limit, limsup, and liminf. – Dave L. Renfro Nov 07 '18 at 08:09
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    You may have a look at this answer https://math.stackexchange.com/a/1893725/72031 – Paramanand Singh Nov 08 '18 at 13:31

3 Answers3

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One thing that helped me: You could try to prove that if $s$ is the $\limsup$ then for any number $s' < s$, the sequence exceeds $s'$ infinitely often. And for any number $s'' > s$, the sequence exceeds $s''$ only finitely often.

SBK
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  • I tried and I think I proved this. But I was just wondering would s be the unique real number that satisfies these two properties? – gws Nov 07 '18 at 06:38
  • Yes, it is. Any $t< s$ fails to satisfy the property: "for any number $u>t$, the sequence exceeds $u$ only finitely often" (to show that, just pick $u\in(t,s)$, since $u<s$, the sequence must exceed $u$ infinitely often). Analogously, we prove that any $s<t$ fails to satisfy the property: "for any number $u<t$, the sequence exceeds $u$ infinitely often". So $s$ is the only one that satisfies both those properties. – André Porto Nov 07 '18 at 10:25
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One intuition I certainly recommend is that the $\limsup$ is the infimum of the numbers that are passed by only finitely many $a_n$'s. More precisely, $$\limsup a_n=\inf\{\alpha\in\mathbb R:\mbox{the set $\{a_n: a_n>\alpha\}$ is finite}\} $$

Pick $\alpha$ satisfying this. Then, there exists $n\in\mathbb N$ such that for every $k\geq n$, $a_n\leq \alpha$. Therefore, $$ \sup_{k\geq n} a_n \leq \alpha \Rightarrow \limsup a_n = \inf_{n\in\mathbb N}\left(\sup_{k\geq n} a_n\right) \leq \alpha, $$ so $$ \limsup a_n \leq \inf\{\alpha\in\mathbb R:\mbox{the set $\{a_n: a_n>\alpha\}$ is finite}\}. $$

To prove the remaining inequality, note that if $$\beta<\inf\{\alpha\in\mathbb R:\mbox{the set $\{a_n: a_n>\alpha\}$ is finite}\},$$ then certainly $$\beta\notin \{\alpha\in\mathbb R:\mbox{the set $\{a_n: a_n>\alpha\}$ is finite}\},$$ so there are infinite many $a_n$'s greater than $\beta$, so for any $n\in\mathbb N$, $$ \sup_{k\geq n} a_n > \beta, $$ then taking the infimum on the $n$'s, we get $$ \limsup a_n = \inf_{n\in\mathbb N}\left(\sup_{k\geq n} a_n\right) \geq \beta. $$ Since this happens for any $\beta<\inf\{\alpha\in\mathbb R:\mbox{the set $\{a_n: a_n>\alpha\}$ is finite}\}$, we get that $$ \limsup a_n = \inf_{n\in\mathbb N}\left(\sup_{k\geq n} a_n\right) \geq \inf\{\alpha\in\mathbb R:\mbox{the set $\{a_n: a_n>\alpha\}$ is finite}\}. $$

André Porto
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Yes it can be proved that if we consider the set $S\subseteq \mathbb{\bar R}$ of all the limits of the subsequences of $a_n$ we have that

$$\max\{S\}=\limsup a_n \in \mathbb{\bar R}$$ $$\min\{S\}=\liminf a_n\in \mathbb{\bar R}$$

extending the notation/definition also to the infinity cases.

That property with bounding evauation is used to prove what $\limsup$ and $\liminf$ are.

Refer also to the related

user
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