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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

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  • $\begingroup$ Well both lim sup and lim inf can be infinite, so perhaps you should say "extended real number." $\endgroup$ – saulspatz Nov 7 '18 at 5:10
  • $\begingroup$ yes, that's what I meant. Thanks for pointing out. $\endgroup$ – davidh Nov 7 '18 at 5:14
  • $\begingroup$ Which is the definition of $\limsup$ that you are using? $\endgroup$ – André Porto Nov 7 '18 at 5:14
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    $\begingroup$ 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. $\endgroup$ – Michael Nov 7 '18 at 5:27
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    $\begingroup$ You may have a look at this answer math.stackexchange.com/a/1893725/72031 $\endgroup$ – Paramanand Singh Nov 8 '18 at 13:31
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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.

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  • $\begingroup$ 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? $\endgroup$ – davidh Nov 7 '18 at 6:38
  • $\begingroup$ 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. $\endgroup$ – André Porto Nov 7 '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 $$ \limsup a_n = \inf_{n\in\mathbb N}\left(\sup_{k\geq n} a_n\right) \geq \inf\{\alpha\in\mathbb R:\mbox{the set $\{a_n: a_n>\alpha\}$ is finite}\}. $$

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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

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