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I'm working through Rudin, and the first instance I see this notation used is Definition 3.16. The definition states:

Let $\{s_n\}$ be a sequence of real numbers. Let $E$ be the set of numbers $x$ (in the extended real number system) such that $s_{n_x}\rightarrow x$ for some subsequence $\{s_{n_k}\}$. This set $E$ contains all subsequential limits as defined in definition 3.5, plus possibly the numbers $+\infty$ and $-\infty$

We now recall Definitions 1.8 and 1.23 and put:

$$s^* = \sup E$$ $$s_* = \inf E$$.

The numbers $s^*$ are called the upper and lower limits of $\{s_n\}$; we use the notation

$$\lim_{n\to\infty}\sup s_n=s^*, \lim_{n\to\infty}\inf s_n=s_*$$

The concept of $\inf$ and $\sup$ made sense to me in chapter 1/2 in the context of sets, and the first pair of definitions, with regards to the set $E$ makes sense. Based off of those definitions, I interpret $s^*$ as the largest convergent value you can create from a subsequence of $\{s_n\}$; similarly, $s_*$ is the smallest convergent value you can create from a subsequence of $\{s_n\}$ (although those definitions might be inaccurate).

The lower notation, on the other hand, confuses me. As you're taking the limit, It seems to me that you are no longer dealing with a set, just individual set elements. In that context, I can't distiguish $\lim\sup s_n$ or $\lim\inf s_n$ from simply $s_n$. How does this pair of definitions for $s^*$ and $s_*$ make sense?

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Even though $\limsup$ and $\liminf$ are written with spaces, they are single symbols.

They should be read $$ \limsup_{n\to\infty} s_n = s^* $$ and not $$ \lim_{n\to\infty} \Big( \sup s_n \Big ) = s^*$$ which doesn't really make sense (it's trying to take the supremum of a single number at a time, which has no defined mening).

However, if you consider "$\limsup$" to be a single symbol with a definition separate from the meanings of $\lim$ and $\sup$ by their own, then it's just notation, and what you have there is a definition of what that notation means.

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  • $\begingroup$ Oh that does make more sense, thanks for clarifying about the notation. $\endgroup$ Jul 11, 2018 at 23:49
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The confusion might dissipate if you write it out on one line. That is, $$ \limsup_{n\rightarrow\infty}s_{n} \equiv\sup\left\{ s\in\mathbb{R}\colon\text{there exists a subsequence }\{s_{n_{k}}\}\text{ of }\{s_{n}\}\text{ such that }s_{n_{k}}\rightarrow s\right\} . $$ The set that we are taking the supremum of is what Rudin calls $E$. The supremum itself is what Rudin calls $s^{*}$. $\liminf$ is defined similarly. Does that help?

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An equivalent definition is given by $$ \limsup_{n \to \infty} s_n \equiv \lim_{n \to \infty} \sup_{k \geq n} s_k \equiv \lim_{n \to \infty} \sup \{s_k : k \geq n\} \, .$$ The right-hand side is a limit of suprema of sets (which you know), and the first two expressions are just notation. A similar definition applies for $\liminf$.

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There are many equivalent definitions, one being using least upper bonds ands greatest lower bounds. You can replace these with "sup" and "inf" if you desire, as their definitions carry over to sets:

$\lim\sup_{n\rightarrow\infty} a_n = \lim_{n\rightarrow\infty} \mbox{lub}(\{a_n,a_{n+1},...\})$

and similarly with liminf, with glb. This is closest to your desire for a sequential-limit definition, as now it's clearly a limit of a sequence of bounds.

Also, you can readily show that if $s=\lim\sup_{n\rightarrow\infty} a_n$, then there must exist a subsequence $a_{n_k}$ that converges to $s$.

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Essentially $\limsup_{n \rightarrow \infty} \{a_n\} $ always exists in the extended real even if $\lim \{a_n\}$ doesn't. In particular, $\limsup$ is the largest limit a subsequence can achieve. While $\liminf$ is the smallest limit a subsequence can achieve. In particular, the limit exists if and only if $\limsup = \liminf$.

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