Rudin's Chapter 3: Numerical sequences and series In the Rudin's Principles of Mathematical Analysis 3rd edition, Chapter 3, page 56, there is a definition of a set $E$ that, in my point of view, is very doubtful. What is the set $E$? I couldn't also to understand the proof of the theorem just below the definition. (I boldfaced the points of great doubt):
"3.16 Definition: Let $\{s_n\}$ be a sequence of real numbers. Let $E$ be the set of number $x$ (in the extended real number system) such that $s_{n_k}\to x$ for some subsequence $\{s_{n_k}\}$. This set $E$ contains all subsequential limits as defined in Definition 3.5, plus ??possibly?? the numbers $+\infty,-\infty$.  We now recall Definitions 1.8 and 1.23 and put $$s^*=\sup E$$  $$s_*=\inf E.$$ The numbers $s^*,s_*$ are called the upper and lower limits of $\{s_n\}$; we use the notation  $$\lim_{n\to \infty}\sup s_n=s^*, \lim_{n\to \infty}\inf s_n=s_*.$$
3.17 Theorem Let $\{s_n\}$ be a sequence of real numbers. Let $E$ and $s^*$ have the same meaning in Definition 3.16. Then $s^*$ has the following porperties:
(a)$s^*\in E.$
(b)If $x>s^*$, there is an integer $N$ such that $n\geq N$ implies $s_n\leq x$.
Moreover, $s^*$ is the only number with the poperties (a) and (b).
Of course an analogous result is true for $s_*$.
Proof
(a) If $s^*=+\infty$, then $E$ is not bounded above; hence $\{s_n\}$ is not bounded above, and there is a subsequence $\{s_{n_k}\}$ tal que $s_{n_k}\to +\infty$.
If $s^*$ is real, then $E$ is bounded above, and at least one subsequential limit exists, so that (a) follows from Theorems 3.7 and 2.28.
If $s^*=-\infty$, then $E$ contains only one element, namely $-\infty$, and there is no subsequential limit. Hence, for any real $M$, $s_n>M$ for at most a finite number of values of $n$, so that $s_n\to -\infty$. (...)"
Has anybody some clarification of what the set $E$ is, or some another source (site, book, etc) that defines $E$ more precisely? And about this proof? I cannot understand! Need some help!
 A: The extended real numbers are precisely the set $\mathbb{R}$ with two additional points that we call $-\infty$ and $\infty$. They behave in the following way:


*

*$-\infty < a$ for all $a \in \mathbb{R}$

*$\infty > a$ for all $a \in \mathbb{R}$. 

*Neighborhoods of $-\infty$ are open sets of the form $(-\infty, a)$

*Neighborhoods of $\infty$ are open sets of the form $(a, \infty)$. 


Rudin then defines $E$ as the subset of the extended real numbers that appear as limit points of the sequence of numbers $\{s_n\}$. For example, $\infty \in E$ iff there is a subsequence $\{s_{n_k}\} \to \infty$. If you feel so inclined, you can check that this agrees with your normal idea of a limit point when combined with the four behaviours described above. 
A: By the definition, $E=\{ x : x=\lim\limits_{k\rightarrow \infty} s_{n_k}\}$. In other words, it is the set of all limits of any subsequence of $\{s_n\}$.
Consider the sequence given by the recursive definition $s_n = (-1)^n$. The set $E$ for this sequence would consist of two elements, $1$ and $-1$. These are the limits of the sequences for $n=2k$ and $n=2k+1$. Thus, the supremum of $E$ and infimum of $E$ are the obvious ones, and we have the upper and lower limits from this set.
Note: For more crazy sequences, this set is sure to have more elements.
As for the proof,
If $\sup E = -\infty$, then is it possible to have an element such that $x<\sup E$?
We may prove the final claim with contradiction:
Assume, for a contradiction, there are an infinite number of terms $s_n$ such that $s_n > M$. Then we may construct a monotone sequence (such as in the bounded sequence theorem earlier) which has a limit $L$ such that $L\geq M$. But this contradicts that $\sup E = - \infty$! Therefore there are only a finite number of such $s_n$. 
Hope this clears things up, cheers.
