What does "$E$ is not bounded above" mean? I am confused. "Principles of Mathematical Analysis" by Walter Rudin Theorem 3.17. I am reading Walter Rudin's "Principles of Mathematical Analysis".  
There are the following definition and theorem and its proof in this book.  
Definition 3.16: 

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 $s_{n_k} \rightarrow x$ for some subsequence $\{s_{n_k}\}$. This set $E$ contains all subsequential limits, plus possibly the numbers $+\infty$, $-\infty$. 
Put $$s^* = \sup E,$$ $$s_* = \inf E.$$

Theorem 3.17:  

Let $\{s_n \}$ be a sequence of real numbers. Let $E$ and $s^*$ have the same meaning as in Definition 3.16. Then $s^*$ has the following two properties: 
(a) $s^* \in E$. 
(b) If $x> s^*$, there is an integer $N$ such that $n \geq N$ implies $s_n < x$. 
Moreover, $s^*$ is the only number with the properties (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 $\{s_{n_k}\}$ such that $s_{n_k} \to +\infty$.  
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 $s_n \to -\infty$.  
This establishes (a) in all cases.  

I cannot understand the following argument:  

(a)
  if $s^* = +\infty$, then $E$ is not bounded above; hence $\{s_n\}$ is not bounded above, and there is a subsequence $\{s_{n_k}\}$ such that $s_{n_k} \to +\infty$.  

What does "$E$ is not bounded above" mean?
p.12, Rudin wrote "It is then clear that $+\infty$ is an upper bound of every subset of the extended real number system".
And $E$ is a subset of the extended real number system.  
Does this mean "$E \cap \mathbb{R}$ is not bounded in $\mathbb{R}$"?  
Then, for example,
Let $\{s_n\}$ be a sequence such that $s_n = n$.
Then $E = \{+\infty\}$ and $s^* = +\infty$.
And $E \cap \mathbb{R} = \emptyset$.
And $\emptyset$ is bounded above.  
I am very confused.  
I wanna change the above proof to the following proof:  

(a)
  if $s^* = +\infty$, then $\{s_n\}$ is not bounded above, and there is a subsequence $\{s_{n_k}\}$ such that $s_{n_k} \to +\infty$.  

 A: Short answer: I think you are slightly unclear on the definition of an upper limit.
Basically, though, if $\sup(E) = +\infty$, then $E \cap\mathbb{R}$ is not bounded above (in $\mathbb{R}$). We use this to show that $\{s_n\}$ is not bounded above (in $\mathbb{R}$, recall $\{s_n\}$ is a sequence in $\mathbb{R}$, not $\mathbb{R} \cup\{\pm\infty\}$).
We use the fact that $\{s_n\}$ is not bounded above, to show that there exists a subsequence which diverges to $+\infty$. So, $+\infty\in\mathbb{R}.$
Long answer:
Let $(a_n)$ be a sequence in $\mathbb{R}.$ Define $A$ to be the subset $A$ of $\mathbb{R} \cup\{\pm\infty\}$ consisting of all subsequential limits of $(a_n)$ and, also, $\pm\infty\in A$ if there exists a subsequence of $(a_n)$ which diverges to $\pm\infty$, respectively.
$a^*$ is merely a shorthand $\limsup_{n\to\infty} a_n$. The definition of $\limsup_{n\to\infty} a_n$ is $\sup(A)$ where $A$ is defined as above. (This is what definition 3.16 means).
Therefore, $\sup(A) = +\infty$ means, by definition of supremum, that if $x\in\mathbb{R} \cup\{\pm\infty\}\ni x<+\infty$ then $x$ is not an upper bound of $A$.
In other words, no real number is an upper bound of $A$. In other words, as subset of $\mathbb{R}$, $A$ is not bounded above, by the definition of ''bounded above'' (see definition 1.7).
Moreover, $A\cap\mathbb{R}\neq\varnothing,$ since $\sup(\varnothing) = -\infty$ (see here). Therefore, we avoid vacuous statements.
In other words, since $A$ is not bounded above in $\mathbb{R},$
$$
\forall M\in\mathbb{R},\hspace{1mm}\exists\hspace{1mm}
a\in A \cap\mathbb{R}\ni a> M.
$$
In other words, viewing the definition of $A$, for all $M\in\mathbb{R}$ there exists a real number $a>M$ for which there exists a subsequence of $(a_n)$ which converges to $a$.
Therefore,
fix an arbitrary $M \in\mathbb{R}$
and consider the real (i.e., $a \neq +\infty$) $a>M$ for which there exists a sub-sequence $(a_{n_k})$ of $(a_n)$ such that $a_{n_k} \to a$.
Consider, also, this sub-sequence $(a_{n_k})$.
Using the definition of convergence of a sequence,
we have that for the positive quantity $a -M$,
there exists an $N\in\mathbb{Z}\ni k \geq N \implies |a - a_{n_k}| <a -M$.
Consider any integer $k \geq N$.
So, now we have fixed a $k\in\mathbb{Z}$ such that $|a - a_{n_k}| <a -M$.
It folows (since $a - a_{n_k} \leq |a - a_{n_k}| < a - M$) that $a_{n_k} > M$.
Now, forget that we fixed $M$.
This shows that $\forall M \in\mathbb{R}$,
some term of $(a_n)$ is greater than $M$.
Hence, $(a_n)$ is not bounded above.
With this in hand, it is relatively easy to construct a sub-sequence of $(a_n)$ which diverges to $+\infty.$
This, I leave to you.
