Understanding Theorem 2.28 from Rudin's Real Analysis I don't really understand the wording in this theorem, i.e. what is the hypothesis and what I need to prove.
Theorem 2.28: Let $E$ be a nonempty set of real numbers which is bounded above. Let $y = supE$. Then $y$ is in the closure of $E$. Hence $y$ is in $E$ if $E$ is closed.
I need help understanding what the theorem says and some tips on the proof.
 A: The reals have the least upper bound property.
So if $E$ is bounded above then $y = \sup E$ must exist.
Now there are three possibilities. 
i) $y \in E$.
ii) $y \not \in E$ but $y$ is kind of close to $E$.
iii) $y$ is a million freaking miles away from $E$ and has nothing to do with $E$.
This theorem says:
iii) is not true.  $y = \sup E$ will always be in the closure of $E$.
Furthermore, if $E$ is closed, then $y = \sup E$ will be in $E$.
Another way of putting this:  $\sup E$ is always a limit point of $E$.   
(Hmmm, come to think of it, I wonder we they don't put it that way.)
So
Thereom:  If $E$ is bounded above, then $y= \sup E$ is a limit point of $E$.
Corollary 1: If $E$ is closed and bounded above, then $y = \sup E \in E$.
Corollary 2: If $E$ is bounded above, then $y =\sup B \in \overline {E} = E \cup E'$ (where $E' = \{\text{limit points of }E\}$.
Pf:  $E$ is bounded above so $y= \sup E$ exists.  Let $B_{\epsilon}(y) = \{x\in \mathbb R| d(x,y) < \epsilon\} = (y - \epsilon, y+ \epsilon)$.
$y - \epsilon < y = \sup E$ so $y - \epsilon $ is not an upper bound of $E$.
So there is an $k \in E$ so that $y-\epsilon < k \le y$.  So $k \in B_{\epsilon}(y)$.
So $y$ is a limit point of $E$.
QED
Corollary 1 and 2: follow directly as the definition of limit points and closed sets.
So Theorem 2.28 is basically the theorem and corollaries proven above.
A: The part in blue below isthe hypothesis, and the part in red is what you need to prove.
Theorem 2.28: $\color{blue}{\text{Let } E \text{ be a nonempty set of real numbers which is bounded above. Let }  y = \sup E}.$ $\color{red}{ \text{Then } y \text{ is in the closure of }E. \text{ Hence } y \text{ is in }E\text{ if }E \text{ is closed.}}$
In a theorem, the word "let", "suppose", "assume" let us introduce hypothesis. 
You are given a non-empty set  of real numbers, $E$, and it is bounded above. That is $E \neq \emptyset$  and we can find an integer $M$ such that $\forall x \in E, x < M$. 
Let $y = \sup E$. You have to prove that $y$ is in the closure of $E$, it means you have to show that $y$ is in $E$ or $y$ is a limit point of $E$. In other words, if $y$ is not in $E$, show that $y$ must be a limit point of $E$. Suppose not $\ldots$, try to use the definition of supremum to get a contradiction.
A: HINT:
$y=\sup E$ is the smallest upper bound of $E$. So for every $y'< y$, $y'$ is not an upper bound of $E$. So, there exists $e \in (y',y]$.  That shows that every open interval around $y$ contains points of $E$, and therefore $y \in \bar E$. 
A: $s=\sup E$ is either in $E$ or a limit point of $E$.  Proof:  Assume $\sup E$ is not in $E$.  Then by definition of $\sup$, there is no  $t < s$ such that $\forall e\in E:t \ge e$.  That implies for every $\epsilon > 0$ there is some "test" element $u\in E$ that is greater than $s -\epsilon$; otherwise,  $s-\epsilon$  is an upper bound of $E$ which is less than $\sup E$, contradicting the definition. Lets call such an element $\tau(\epsilon)$.
Now for some given $\epsilon$ and corresponding "test" element $u > s-\epsilon$, consider the sequence $u_0 = u, u_1 = \tau((s-u_0)/2), \cdots u_k = \tau((s-u_{k-1})/2)$.  That sequence demonstrates that $s$ is a limit point of $E$.
Therefore $s$ is in the closure of $E$, and by definition of a closed set, $s$ is in $E$ if $E$ is closed.
