Upper bound of a set (equivalence problem) I took a real analysis course three years ago and I unfortunately didn't get all of it, starting with basics.

Question: Let $E$ be a set of real numbers. Show that $x$ is not an upper bound of $E$ if and only if there exists a number $e \in E$ such that $e>x$.

Relevant definition given a few pages back:

Definition (Upper bound): Let $E$ be a set of real numbers. A number $M$ is said to be an upper bound for $E$ if $x \leq M$ for all $x \in E$.

I feel the problem is tautological, but I tried writing a proof. 

Proof: If $x$ is not an upper bound for $E$ then by the definition exists $e \in E$ such that $e>x$. If $x$ is an upper bound for $E$ then again by the definition we have $e \leq x$ for all $e \in E$, therefore no $e \in E$ exists such that $e >x$.

I mentioned the definition twice, which makes me feel like I'm begging the question (or worse, using what is meant to be proven). It is from this book: Elementary Real Analysis - Thomson.
 A: Suggestion: when you are trying to prove statements in real analysis (or elsewhere, come to that), think about the logic first.  It's not unusual for that to be the main difficulty.  In this case I think the definition in your text is poorly phrased - the role of the word "if" is not very clear.  You could rewrite it this way: "$M$ is an upper bound for $E$" means "for all $x\in E$ we have $x\le M$".
So, what does it mean to say that $M$ is not an upper bound for $E$?  It means that the statement "forall. . ." is false.  Now (logic again!) the negation of
$$\hbox{for all $x\in S$ ( . . . )}$$
is
$$\hbox{there exists $x\in S$ such that not ( . . . ) .}$$
For example, if someone says to you, "all proofs in real analysis are easy" and you don't agree, then what you are saying is that "there exists a proof in real analysis which is not easy".
So to say that $M$ is not an upper bound for $E$ means
$$\hbox{there exists $x\in E$ such that not ($x\le M$)}$$
or in other words
$$\hbox{there exists $x\in E$ such that $x>M$.}$$
And that's pretty much done!  Your question uses different names for the variables, which hopefully is something you are comfortable with - though it does make it seem that the text writers have gone out of their way to be confusing.  They are asking what it means for $x$ to be not an upper bound, so we have to replace $M$ by $x$, and then we have to replace $x$ by another letter, say $e$.  So we get: $x$ is not an upper bound for $E$ means
$$\hbox{there exists $e\in E$ such that $e>x$}$$
which is exactly what you were asked.
A: In your original proof, it isn't totally clear exactly where the definition was being applied, and it isn't totally clear which direction(s) of the "if and only if" was being demonstrated.
You can expand out the proof by being more explicit in your negations. We need to negate both a quantifier and an order relation. Ordinarily you would gloss over both negations at once without any fanfare, but the question is so elementary that it's better to proceed in painstaking detail, negating each clause in turn:

First, recall that for real numbers $e$ and $x$, the negation of $e\leq x$ is equivalent to $e>x$. This is worth proving by itself, since it's a nontrivial property of the real numbers. Let's assume that we've already proved it.
Now by definition, $x$ is not an upper bound of $E$ iff it is not the case that ($e\leq x$ for all $e\in E$), which holds iff there exists some $e\in E$ such that (it is not the case that $e\leq x$), which holds iff there exists some $e\in E$ such that $e>x$. Collapsing the "iff"s results in the desired statement: $x$ is not an upper bound of $E$ iff there exists some $e\in E$ such that $e>x$.

If you prefer, you can write the proof more compactly with the symbols $\forall, \exists, \neg,$ and $\iff$.
One final note: the proof is probably incorrect in some non-classical logics. Goodness knows I'm not an expert there, so don't take my word for it. But the very thought should give you pause and make the theorem feel less like a tautology.
