I'm not sure if I did the first part of the proof correctly.
Proof: Suppose there exists a number $e∈ E$ such that $e>x$. Then by Definition 1.9 we say that $e$ is a least upper bound of $E$ or $\sup E$. Since $\sup E > x$ for all $x∈E$, it then follows that $x$ is not an upper bound for $E$, hence this is a contradiction.
I'm stuck in figuring out the rest of the proof and I know its a biconditional but not sure if I should write out both conditionals to finish the proof.
I'm using Elementary Real Analysis by Brian Thomson and Andrew Bruckner
Definition 1.9: (Least Upper Bound/Supremum) Let E be a set of real numbers that is bounded above and nonempty. If M is the least of all the upper bounds, then M is said to be the least upper bound of E or the supremum of E and we write M = supE.
Best Answer
Note that your problem is true, just by the definition of upper bound: "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∈ E such that e > x". Formally this just means for a fixed set $E$: $$ \neg (x\text{ is upper bound of }E) \leftrightarrow \exists e \in E: e > x $$ As $(A \leftrightarrow B)$ is equivalent to $((\neg A) \leftrightarrow (\neg B))$ your problem is equivalent to the statement: $$ (x\text{ is upper bound of }E) \leftrightarrow \neg\exists e \in E: e > x $$ which itself is again equivalent to the definition of upper bound $$ (x\text{ is upper bound of }E) \leftrightarrow \forall e \in E: e \le x $$ Because $\neg \exists\ldots$ is equivalent to $\forall \neg\ldots$ and because $e \not> x$ is equivalent to $e\le x$ as the real numbers are totally ordered.