# 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

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.

• Careful, sup E is the least of the upper bounds, and is unique if it exists.. You seem to be claiming any upper bound is sup E. – Pedro Tamaroff Dec 30 '14 at 0:16
• But what is this proof supposed to prove? – Bernard Dec 30 '14 at 0:16
• I don't think you've stated your problem right. Also, you have to state the definitions you're using on SE since people are unlikely to have the books on hand. – Batman Dec 30 '14 at 0:21
• I think you question is an alternative form of the upper bound's definition, so checking the book may help~ – robit Dec 30 '14 at 0:24
• I think that the claim of the post is THE definition of not being an upper bound... – Amitai Yuval Dec 30 '14 at 0:30

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.

• ... and because the reals are totally ordered, so $\neg(e>x)$ is the same as $e\le x$. – Henning Makholm Dec 30 '14 at 0:41
• Oh, you are right. Thank you! – Thorsten Dec 30 '14 at 0:42
• I'm just confused over e > x because I'm not sure if that means that its a upper bound and a supremum. I know the supremum requires that it be the least of the upper bounds but the greater than inequality sign confuses me. – Alexander King Dec 30 '14 at 0:59
• If you pattern match $e > x$ against the definition of upper bound or supremum, you will notice that a $>$-sign is not mentioned at all! Luckily you can use the logical equivalences as above to turn it into some $\le$ which matches the definition of upper bound. (But it is not enough for supremum). – Thorsten Dec 30 '14 at 1:18