Show that $\ M$ =sup $\ E$ iff for all $\varepsilon \gt$ 0 there is an $\ x$ in $\ E$ such that $\ M$ $\ - \varepsilon \lt x \le M$ Let E be a bounded set and M be an upper bound for E. Show that M=sup E iff for all $\varepsilon \gt 0$ there is an x in E such that $\ M - \varepsilon \lt x \le M$
so I know we have to prove it both ways, so we must prove:
for the forwards direction:
$\ M = $ sup E $\ \Rightarrow \forall$ $\varepsilon \gt 0$ $\exists$ $\ x \in E \ni M - \varepsilon \lt x \le M$
and for the backwards direction:
$ \forall$ $\varepsilon \gt 0$ $\exists$ $\ x \in E \ni M - \varepsilon \lt x \le M$ $\Rightarrow$ M = sup E
I figured for the forward direction that you could suppose x $\gt$ M which would imply that there is an $\ m \in \mathbb{R} \ni$ m $\lt$ M and also an upper bound for E but I'm not sure where that gets me in regards to proving the statement..
I was playing around with the backwards direction and this is what i came up with:
suppose $\ M \neq$ sup E, then $\ \exists \varepsilon \gt 0 \ni \forall x \in E$, $\ M - \varepsilon \ge x \gt M $
$\Rightarrow$ M is in E and there are elements of E that are greater than M. This means M could not be an upper bound for E which contradicts what we are given about M (that M is an upper bound for E).
Thus our assumption is false and therefore M = sup E.
I guess I'm looking for verification of the backwards direction of my proof and help in proving the forward direction of the statement (and the backwards direction if I am wrong).
Thanks in advance.
 A: hint for farward
Le $ P,Q $ and $ R$  be the propositions
$$P : M=\sup E$$
$$Q : \forall x\in E \; x\le M$$
$$R : \forall \epsilon>0\; \exists x\in E : \; x>M-\epsilon $$
You want to prove that
$$P \implies Q \wedge R$$
$$P\implies M \text{ is an upperbound of } E$$
$$\implies \forall x\in E \; x\le M$$
Now let us prove that
$$P \implies R$$
by proving the contrapositive
$$\text{ not } R\implies \text{ not } P.$$
$$\text{not } R \implies$$
$$\exists \epsilon>0 : \forall x\in E\; x\le M-\epsilon$$
$$\implies M-\epsilon \text{ is an upperbound of } E$$
$$\implies M-\epsilon \text{ is an upperbound of } E \text{ smaller than } M$$
$$\implies M\ne \sup E$$
done.
A: I guess that you are defining $\sup E$ as the least upper bound of $E$.
Suppose $M=\sup E$. Then, if $\varepsilon>0$, the number $M-\varepsilon$ is not an upper bound of $E$, hence there exists $x\in E$ such that $M-\varepsilon<x$. The fact that $x\le M$ follows from $M$ being an upper bound of $E$.
Suppose the stated condition holds and let $N$ be an upper bound of $E$. We need to prove that $N\ge M$. By way of contradiction, suppose $N<M$. Then $\varepsilon=M-N>0$ and so there exists $x\in E$ with
$$
M-\varepsilon <x\le M
$$
On the other hand $M-\varepsilon=M-(M-N)=N$, so we have $N<x$: a contradiction to $N$ being an upper bound of $E$.
Now compare this argument to yours to see whether your proof is good.
A: The definition of $\sup E= k$ is 1) $k$ is an upper bound of $E$ and 2) if $a < k$ than $a$ is not an upper bound.
We are given that $M$ is an upper bound if $E$.
ONE:  $\sup E = M\implies$ for all $\epsilon >0$ then there is an $x\in E$ so that $M-\epsilon < x \le M$.
Pf:  $M=\sup E$.  For all $\epsilon > 0$ then $M - \epsilon < M$. So $M-\epsilon$ is not an upper bound of $E$.
So $M-\epsilon$ is not $\ge x$ for all $x \in E$ so there must exist an $x \in E$ so that $x > M-\epsilon$.
And as $M$ is an upper bound of $E$ we know $M \ge x$.
So $M-\epsilon < x \le M$.
TWO:  If for all $\epsilon >0$ then there is an $x: M-\epsilon < x \le M$ then $M = \sup E$.
Proof:  Fo $M = \sup E$ two thing must happen.  1) $M $ must be an upper bound of $E$.  It is.


*If $a < M$ then $a$ can not be an upper bound of $E$.

Let $a < M$.  Then if we let $\epsilon = M -a > 0$ then we have that there is an $x \in E$ so that
$a = M -\epsilon < x \le M$.  So $x > a$ and so $a$ is not an upper bound of $M$.
So the second condition holds.  Therefore $M = \sup E$.
