Show that $\ M$ =sup $\ E$ iff for all $\varepsilon \gt$ 0 there is an $\ x$ in $\ E$ such that $\ M$ $\ - \varepsilon \lt x \le M$

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.

2. 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 $M = \sup E$.