Can You Solve a Problem Using the Definition of Supremum?

[member 731016]

Homework Statement

I have solved the problem below; however, I am unsure whether it is correct (I list the step below which I am not sure whether I am allowed to do).

Relevant Equations

$M = sup S$
$s > M - \epsilon$

For this problem,

My solution:
Using definition of Supremum,
(a) $M ≥ s$ for all s
(b) $K ≥ s$ for all s implying $K ≥ M$

$M ≥ s$
$M + \epsilon ≥ s + \epsilon$
$K ≥ s + \epsilon$ (Defintion of upper bound)
$K ≥ M ≥ s + \epsilon$ (b) in definition of Supremum
$M ≥ s + \epsilon$

Now we have two cases for the inequality:

(1) $M = s + \epsilon$
(2) $M > s + \epsilon$

For case (1), $M ≠ s + \epsilon$ from definition of Supremum so,

$M > s + \epsilon$
$M - \epsilon > s$

I am unsure why I get the wrong inequality direction. However, apart for that, it seems correct. I would appreciate any help.

Thanks for any help - Chiral.

 

[docnet]

I wonder if you could use a proof by contradiction., i.e.,

Assume there exists an $\epsilon>0$ such that there does not exist a $s\in S$ such that$s>M-\epsilon$, then it must be that $M \neq \text{sup}(S)$.

 

[FactChecker]

Homework Statement: I have solved the problem below; however, I am unsure whether it is correct (I list the step below which I am not sure whether I am allowed to do).
Relevant Equations: $M = sup S$
$s > M - \epsilon$

For this problem,
View attachment 341649
My solution:
Using definition of Supremum,
(a) $M ≥ s$ for all s
(b) $K ≥ s$ for all s implying $K ≥ M$

$M ≥ s$

Is this any particular s? Suppose it is thousands below M. What will that prove?

$M + \epsilon ≥ s + \epsilon$

Is this any particular $\epsilon$? What if it is huge?

$K ≥ s + \epsilon$ (Defintion of upper bound)

No. If s is close enough to M and $\epsilon$ is huge, this is not true.
I think this approach is doomed to fail. You need to rethink it.