Proof using Supremum

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.

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)##.

ChiralSuperfields said:

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?

ChiralSuperfields said:

##M + \epsilon ≥ s + \epsilon##

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

ChiralSuperfields said:

##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.