Question

5. Let S be a non-empty bounded subset of R. If a > 0, show that sup (aS) = a sup S where aS = {as : s E S}. Let c = sup S, show ac = sup (aS). This is done by showing: (a) ac is an upper bound of aS.

Answer 1

please keep in mind that the by showing part a and part b we are actually proving that ac is supremum because if you notice the definition of supremum of a set then this is same as showing part a and part b if you still dont understand something you can comment down i will explain more explicitly

Goin? poin: We have axo we need to show ac = suplas) where ca sups @ fiestly we will show ac is an upper bound of as= {as/ses} 4G is the supremum of s so c is an upper bound 1>) + Xes → XEC ə ax sac if azo ty - an sac aty e as we have y sac - thus ac is an upper bound. 6 Now supposer is an upper bound of as ses we have as ar ssr Z is upper bound of s ř >c since is the supreman ogs → rzac » ac sr thus ac is the supremum of as from @80 | १२१२