Question

 

if S is non empty and it has a lower bound, then it has an infimum, inf S

Start of proof:

b<=s for all s in S

LB: {y in R:y<=S, for all s in S}

LB is bounded above by X and LB is non empty since b is in LB

Find supLB=infS, thats what needs to be completed

Answer 1

As you have not clearly mentioned the nature of S so I am answering for two different cases with two different results

1) S is a set of rational Numbers

Lets define S as

Here S is non empty as 4,9 etc belong to S

Also it is bounded below as rational numbers like 1,1.2 etc exist which are its lower bounds

However S does not have an infimum as √2 is an irrational number and thus we can not find any unique rational number which can be greatest member of set of lower bouds of S

So the statement in the question is not correct in this case

2) S is a set of real numbers (This is most likely the case of your question as in your solution you write y in R)

In this case the statement is true and is the outcome of Axiom called

Order-Completeness property of R : The set of upper bounds of a non empty set of real numbers which is bounded above has a smallest member or equivalently every non empty sub set of real numbers which is bounded above has a least upper bound

To use this property we can define a new set S' = -S

As S is bounded below S' will be bounded above

Therefore by above property S' has a supremum

Consequently S has an infimum