Answer:
See proof below
Step-by-step explanation:
Remember, an upper bound of an ordered set A is an element u∈A such that u≥a for all a∈A. A is bounded above when A has an upper bound. The supremum of A exists when A is bounded above (careful, this is only true for subsets of real numbers!). It is the least upper bound of A, that is, supA≤u for all upper bounds of A, u.
Since we want to prove an "if and only if" statement, we have to prove the following to implications:
First, asume that S has a maximum. Denote it by M. Then M is an upper bound of S, thus supS exists and M≤supS. However, M∈S thus, by definition of upper bound, supS≤M. Combining these inequalities, M=supS, thus S is bounded above and M=supS∈S.
For the second implication, assume that S is bounded from above and sup S belongs to S. supS is an upper bound, and by assumption, supS∈S. Then, according to your definition, supS is the maximum of S, hence S has a maximum.
This proves the statement.
