Let S be nonempty and bounded above subset of the reals.
Prove sup(S)=max{closure of S}.
Say b=max{closure of S}
I know how to prove b is an upper bound for S. It follows easily in a proof by contradiction.
I'm confused about how to prove b is the LEAST upper bound.
I thought at first I should try to do proof by contradiction again.
But I get stuck.
For instance...
Assume there exists a real number c < b so that c>=s for all s in S
If b is in S then c>=b and c<b. Contradiction.
If b is not in S then b must be on the boundary of S, which I'll denote dS. (Because b=max{closure of S}=max{S union dS})
Its at this point I'm not sure how to proceed.
I was thinking to break into cases of whether or not c is in S or S complement.
But if c is in S then no contradiction follows. Since c<b.
Any ideas??
Thanks