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