1. ## Relative supremum proof

If sup B > sup A, then prove that there is an element b of B that is an upper bound for A.

My work: I split this into two cases, that where B has a maximum, and that where it doesn't. If it has a maximum, then we can set b = sup B = max B, and clearly this is an upper bound for A since sup B > sup A. But I'm lost for the second case...

In general, I'm having trouble with these supremum proofs when it comes to the case where the set doesn't have a maximum. Any help?

2. sup A can't be an upper bound for B anymore, because it is smaller than sup B (which is already the SMALLEST upper bound of B).
So therefore NOT all b are <= sup A.
=>
There exists a b , with b>sup A.
sup A is an upper bound of A, so b must also be an upper bound of A.

3. In your second case, you know that there are b arbitrarily close to sup(B), so then there must be a b such that sup(B) > b > sup(A), for if sup(B) > sup(A) > b for all b, then we cannot find b's that are arbitrarily close to sup(B).