401.2A Proof using supremum

[@lemyul]

where is part 2?

[@vampirodemente]

Hi! I just wanted to follow up on this exercise because I'm trying to freshen up my real analysis proofs. So if we proof by contradiction, assuming supF>supE then we can say, if supF=m and supE=M, that m is now an upper bound for E. Because m>M there is a real number x s.t. m>x>M (that's an axiom right?), therefore x is an element of F due to it being below supF. However that means x is not an element of E which contradicts the fact that F is a subset of E, implying our original assumption, supF>supE, which means the negation must be true: supF

[@cristhian8790]

+1 sub! :D

[@albinsopaj]

The explanation was complicated and wasn't delivered clearly.