Proving Existence of an Upper Bound for A Given Sup A < Sup B

[Enjoicube]

Homework Statement

If Sup A<Sup B Then show that there exists a b\inB which serves as an upper bound for A.
First off, I am not looking for a complete solution but rather a hint.

Homework Equations

SupA-\epsilon<a for some a\inA
SupB-\epsilon<b for some b\inB

The Attempt at a Solution

The only thing I have succeeded in so far is "locating one element of each set"
SupA-\epsilon<a\leqSupA for some a\inA
SupB-\epsilon<b\leqSupB for some b\inB

I know that if I can demonstrate that SupA\leqSupB-\epsilon, then I can be done. Equivalently, if I can show that b\geqSupA for some b\inB I will also be done. However, right now this has me caught.

 

[Enjoicube]

I think I might have a solution. Redefine Epsilon so that it is less than or equal to SupB-SupA. Then SupA\leqSupB-\epsilon. We can still guarantee an element of b SupB-\epsilon and Sup B (same for a). Thus this element of b\geqSupA and so b is an upper bound. If this incorrect, please tell me. Sorry to those of you who thought long and hard.

 

[Unco]

Firstly I assume we're talking about real numbers.

The key property we want to use: for every \varepsilon&gt;0 there is a b\in B such that \sup{B} - \varepsilon &lt; b \leq \sup{B}.

Now what positive number to choose for epsilon? It's given (albeit subtly) in the question: we know that sup(B)-sup(A) > 0.

 

[Enjoicube]

Many thanks, but I think I realized this just before you answered (in my post). Yeah, it was about the reals; what a great feeling when I recognized this. Again, Many thanks for being the first to help. Thread closed, because my homework has been turned in.