Proving Suprema & Infima: Showing Sub B is an Upper Bound of A

[lokisapocalypse]

I need to show that if sup A < sup B, then there is an element in B that is an upper bound of A.

Well I know that if sup B is in B, then sub B is the element in B that is an upper bound of A. But I don't know how to show this is true if sup B is not in B.

 

[Hurkyl]

You try a proof by contradiction?

 

[AKG]

You need to prove simply that $\exists b \in B$ such that $\sup A \leq b \leq \sup B$. Assume not: then $\sup A$ is an upper bound of B, which contradicts the fact that $\sup A < \sup B$, and you're done.

 

[lokisapocalypse]

AKG--

I'm not sure I understand your reasoning. What you say makes sense but that doesn't prove that such a b exists. Or does it and I just don't get it which is equally as possible.

 

[Hurkyl]

Suppose b doesn't exist. Can you prove a contradiction? (such as sup A = sup B)

 

[lokisapocalypse]

But what about the case where sup B is not in B? Then I don't think I can get that contradiction.

 

[Hurkyl]

Then, try to find two sets A and B where b doesn't exist, but sup A < sup B!

We're lucky, we already know this will be impossible, but it still might give you insight.

(As always, it often helps to write out what the definitions say too)

 

[lokisapocalypse]

Okay solid. Thanks.