(a) If sup A < sup B, show that there exists an element of $b \in B$ that is an upper bound for $A$.
I have argued that if sup A $\lt$ sup B, then choose an $\epsilon>0$ such that sup A +$\epsilon \in B$. Since $a \le $ sup A for all $a \in A$, it follows $a \lt $ sup A + $\epsilon$ hence sup A + $\epsilon $ is an upper bound of A as well as an element of B.
(b) Give an example to show that this is not always the case if we only assume sup A $\le$ sup B.
I am having trouble extending my argument into an example which makes me think that it may not be correct.
Question: Is my argument for (a) correct? If not how would one show this fact? Can you give an example asked for in (b)? Thanks in advance.
I need to understand how I am assuming what I am supposed to prove in my approach. Can someone elaborate on that please? The other question doesn't really address this issue.
Best Answer
The reason your argument is circular is that you have asserted the existence of your $\epsilon$ without any supporting argument. This is really the crux of the proof, as once you have found $b\in B$ with $b > \sup A$ it follows quickly that $b$ is an upper bound for $A$. I would suggest the following type of argument for the first part:
Since $\sup A < \sup B$, $\sup A$ is not an upper bound for $B$. It follows that there exists $b\in B$ such that $b > \sup A$.
The first sentence is justified by the definition of the supremum, and the second by negating the definition of an upper bound. One more sentence should be enough to show that $b$ is an upper bound for $A$.
For the second part: Relaxing the strict inequality only allows for one new possibility: $\sup A = \sup B$. So what can go wrong in this case? Keep in mind that in general, a set need not contain its supremum.