Let A and B be two nonempty sets that are bounded from above.
1) lets suppose that that there is $\epsilon>0$ such that for each $a\in A$ there is b such that $a + \epsilon < b$. prove that $supA<supB$
2) lets suppose now that for every $a\in A$ there is $\epsilon<0$ and $b\in B$ such that $a + \epsilon < b$. prove or disprove that $supA<supB$
By definition $\sup A-\epsilon$ is not an upper bound of A, $\exists a\in A$ s.t. $a>\sup A-\epsilon$, by assumption, $\exists b$ such that $a + \epsilon < b$, hence $b>a + \epsilon>\sup A$. Hence $\sup B>\sup A$.
The second is wrong. Let $A=B=(0,1)$ satisfies the condition, $\sup A=\sup B=1$