Let A and B be nonempty and upper bounded real sets.
Suppose That exists $\epsilon > 0$ so for all $ a \in A$ exists $ b \in B$ so that $a + \epsilon < b$
Prove that $\sup A < \sup B $
and if we suppose that for every $ a \in A$ exists $\epsilon > 0$ and $ b \in B$ such that $a + \epsilon < b$. Will necessarily be $\sup A < \sup B $ ?
I think that the first one is false because let $ A = B = (0,1) $ so there is $\epsilon > 0$ so for all $ a \in A$ exists $ b \in B$ so that $a + \epsilon < b$ but $\sup A = \sup B =1 $.
Is that correct ? if not how to prove it ? and about the second one is that true or false and why?
thanks
Your example works for the second question but not for the first: if we take $a =1-\epsilon$ the there is no $b \in B$ with $a+\epsilon =1 <b$.
To prove the first result note that for all $a \in A$ there exist $b \in B$such that $a <b-\epsilon $. This implies that $a <\sup B-\epsilon $. Since this holds for all $a \in A$ we get $sup A \leq \sup B-\epsilon $ Hence $sup A <\sup B$