It's given that: There is an $\epsilon>0$, such that for each $a\in$A, there is a $b\in$B such that $a+\epsilon < b$, Prove that $supA<supB$.
Note: Sup means supermum.
Here is what I did, and I hope you guys give me a Hint for how to continue from here:
$SupA-\epsilon_1<a_1$ (which means that if we substract a little amount $\epsilon_1>0$ from $SupA$ there must be a member $a_1$ in $A$ that is bigger than $SupA-\epsilon_1$.
By using the information they gave us, by substituting $a_1$, we get that there is an ($\epsilon>0$) such that $a_1+\epsilon<b_1$. (where $b_1$ is a certain member in $B$). and from this inequality w can deduce that $a_1<b_1$.
thus, we get that $SupA-\epsilon_1<b_1-\epsilon$
and I also know that $SupB+\epsilon>b_1$ so it means that:
- $SupA-\epsilon_1<b_1$
- $SupB>b_1-\epsilon$
but that inequality doesn't prove that $SupA<SupB$, what did I do wrong? and how do I continue from here?
From here
$$\sup A-\epsilon_1<b_1-\epsilon,$$
try not to throw away that $\epsilon$. As $b_1\in B$, $b_1 \le \sup B$ and so
$$\sup A-\epsilon_1<\sup B-\epsilon. $$
Now this inequality holds for all $\epsilon_1 >0$. Thus we have
$$\sup A \le \sup B-\epsilon.$$
(Try to prove that by assuming the contrary that $>$ holds). Now as $\epsilon >0$, we have
$$\sup A \le \sup B-\epsilon < \sup B,$$
which is what you want.
Remark I hope this is clear that if you throw away that $\epsilon$ too early, you could at best get $\sup A \le \sup B$ (instead of the strict inequality $<$).