# Thread: Analysis - supA<sup B

1. ## Analysis - supA<sup B

If sup A<sup B, then show that there exists an element bB that is an upper bound for A.
This one is really frustrating me. I'm having trouble even beginning.
I know sup A implies s<=b where c is an upper bound for A.
Similarily sup B implies s<= d where d is an upper bound for B.

2. Let $\displaystyle \alpha = \sup (A)$ and $\displaystyle \beta = \sup (B)$
$\displaystyle \alpha < \beta \Rightarrow \alpha \not= \beta$
$\displaystyle \rightarrow B\setminus A \not= \emptyset$
$\displaystyle \rightarrow \exists b \in B : b \notin A$
$\displaystyle \rightarrow \alpha \le b$
$\displaystyle \rightarrow \forall a \in A ; a \le b$

$\displaystyle \rightarrow \exists b \in B : b \notin A$ to
$\displaystyle \rightarrow \alpha \le b$
What if $\displaystyle A=(1,2)$ and $\displaystyle B=(0,3)$? Then $\displaystyle 0.5\in B$, $\displaystyle 0.5\not\in A$, but $\displaystyle 0.5<a$ for all $\displaystyle a\in A.$ Your proof is the right outline. You just need to do a bit more to ensure the second line I've quoted.
4. Yes sorry i was referring to the interval $\displaystyle \ (\alpha , \beta)$