# Analysis - supA<sup B

• Sep 5th 2010, 07:01 PM
kathrynmath
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.
• Sep 5th 2010, 09:34 PM
bubble86
Let $\alpha = \sup (A)$ and $\beta = \sup (B)$
$\alpha < \beta \Rightarrow \alpha \not= \beta$
$\rightarrow B\setminus A \not= \emptyset$
$\rightarrow \exists b \in B : b \notin A$
$\rightarrow \alpha \le b$
$\rightarrow \forall a \in A ; a \le b$
• Sep 6th 2010, 05:43 AM
Ackbeet
$\rightarrow \exists b \in B : b \notin A$ to
$\rightarrow \alpha \le b$
What if $A=(1,2)$ and $B=(0,3)$? Then $0.5\in B$, $0.5\not\in A$, but $0.5 for all $a\in A.$ Your proof is the right outline. You just need to do a bit more to ensure the second line I've quoted.
Yes sorry i was referring to the interval $\ (\alpha , \beta)$