 sup C = sup A + sup B ?
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MHB Craftsman
#1
October 19th, 2016,
09:59
Let $A$ and $B$ be sets of real numbers and write $$C=\{x+y:x\in A,y\in B\}.$$ Find a relation among $\sup A$, $\sup B$, and $\sup C$.
My attempt:
I'm assuming the answer is $\sup C=\sup A+\sup B$.
$x\le \sup A\ \forall x\in A$
$y\le \sup B\ \forall y\in B$
$\implies x+y\le \sup A+\sup B$ $\forall x\in A,\ y\in B$
So, $\sup A+\sup B$ is an upper bound for $C$.
Suppose $\sup C\neq\sup A+\sup B$.
$\implies \exists l<\sup A+\sup B$ such that $x+y\le l$ $\forall x\in A,\ y\in B$
$\implies x\le ly\ \forall x\in A,\ y\in B$
So, $ly$ is an upper bound of $A\ \forall y\in B$
I feel that I'm on the right track but I don't know how to get a contradiction. Any suggestions?
Last edited by Alexmahone; October 19th, 2016 at 10:05.

#2
October 19th, 2016,
11:07
Prove that $\sup A+\sup B\varepsilon$ is not an upper bound of $C$ for any $\varepsilon>0$. For this find a $c\in C$ such that $c>\sup A+\sup B\varepsilon$.
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