Let $A$ be a subset of $\mathbb{R}$ such that $A$ is bounded below with inf $A = L > 0$. Prove that if $a,b < 0$, then the set $B = {ax^2 + bx} : x \in A$ is bounded above and sup $B = aL^2 + bL$.
Suppose $k \in B$. Then $k = ax^2 + bx$ for some $x \in A$.
inf $A = L > 0$, $x \geq L$ for all $x \in A$. Hence $k \leq aL^2 + bL$ since $a,b < 0$ for all $k \in B$. Therefore $B$ is bounded above.
Next we show that sup $B = aL^2 + bL$.
Since sup $B$ is the least upper bound for $B$, sup $B \leq aL^2 + bL$.
Let $\epsilon > 0$. By the characterization of supremum, $\exists$ $k = ax^2 + bx \in B, x \in A, a,b < 0$ such that $k > aL^2 + bL – \epsilon$.
This implies that $aL^2 + bL – \epsilon < aL^2 + bL \leq$ sup $B$.
Hence $aL^2 + bL <$ sup $B + \epsilon$. Since this holds for every $\epsilon > 0$, we have sup $B = aL^2 + bL$.
Is the proof above correct?
Best Answer
Since, $\inf A = L > 0 $ , $ A $ is the set of positive reals, $(-a)x^2 + (-b)x > (-a)L^2 + (-b)L > 0 $
implies $ax^2 + bx < aL^2 + bL < 0 $
So, $aL^2 + bL $ is a upper bound for $B $. ( $B $ is the set of negative reals)
Now, we have to show that $aL^2 + bL $ is a least upper bound.
If possible let there is some upper bound $aM^2 + bM $ of $B $ such that $aM^2 + bM < aL^2 + bL < 0 $
this implies $(-a)M^2 + (-b)M > (-a)L^2 + (-b)L > 0 $
from these we can deduce that $ M > L $ and $ M $ is a lower bound of $ A $.(because if $ M $ is not a lower bound of $ A $, then $aM^2 + bM $ can't be a upper bound of $ B $).
So, clearly from $ M > L $ , we get a contradiction to the fact that $ L $ is greatest lower bound of $ A $.
So, our assumption is wrong.
Hence, $ aL^2 + bL = \sup B$