Mathematics Asked by Outlier on July 26, 2020

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?

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$

Answered by Subhajit on July 26, 2020

No it's not correct. This inequality, $k > aL^2 + bL - epsilon$, cannot be derived for all $epsilon$. The respective inequality can be only derived for $sup B$, since that is the supreme.

You can assume thou that if $aL^2 + bL > sup B$ then by setting $epsilon_0= aL^2+bL-sup B>0$, you have that, there exist $x_0$ such that, $0>aL^2+bL geq ax_0^2+bx_0geq sup B-epsilon_0=-(aL^2+bL)> 0$, which is absurd

Answered by random certainty on July 26, 2020

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