# Find the supremum of ##Y## if it exists. Justify your answer.

• chwala
chwala
Gold Member
Homework Statement
Let ##X## be a set of positive real numbers with infimum ## inf(x)=α>0##.
Let ##Y=\{\sqrt 2 -x^3| x\in X\} ##.

Relevant Equations
Analysis
Refreshing on old university notes...phew, not sure on this...
Ok in my take, ##x>0##, and ##\dfrac{dy}{dx} = -3x^2=0, ⇒x=0## therefore, ##(x,y)=(0,\sqrt2)## is a critical point. Further, ##\dfrac{d^2y}{dx^2}(x=0)=-6x=-6⋅0=0, ⇒f(x)## has an inflection at ##(x,y)=(0,\sqrt2)##.
The supremum of ##Y## is ##\sqrt{2}.##

Last edited:
If I'm understanding correctly, you can't have ##x=0## because ##0## is not in ##X##.

PeroK
docnet said:
If I'm understanding correctly, you can't have ##x=0## because ##0## is not in ##X##.
True, but i need to analyse behavior of the function at critical point. Or not necessary?

chwala said:
True, but i need to analyse behavior of the function at critical point. Or not necessary?
No. You are asked for the supremum of a set.

docnet
PeroK said:
No. You are asked for the supremum of a set.
Okay, i noted that for every increasing values of ##x\in X## with ##x## being positive- i.e (monotonically increasing).. the value of ##Y## is decreasing....where ##(-∞, \sqrt 2)## being my lower and upper bound.

Try choosing the smallest element ##x^*\in X## (it's not necessarily the infimum of ##X##) and plugging it into ##\sqrt{2}-x^3##. Then, see if you can make out the supremum of ##Y## any larger than that number.

docnet said:
Try choosing the smallest element ##x^*\in X## (it's not necessarily the infimum of ##X##) and plugging it into ##\sqrt{2}-x^3##. Then, see if you can make out the supremum of ##Y## any larger than that number.
Let ##f(x)=\sqrt2 -x^3##
##f(0.5)=1.28, f(1)=0.414, f(100)=-999,998.58, f(0.01)=1.4142##... clearly the values of ##f(x) <\sqrt{2}##at ##x>0##.

PeroK
chwala said:
Okay, i noted that for every increasing values of ##x\in X## with ##x## being positive- i.e (monotonically increasing).. the value of ##Y## is decreasing....where ##(-∞, \sqrt 2)## being my lower and upper bound.
The supremum of ##Y## is not ##\sqrt 2##.

If you don't want to study analysis, which is the foundation of calculus, then do some calculus problems. But, you can't use calculus (whether correctly or incorrectly) to solve a problem like this.

docnet said:
Try choosing the smallest element ##x^*\in X## (it's not necessarily the infimum of ##X##) and plugging it into ##\sqrt{2}-x^3##. Then, see if you can make out the supremum of ##Y## any larger than that number.
There is no guarantee that ##X## has a smallest element.

docnet
PeroK said:
The supremum of ##Y## is not ##\sqrt 2##.

If you don't want to study analysis, which is the foundation of calculus, then do some calculus problems. But, you can't use calculus (whether correctly or incorrectly) to solve a problem like this.
Maybe i was pointed in the wrong direction...then let me check my notes again. Nothing difficult for me... i just need direction. Cheers.

PeroK said:
There is no guarantee that ##X## has a smallest element.
OH! you're right. Forgive my mistake.

I want to see if @PeroK approves of my reasoning because he is a fountain of mathy wisdom, and to see if I have gotten any better at analysis.

@@@@ possible spoiler alert. chwala don't look. @@@@

If the smallest element in ##X## is its infimum ##\alpha##, then the answer is is ##\sqrt{2}-\alpha^3##. There is no least number that is greater than a number, so the supremum of ##Y## is in ##Y##.

If ##\alpha<x## for all ##x\in X##, then ##y<\sqrt{2}-\alpha^3## for all ##\in Y##. And the supremum of ##Y## is ##\sqrt{2}-\alpha^3##, since any number less than ##\sqrt{2}-\alpha^3## is less than some element in ##Y##, and any number in ##Y## that is greater than ##\sqrt{2}-\alpha^3## necessitates ##x<\alpha##.

chwala
docnet said:
I want to see if @PeroK approves of my reasoning because he is a fountain of mathy wisdom, and to see if I have gotten any better at analysis.

@@@@ possible spoiler alert. chwala don't look. @@@@

If the smallest element in ##X## is its infimum ##\alpha##, then the answer is is ##\sqrt{2}-\alpha^3##. There is no least number that is greater than a number, so the supremum of ##Y## is in ##Y##.

If ##\alpha<x## for all ##x\in X##, then ##y<\sqrt{2}-\alpha^3## for all ##\in Y##. And the supremum of ##Y## is ##\sqrt{2}-\alpha^3##, since any number less than ##\sqrt{2}-\alpha^3## is less than some element in ##Y##, and any number in ##Y## that is greater than ##\sqrt{2}-\alpha^3## necessitates ##x<\alpha##.
I am looking through the literature now...actually now at archimedean property of ##\mathbb{R}##...self studying... cheers man.

docnet said:
I want to see if @PeroK approves of my reasoning because he is a fountain of mathy wisdom, and to see if I have gotten any better at analysis.

@@@@ possible spoiler alert. chwala don't look. @@@@

If the smallest element in ##X## is its infimum ##\alpha##, then the answer is is ##\sqrt{2}-\alpha^3##. There is no least number that is greater than a number, so the supremum of ##Y## is in ##Y##.

If ##\alpha<x## for all ##x\in X##, then ##y<\sqrt{2}-\alpha^3## for all ##\in Y##. And the supremum of ##Y## is ##\sqrt{2}-\alpha^3##, since any number less than ##\sqrt{2}-\alpha^3## is less than some element in ##Y##, and any number in ##Y## that is greater than ##\sqrt{2}-\alpha^3## necessitates ##x<\alpha##.
My bad, i did not realize that the infimum was given that is inf##(X) = α##. I overlooked this aspect.

docnet said:
I want to see if @PeroK approves of my reasoning because he is a fountain of mathy wisdom, and to see if I have gotten any better at analysis.

@@@@ possible spoiler alert. chwala don't look. @@@@

If the smallest element in ##X## is its infimum ##\alpha##, then the answer is is ##\sqrt{2}-\alpha^3##. There is no least number that is greater than a number, so the supremum of ##Y## is in ##Y##.

If ##\alpha<x## for all ##x\in X##, then ##y<\sqrt{2}-\alpha^3## for all ##\in Y##. And the supremum of ##Y## is ##\sqrt{2}-\alpha^3##, since any number less than ##\sqrt{2}-\alpha^3## is less than some element in ##Y##, and any number in ##Y## that is greater than ##\sqrt{2}-\alpha^3## necessitates ##x<\alpha##.
That's a good outline of a proof. Generally, we should show that ##U = \sqrt{2}-\alpha^3## is an upper bound for ##Y##. Then show either that every upper bound is greater than or equal to ##U##; or, that if ##V < U##, then ##V## is not an upper bound for ##Y##.

docnet and chwala
What if the infimum inf##(X) = α## was not given; can one not use calculus going with the examples that i have seen?

PeroK said:
That's a good outline of a proof. Generally, we should show that ##U = \sqrt{2}-\alpha^3## is an upper bound for ##Y##. Then show either that every upper bound is greater than or equal to ##U##; or, that if ##V < U##, then ##V## is not an upper bound for ##Y##.

If the smallest element in ##X## is its infimum ##\alpha##, then the answer is is ##\sqrt{2}-\alpha^3##. There is no least number that is greater than a number, so the supremum of ##Y## is in ##Y##.

If there is no smallest element in ##X##, then ##x>\alpha##, and ##Y ## is a set of numbers ##y## such that ##y = \sqrt{2}-x^3<\sqrt{2}-\alpha^3=u## and ##x>\alpha##. And so ##y < u## for all ##y## in ##Y##, i.e., ##u ## is an upper bound for ##Y##.

And if there is a number ##v## such that ##v<u##, then ##y^* = v + \frac{|u-v|}{2}## is in ##Y## and ##v<y^*##, and hence ##v## cannot be an upper bound of ##Y##.

And so least upper bound of ##Y## is ##\sqrt{2}-\alpha^3##.

chwala
I would do it more like this.

Note that ##\forall x, y \in \mathbb R: x \le y \iff x^3 \le y^3##.

First, we show that ##U = \sqrt 2 - \alpha^3## is an upper bound for ##Y##.

Let ##y \in Y##. Then ##y = \sqrt 2 - x^3## for some ##x \in X##. As ##x \ge inf(X) = \alpha##, we have ##x^3 \ge \alpha^3## and ##y \le \sqrt 2 - \alpha^3##. Hence, ##U## is an upper bound for ##Y##.

Next, we show that ##U## is the least upper bound.

Let ##V## be an upper bound for ##Y##. Let ##x \in X##. Then, ##\sqrt 2 - x^3 \in Y##. Hence ##V \ge \sqrt 2 - x^3##, ##x^3 \ge \sqrt 2 - V##, and ##x \ge \sqrt[3]{\sqrt 2 - V}##.

Hence ##\sqrt[3]{\sqrt 2 - V}## is a lower bound for ##X## and ##\sqrt[3]{\sqrt 2 - V} \le \alpha##.

It follows that ##\sqrt 2 - V \le \alpha^3## and ##V \ge \sqrt 2 -\alpha^3 = U##.

This shows that ##U## is the least upper bound for ##Y## and completes the proof.

docnet and chwala
PeroK said:
I would do it more like this.

Note that ##\forall x, y \in \mathbb R: x \le y \iff x^3 \le y^3##.

First, we show that ##U = \sqrt 2 - \alpha^3## is an upper bound for ##Y##.

Let ##y \in Y##. Then ##y = \sqrt 2 - x^3## for some ##x \in X##. As ##x \ge inf(X) = \alpha##, we have ##x^3 \ge \alpha^3## and ##y \le \sqrt 2 - \alpha^3##. Hence, ##U## is an upper bound for ##Y##.

Next, we show that ##U## is the least upper bound.

Let ##V## be an upper bound for ##Y##. Let ##x \in X##. Then, ##\sqrt 2 - x^3 \in Y##. Hence ##V \ge \sqrt 2 - x^3##, ##x^3 \ge \sqrt 2 - V##, and ##x \ge \sqrt[3]{\sqrt 2 - V}##.

Hence ##\sqrt[3]{\sqrt 2 - V}## is a lower bound for ##X## and ##\sqrt[3]{\sqrt 2 - V} \le \alpha##.

It follows that ##\sqrt 2 - V \le \alpha^3## and ##V \ge \sqrt 2 -\alpha^3 = U##.

This shows that ##U## is the least upper bound for ##Y## and completes the proof.
Impressive!

PeroK
Note that we didn't need the condition ##\alpha > 0##.

Also, a general point. I've done a detailed proof for this specific case. Sometimes, such a proof is more complicated than the general case. It's not hard to prove that in general::
$$sup(-X) = -inf(X)$$Moreover, if we have a monotonic increasing function ##f## defined on a set ##X##, then, it's not hard to show that:
$$inf(f(X)) = f(inf(X))$$You could prove that as easily as I proved the one specific case. And, if we put these two results together, then the problem is just a specific case of:
$$sup(-f(X)) = -f(inf(X))$$If you'll allow the informal notation.

And, just to be precise, ##f## needs to be monotonic increasing on an interval containing ##X##. Or, all of ##\mathbb R##.

chwala and docnet
PeroK said:
Note that we didn't need the condition ##\alpha > 0##.

Also, a general point. I've done a detailed proof for this specific case. Sometimes, such a proof is more complicated than the general case. It's not hard to prove that in general::
$$sup(-X) = -inf(X)$$Moreover, if we have a monotonic increasing function ##f## defined on a set ##X##, then, it's not hard to show that:
$$inf(f(X)) = f(inf(X))$$You could prove that as easily as I proved the one specific case. And, if we put these two results together, then the problem is just a specific case of:
$$sup(-f(X)) = -f(inf(X))$$If you'll allow the informal notation.

And, just to be precise, ##f## needs to be monotonic increasing on an interval containing ##X##. Or, all of ##\mathbb R##.
I will for sure add this to my notebook. Thanks @PeroK

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