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

[chwala]

Homework Statement

Let $X$ be a set of positive real numbers with infimum $inf(x)=α>0$.
Let $Y=\{\sqrt 2 -x^3| x\in X\}$.

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

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 $\sqrt{2}.$

 

[docnet]

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

 

[chwala]

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?

 

[PeroK]

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.

 

[chwala]

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.

 

[docnet]

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.

 

[chwala]

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]

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.

 

[PeroK]

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.

 

[chwala]

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.

 

[docnet]

There is no guarantee that $X$ has a smallest element.

OH! you're right. Forgive my mistake.

 

[docnet]

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]

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.

 

[chwala]

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.

 

[PeroK]

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 $Y$.

 

[chwala]

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

 

[docnet]

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 $\sqrt{2}-\alpha^3$.

 

[PeroK]

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.

 

[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.

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]

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