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

docnet said:

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

chwala said:

True, but i need to analyse behavior of the function at critical point. Or not necessary?

PeroK said:

No. You are asked for the supremum of a set.

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.

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.

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.

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.

PeroK said:

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

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##.

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##.

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##.

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##.

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.

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##.