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Find Infimum, Supremum, Minimum, Maximum, if those exist, for this subset of $\mathbb{R}$

$$C = ( \frac{xy}{x^2 + y^2} : 0 < x,y \in \mathbb{R} ) $$

The Infimum is obviously $0$. There is no Minimum because $0 < x, y$ so the value of $0$ is never really reached.

The Maximum is reached for $x = y$ which gives

$$\frac{x^2}{2x^2} = \frac{1}{2}$$

First, is this correct until now ?

For the Supremum, I'm a bit confused. As $x$ and $y$ increase, the value of $x^2 + y^2$ in the denominator gets larger than $xy$ in the numerator, so over time it will again tend toward zero as $x,y \to \infty$, isn't it ?

wengen
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    Over the line $x=y$ the fraction tends to $1/2$. Over the line $y=2x$ the fraction tends to $2/5$. – kmitov Oct 01 '23 at 13:28
  • @kmitov yes in fact and ? – wengen Oct 01 '23 at 13:32
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    You are correct that $C$ has a maximum of $1/2$, thus supremum is also $1/2$. (Because if a set has a maximum, then the maximum is also a supremum, see here for example) – Mengchun Zhang Oct 01 '23 at 13:32
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    Polar coordinates make things easier. – Anne Bauval Oct 01 '23 at 13:33
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    You wrote that it tends to 0. – kmitov Oct 01 '23 at 13:35
  • @MengchunZhang Uh yes, in fact, I did not know that if a set has a maximum, it's also the supremum. It makes intuitive sense actually – wengen Oct 01 '23 at 13:44
  • @AnneBauval But this would give $\cos(\theta) \cdot \sin (\theta)$ in the numerator ? Because $\frac{r \cos(\theta) \cdot r \sin(\theta)}{(r \cos(\theta))^2 + (r \sin(\theta))^2} = \frac{r^2 \cdot (\cos(\theta) \cdot \sin(\theta))}{r^2 \cdot (\cos(\theta)^2 + (\sin(\theta)^2)} = \cos(\theta) \cdot \sin (\theta)$ – wengen Oct 01 '23 at 13:44
  • @kmitov But if both $x$ and $y$ get larger, it still goes to zero, isn't it ? – wengen Oct 01 '23 at 13:45
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    No, it is not true. – kmitov Oct 01 '23 at 13:49
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    If you use polar coordinates, you have to consider $\frac{\sin 2\theta}{2}$ for $\theta \in (0,\pi/2)$ because $x,y>0$. – kmitov Oct 01 '23 at 13:53
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    @wengen (answer to your comment of 9 min ago) Indeed, and the max and inf of $\cos\theta\sin\theta=\frac{\sin(2\theta)}2$ for $\theta\in(0,\pi/2)$ are easy to find! – Anne Bauval Oct 01 '23 at 13:54
  • @kmitov yes in fact, so the minimum would be $0$ with $\theta = 0$ and the maximum is $1/2$ for $\theta = \pi/2$. If you want to post an answer, feel free to do so (so the question doesn't appear in the unanswered section anymore) – wengen Oct 01 '23 at 18:49
  • @AnneBauval Uh yes in fact, it makes more sense now thank you. So the maximum would obviously be for θ=π/2 which gives 1/2. If you want to post an answer, feel free to do so – wengen Oct 01 '23 at 18:50

2 Answers2

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Using polar coordinates, your subset $C$ becomes $$\begin{align}C&=\left\{\cos\theta\sin\theta\mid0<\theta<\frac\pi2\right\}\\ &=\frac12\left\{\sin(2\theta)\mid0<\theta<\frac\pi2\right\}\\ &=\frac12\left\{\sin t\mid0<t<\pi\right\}\\ &=\frac12(0,1]\\ \end{align}$$ Hence $\inf(C)=0\notin C$ ($C$ has no minimum) and $\sup(C)=\max(C)=\frac12.$

Anne Bauval
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  • Merci beaucoup pour votre aide ! J'ose vous demander si vous auriez le temps ces prochains jours pour revoir ces deux questions que j'ai posé hier, je peine un peu à comprendre. Seulement si vous avez le temps bien sûr https://math.stackexchange.com/questions/4778401/prove-a-subset-e-of-mathbbr3-is-exactly-a-plane-if-there-are-vectors-u https://math.stackexchange.com/questions/4778008/let-p-q-r-in-mathbbr3-be-three-points-not-on-a-line-prove-there-is-exa – wengen Oct 01 '23 at 20:27
  • Hi! I'd just like to inform you that the user whose wrong answer on the other thread was deleted has actually reposted it in a question (id 4778597), stuffing it in as if it it is a fact! Would you like to close-vote too? – user21820 Oct 02 '23 at 06:10
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If the maximum of a set exists, then the supremum is the maximum. You correctly identified $\frac{1}{2}$ as the maximum, so the supremum is also $\frac{1}{2}$.