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is there any way to prove with epsilon \ other way that the set is unbounded?
$$D=\{(x,y)\in R^2\mid xy=1\} = \{(x,y)\in R^2\mid y=\frac 1 x\}$$ Here is a proof my exerciser at class did ( which is not enough proof for me, If there is way to prove with epsilon, I will be glad if there is ):
let $M\in R$, happens: $(M, \frac 1 M )\in D$ Thus:
$||M + \frac 1 M|| = (M^2 + (\frac 1 M)^2)^{(\frac 1 2)} \ge M$
Thus, not bounded. Is there any other proof for it? something with $\epsilon$, any other way which is more friendly?

Asaf Karagila
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    What definition of boundedness are you working with? Does it have anything to do with $\epsilon$? – L. F. Nov 22 '22 at 20:29
  • Nope. But there is a way to show with epsilon or something similliar no? almost all proofs was with epsilon. but still, even if there is no epsilon, how can I prove it differently than my exerciser? cus I didnt understand his proof –  Nov 22 '22 at 20:31
  • Is there any other proof with is more "trivial" and understandable? –  Nov 22 '22 at 20:31
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    $\epsilon$ usually denotes a small positive number, which has nothing to do with boundedness whatsoever. Again, can you show the definition of boundedness that you are working with? – L. F. Nov 22 '22 at 20:32
  • Oh one minute, sorry –  Nov 22 '22 at 20:32
  • $D$ is bounded if there exists $M>0$ such that $D$ is contained in the ball of radius $M$. –  Nov 22 '22 at 20:33
  • Good. We are trying to prove that $D$ is not bounded. According to the definition you gave, this is the same as showing that there does not exist $M > 0$ such that $D$ is contained in a ball of radius $M$. By De Morgan's laws for quantifiers, this is the same as showing that, for all $M > 0$, $D$ is not contained in a ball of radius $M$. – L. F. Nov 22 '22 at 20:39
  • @L.F. Yea, the problem is, the prove above my exercises gave, is not related to a ball. I dont know which ball he gave, and just something in general, is there any "common" balls I should think of? –  Nov 22 '22 at 20:40
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    Your teacher did exactly what was required: he took an arbitrary $M>0$ and proved that $D$ is not contained in "the" ball of radius $M$ (more precisely: the open ball of radius $M$ and center $(0,0)$), by showing some pair which belongs to $D$ and not to that ball. But he certainly did not write $|M+\frac1M|,$ which means nothing, instead of $|(M,\frac1M)|,$ which is indeed equal to $(M^2 + (\frac 1 M)^2)^{(\frac 1 2)}.$ – Anne Bauval Nov 22 '22 at 20:45
  • I can take which ball I like? So the norm, is basically $||(x,y) - (x_0,y_0)||$, when x,y is the D part, minus the $x_0, y_0$ of ball? ( which are both equal to 0 ). And just a question, how should I think which balls to assign? I am having trouble ( as you just see ) –  Nov 22 '22 at 20:48
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    According to the definition you gave for boundedness, you cannot "take which ball you like": by the reasoning detailed above by L.F. you must show that for all $M>0,$ $D$ is not contained in "the" ball of radius $M,$ which, because of your "the" and not "a", tacitely means: the open ball of radius $M$ and center $(0,0).$ – Anne Bauval Nov 22 '22 at 20:53
  • About the norm you just did, sorry, yea, I just noticed it. ,my bad. Regaridng the solution itself, when he received the expression in sqrt and saw it was bigger then M, why does it prove its not bounded? because all I am saying is for All $M>0$, the set is not in the ball. but I am not saying the set is bounded by M –  Nov 22 '22 at 20:55
  • About what you wrote, got it, thanks :) And one last thing regarding what I just wrote is my final question, then I understood completely :) –  Nov 22 '22 at 20:55

1 Answers1

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Notice that : $$\forall n \in \mathbb{N}^*, \left(n, \dfrac{1}{n}\right) \in D$$ and : $$\left\| \left(n, \dfrac{1}{n}\right) \right\|^2 = n^2 + \dfrac{1}{n^2} \to +\infty$$ then $D$ is unbounded.

Essaidi
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  • Thanks for the answer. I dont have to show that for $M>0$? is it any different from $N$ belong to the natural teams? or it is the same. and just a question, what does it mean theoretically that the limit goes to infinity? what does it say about the set? –  Nov 22 '22 at 21:05
  • $x_n \to +\infty$ means that $x_n$ becomes big enough when $n$ is big. Which means that $D$ contains elements with big enough norm so $D$ can't be bounded. – Essaidi Nov 22 '22 at 21:12
  • I see, thank you very much :) –  Nov 22 '22 at 21:14
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    You're welcome. – Essaidi Nov 22 '22 at 21:14