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Let A be a convex subset of $$R^{2}$$ containing the origin and possessing the following property: given any constants $$\alpha_{1}, \alpha_2\in R$$ such that $$\vert \alpha_1 \vert + \vert \alpha_2 \vert > 0,$$ the subset $$\{ (t \alpha_1,t\alpha_2 )\in R^2 ;t\geq0\}$$ contains at least one point that does not belong to A. How to show that A is bounded?

Ruy
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JHY
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  • Just a thought - show that you can trap the set by a circle. – Sean Roberson May 23 '22 at 02:10
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    You are missing the condition that $A$ is closed, otherwise, the claim is false. – Moishe Kohan May 23 '22 at 02:41
  • @MoisheKohan, Do you have a counter-example when $A$ is not closed? – Sangchul Lee May 23 '22 at 02:51
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    A counter example is ${(x,y): x>0,\ y>x^2}\cup{(0,0)}$. – Ruy May 23 '22 at 02:52
  • @Ruy, Ah, that makes sense. I was thinking about the case where $0$ is an interior point of $A$. – Sangchul Lee May 23 '22 at 02:57
  • If 0 is assumed to be an interior point, then the question is a duplicate of this one (in view of the accepted answer). However, what's missing in the question is OP's thoughts and attempts to solve the problem. – Moishe Kohan May 23 '22 at 04:06
  • If $A \subset \mathbb{R}^n$ is convex and unbounded, it always contains a ray, The ray need not be based at the origin, of course (hence the deletion of my incorrect answer). – copper.hat May 23 '22 at 04:27
  • @copper.hat: This is what's proven in the answer to the question I linked above. – Moishe Kohan May 23 '22 at 04:29
  • To OP: Welcome to math.SE! Since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Please take moment to give this posting a read to learn how to ask a good question. – Sangchul Lee May 23 '22 at 04:31
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    @MoisheKohan It is a standard result in convex analysis, see Rockafellar's Convex Analysis Theorem 8.4, for example (it deals with a closed convex set, but $A$ is bounded iff $\overline{A}$ is bounded). – copper.hat May 23 '22 at 04:35
  • @copper.hat: of course. – Moishe Kohan May 23 '22 at 04:37
  • @MoisheKohan. The proof in the answer you linked to may be easily adapted to prove the present question, without the assumption that 0 is an interior point, but assuming the convex set is closed. Actually the proof becomes even simpler as it is obvious that $Rx$ (see notation there) is in the convex set. – Ruy May 23 '22 at 04:58
  • I find it very hard justifying a vote to close a question with so much activity! – Ruy May 23 '22 at 05:03
  • @Ruy: In the case when $0$ is an interior point, you do not have to adopt the linked proof, it goes through verbatim. The closed case is, of course, easy as well. As for your last comment, read here. – Moishe Kohan May 23 '22 at 08:15
  • the answer here https://math.stackexchange.com/questions/1686487/to-show-a-closed-convex-set-s-subseteq-rn-is-bounded-if-and-only-if-s-cont?noredirect=1&lq=1 is much clearer than the one above, as it does not discuss interior points (which is not necessary). vote to close – daw May 23 '22 at 13:27
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    Does this answer your question? To show a closed convex set $S \subseteq R^n$ is bounded if and only if $S$ contains no rays. – daw May 23 '22 at 13:27

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How about the set $\{(x, y) | (x \ge 0) \land (|y| \le \sqrt{x})\} $?

This is convex and all lines through the origin with non-zero slope eventually leave the set.

marty cohen
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