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Let $K$ be a convex, closed and unbounded set in $\mathbb{R}^n$. Show that for any $x\in K$ there exists a ray ($\{x+tv:t\ge0\}$ with some $v\in\mathbb{R}^n\setminus 0$) contained in $K$ with start point at $x$. What changes when $K$ isn't closed?

luka5z
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    Look here: http://math.stackexchange.com/questions/312821/there-is-a-ray-from-each-point-of-unbounded-convex-set-that-is-inside-the-set –  Oct 08 '14 at 20:16
  • Thank but can I ask you why we consider in that proof $\mathbb{R}^{n-1}$ and $S^{n-1}$ instead of $\mathbb{R}^n$ and $S^n$?? – luka5z Oct 14 '14 at 15:19

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Here is a counterexample: $$K = \{(0,0)\} \cup \{(s,t) \in \mathbb{R}^2 \mid s>0, \, 0 < t < \arctan(s)\}, \,\, x = (0,0) $$ By following through the proof for closed sets, you can choose longer and longer radial segments that are based in $(0,0)$ by letting their angles approach zero. But, the limiting ray with angle exactly equal to zero is not in the set $K$.

Lee Mosher
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