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Assume we have a convex set $U$. Given $x \in U$, assume there exists a vector $y$ such that $\forall t>0, \ \ x+ty \in U$. I wish to prove that $\forall z \in U,\ \ \forall t>0,\ \ z+t y \in U$ or give a counterexample.

If a set is convex, given any pair of vectors $a,b \in U$, then $a+(t-1)b \in U, \ \ 0\le t \le 1$.

Assume that we have a $z \in U, \ z+ty \in U, \ \ 0<t \le k$ but $z+(k+\epsilon)y \not\in U$.

Then $z+k y \in U$. Convexity gives that $x+ty-(z+ky) \in U, \ \ \forall t>0$. Simplify and we get $x-z+(t-k)y \in U, \ \ \forall t>0$.

What is the angle between $x+ty$ and $x-z+(t-k)y$?

For two vectors $A$ and $B$, $A \cdot B = ||A||||B||\cos(\theta)$. Thus

$\frac{(x+ty)\cdot (x-z+(t-k)y)}{||x+ty|| ||x-z+(t-k)y||} = \cos(\theta)$.

How can I prove that as $t \rightarrow\infty, \ \cos(\theta)\rightarrow 0$? Is there an easier way that proving that the two vectors are parallel at infinity and thus there is no such $k$?

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Try using the fact that $U$ is an intersection of half-spaces. I guess it might help. –  Giuseppe Negro Feb 13 '13 at 23:09
    
Do you mean, for all $x$, there exists such a $y$? Or do you mean, suppose there exists $x$ such that there exists $y$ etc.? –  Jonas Meyer Feb 13 '13 at 23:10
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@JonasMeyer I mean that if there exists such a $y$ for some $x$, then that $y$ has this property with all $z$. –  Valtteri Feb 13 '13 at 23:17
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Here is the counterexample: $$ U = \{(1,0)\} \cup [0,1) \times \mathbb R \subset \mathbb R^2 $$ take $x=(0,0)$, $y=(0,1)$, $z=(1,0)$.

If $U$ is closed you can prove the statement. Take the sequence $x_k = \frac{x+kty}{k}+(1-1/k)z$, it is a convex combination of the points $x+kty$ and $z$, which are points of of $U$. Hence $x_k\in U$. Now notice that $x_k$ converges to $z+ty$ for $k\to \infty$, so $x+ty \in \overline U$.

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Is the statement true for closed convex $U$? –  user7530 Feb 13 '13 at 23:13
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yes, if $U$ is closed you can prove the statement –  Emanuele Paolini Feb 13 '13 at 23:15
    
@EmanuelePaolini If we take these examples, isn't, for example $(0,0)+2*(0,1) = (0,2) \not\in U$? So, can you extend $y$ indefinitely and stay within the set? –  Valtteri Feb 13 '13 at 23:27
    
you mean: $(0,0) + 2\cdot(1,0) = (2,0)\not \in U$. But that is not a convex combination, because $1+2>1$. –  Emanuele Paolini Feb 13 '13 at 23:29
    
@EmanuelePaolini I see what you did with the counterexample, very nice. TBH, I was thinking of closed sets. Still trying to figure how that proof works there. –  Valtteri Feb 13 '13 at 23:38
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