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Let $V$ be a finite dimensional real Euclidean space and $C$ be an open convex cone in $V$. I need to prove that $C = int(\overline{C})$. I proved that $C \subseteq int(\overline{C})$. I have difficulty in proving $int(\overline{C}) \subseteq C$. I tried to prove it as follows.

Let $x \in int(\overline{C})$. Suppose $x \in \overline{C}-C$(that is in the boundary of $C$). Then for every $r>0$, $B(x,r)\cap C$ is non-empty. Now by convexity of $C$, $tx +(1-t)y \in C$ for $t \in (0,1)$ and for any $y \in B(x,r) \cap C$ where $r>0$ . Since $C$ is a cone, we have $x+y\in C$ for any $y \in B(x,r) \cap C$ where $r>0$ .

I don't know how to proceed further. I'm not sure whether the approach is correct or not. I also saw in Wikipedia(Symmetric cone) "An open convex cone coincides with the interior of its closure, since any interior point in the closure must lie in the interior of some polytope in the original cone" which I don't understand.

Please help me to solve it!!

La Rias
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    It's not clear how you know that $tx + (1-t)y \in C$, given that $x \not\in C$. – David Dec 17 '15 at 06:22
  • I don't know how to say it mathematically, but intuitively I feel that since C is convex, y is in C and x in the boundary of C, the line joining x and y(excluding x) should be in C. – La Rias Dec 17 '15 at 06:27
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    Here's an idea. Let $x \in int(\bar{C})$. Then there is some closed tetrahedron (or more generally, simplex) centred at $x$ and contained in $\bar{C}$. The vertices of the tetrahedron are in $\bar{C}$, so there are points in $C$ as close as we like to the vertices. Prove that if the points are taken close enough to the vertices, then $x$ is in the convex hull of those points. – David Dec 17 '15 at 06:36

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