Conical Hull Of A Compact Convex Set

Let $K\subset \mathbb{R}^n$ be a compact convex set not containing the origin. Then the conic hull of $K$ is closed.

### Solution:

We first claim, that the conic hull of $K$ is the set $\bigcup_{z\in K} \mathbb{R}_+ z$: The right-hand side of this equation is definitely contained in the left-hand side by the property of cones (closed under multiplication by non-negative scalars). On the other hand, if we take $x,y\in\bigcup_{z\in K}\mathbb{R}_+ z$, then there are $\alpha_1,\alpha_2\in\mathbb{R}_+$ and $z_1,z_2 \in K$ such that $\alpha_1z_1=x$ and $\alpha_2z_2=y$. Then $x+y=\frac{1}{\alpha_1+\alpha_2}(\frac{\alpha_1}{\alpha_1+\alpha_2}x+\frac{\alpha_2}{\alpha_1+\alpha_2}y)$.The expression in parenthesis is in $K$ and so we proved the claim.

Now, consider the map

(1)
\begin{align} \Phi_c\colon [0,c]\times K \to \mathbb{R}^n \\ (\alpha,z) \mapsto \alpha z \end{align}

The source of this continuous map is compact and therefore its image is compact, too.
Denote by $\delta=\min \{\|x\|\colon x\in K\}$. Then $B_r(0)\cap \operatorname{cone}(K)$ is contained in the image of $\Phi_c$ for $c\geq \frac{r}{\delta}$ and therefore closed as the intersection of two compact sets. This proves that the conic hull of $K$ is closed.

If the origin is allowed to lie in $K$, the statement fails in general. Consider for example the set $K=\operatorname{conv}\{(x,y)\in\mathbb{R}^2\colon x=y^2, x\in [0,1]\}$. Then the conic hull of $K$ is bounded by the half line $\{(x,y)\colon x=y, x\geq 0\}$ and the non-negative $x$-axis. But it does not contain the latter.

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