Is a pointed cone always acute?

Question

how to show ?

A cone C is called acute, if there exists an open halfspace$ Ha = \{x ∈ R^P : \langle x, a \rangle > 0\} $ such that $cl(C) ⊂ Ha ∪{0}$.

Is a pointed cone always acute? What about a convex cone?

A subset$ C ⊆ R^P$ is called a cone, if $αd ∈ C $for all $d ∈ C$ and for all α ∈ R, α > 0.

A cone C in $ R^P $is called

• convex if $αd1 + (1 − α)d2 \in C $for all $d1, d2 \in C $and for all $0 < α < 1$

• pointed if for $d \in C, d\neq 0, −d \neq C, i.e., C ∩ (−C) ⊆ {0}$

$cl(S) = int(S) ∪ bd(S)$ is the closure of S

$int(S)$ is the interior of $S $ and $ bd(S)$ is the boundary of $S $

$\langle x, a \rangle$ is Interior product.

I'm having trouble for solving the exercises of multicriteria optimization m.ehgrott .Or there is a book that can be helpful for better understanding?

thanks

Answer 1

Claim. A pointed cone need not be acute.

Proof. Consider $\Bbb R^2$ and $C=\{\,\alpha u\mid 0<\alpha, u\in\{(1,0),(0,1),(-1,-1)\}\,\}$. $\square$

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In fact, we even have

Claim. A convex pointed cone need not be acute.

Proof. Consider $\Bbb R^2$ and $C=\{\,(x,y)\in\Bbb R^2\mid x> 0\,\}$. $\square$