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We call the set $X$ a cone if $\alpha x\in X, \forall x\in X, \alpha\succeq 0$.

Thinking the cases in $\mathbb{R}, \mathbb{R}^2$, and $\Bbb{R}^3$, if $X$ is a cone, $X^C\cup\{0\}$ is also a cone, isn't it?

ex)

  • $[0,\infty]^C\cup\{0\}=[-\infty,0]$ is a cone
  • Let $X=\{(x_1,x_2)\in\Bbb{R}^2:x_1\ge0,x_2\ge0\}$. Then, $X$ is a cone and $X^C\cup\{0\}=\{(x_1,x_2)\in\mathbb{R}^2:x_1<0 \text{ or } x_2<0\}\cup\{0\}$ is also a cone.
Danny_Kim
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1 Answers1

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In short: Let $x\notin X$. Suppose there is $\alpha> 0$ such that $\alpha x\in X$. Then $\frac1\alpha(\alpha x)\in X$, which is bad.

A more set-theoretically general, yet unnecessarily verbose way:

Consider the equivalence (why?) relation on $\Bbb R^n$ $$a\sim b\iff\exists \alpha>0,\ a=\alpha b$$ Call $\mathfrak Z_n$ the set of its equivalence classes. Notice that $\{0\}\in\mathfrak Z_n$. $S\subseteq \Bbb R^n$ is a cone if and only if there is a subset $\mathfrak C\subseteq \mathfrak Z_n$ such that $\bigcup\mathfrak C=0$ and $\{0\}\in\mathfrak C$. Moreover, given a cone $C$, there is exactly one such subset of $\mathfrak Z_n$: namely $$\mathfrak C_C=\{H\in\mathfrak Z_n\,:\,H\subseteq C\}=\{H\in\mathfrak Z_n\,:\,H\cap C\ne \emptyset\}$$ Now, since the elements of $\mathfrak Z_n$ are pairwise disjoint, for any two $\mathfrak C,\mathfrak B\subseteq \mathfrak Z_n$ it holds $$\left(\bigcup \mathfrak C\right)\setminus\left(\bigcup\mathfrak B\right)=\bigcup(\mathfrak C\setminus\mathfrak B)$$

Therefore, if $C_1,C_2$ are cones in $\Bbb R^n$, then $$(C_1\setminus C_2)\cup\{0\}=\bigcup\left((\mathfrak C_1\setminus\mathfrak C_2)\cup\{\{0\}\}\right)$$