Proof dealing with the union of cones and the intersection of polar cones

Question

I need to solve the following problem:

Let $S_1$ and $S_2$ be two cones. Let $P(S)$ be the notation for Polar Cone (P) of $S$.

Let $S_1, S_2\in \mathbb{R}^n$.

(i) Show that $S=S_1 \cup S_2$ is a cone; and,

(ii) Show that $P(S)=P(S_1)\cap P(S_2)$.

Observation: the problem statement does not assume cones' convexity.

I have a draft for (i) but I have no idea on how to do (ii). My try on (i) might be wrong.

Take $\beta\ge0$ and $x\in S_1\cup S_2$. Hence, $x\in S_1$ or $x\in S_2$.

Since both are cones $\beta x\in S_1$ or $\beta x\in S_2$.

Then, $\beta x\in S_1 \cup S_2$. This is the definition of a cone. So, finally, $S_1 \cup S_2$ is a cone.

Is this proof right? How to do (ii)?

Thanks in advance.

Answer 1

Your proof for (i) is correct. For (ii) observe that the following statements are equivalent:

1. $y \in P(S)$.
2. $\langle y, x \rangle \leq 0$ for all $x \in S$.
3. $\langle y, x \rangle \leq 0$ for each $x \in S_1$ and $\langle y, x, \rangle \leq 0$ for each $x \in S_2$.
4. $y \in P(S_1) \cap P(S_2)$.