Prove that $(C \cap D)^\vee = C^\vee + D^\vee$. Definitions and notations:
Let $M$ to be the $n$-dimensional Euclidean space $\mathbb{R}^n$ and $N$ its dual space $M^\ast = \mathrm{Hom}_\mathbb{R}(M, \mathbb{R}).$ A subset $C \subset N$ is said to be cone if it is spanned by finitely many elements $v_1, \dots, v_s \in \mathbb{Z}^n$;
$$ C = \{ r_1v_1 + \dots + r_sv_s \in N \mid 0 \leq r_1, \dots, r_s \in \mathbb{R} \}. $$
(This is abuse of notation. $\mathbb{Z}^n$ is regarded as a subset of $N$ with an identification $N = \mathbb{R}^n$.)
For a cone $C \subset N$, the subset $C^\vee \subset M$ defined as follows is said to be dual cone of $C$;
$$ C^\vee = \{ \mu \in M \mid \text{$\langle \nu, \mu \rangle := \nu(\mu) \geq 0$ for all $\nu \in C$} \}. $$
Problem: As a homework, I was required to prove the following statement about dual cones.

Let $C, D$ be cones and $C^\vee + D^\vee = \{ \xi + \eta \mid \xi \in C^\vee,\ \eta \in D^\vee \}$.
  Prove that $(C \cap D)^\vee = C^\vee + D^\vee$.

Attempt: I already have proved one inclusion relation $(C \cap D)^\vee \supset C^\vee + D^\vee$. But I have been at a loss how to show the other inclusion relation $(C \cap D)^\vee \subset C^\vee + D^\vee$.  I think I need to decompose every $\mu \in (C \cap D)^\vee$ into $\mu = \xi + \eta$ so that $\xi \in C^\vee$ and $\eta \in D^\vee$. But how can I find out such a nice ones?  
I would greatly appreciated if you gave me a hint rather than a complete answer since this is a homework.
 A: One approach is to first prove $C = C^{\vee\vee}$. Then, with the aid of the fact that $C \subseteq D$ implies $D^\vee \subseteq C^\vee$, the direction you want is made much easier. 
Relevant here is Farkas's lemma. Let me know if these hints are too obscure. 
Edit: Here is some more detail. The easier direction $C^\vee + D^\vee \subseteq (C \cap D)^\vee$ has been proven; we want to show $(C \cap D)^\vee \subseteq C^\vee + D^\vee$. 
Preliminaries: for a finite-dimensional vector space $M$ over $\mathbb{R}$ there is a canonical isomorphism $M \cong \hom(M^\ast, \mathbb{R})$, i.e., an isomorphism $i: M \stackrel{\sim}{\to} M^{\ast\ast}$, where for $v \in M$ we define $i(v)$ by the rule $i(v)(f) := f(v)$ (for any $f \in M^\ast$). 
Define the dual of a cone $D \subseteq M^\ast$ to be $D^\vee = \{v \in M: \; \forall_{f \in D} i(v)(f) \geq 0\}$. 


*

*Step 1: check that if $C_1 \subseteq C_2$ where $C_1$, $C_2$ are cones in $M$, then $C_2^\vee \subseteq C_1^\vee$ (easy). Similarly for cones in $M^\ast$. 

*Step 2: Using the definitions, show $C \subseteq C^{\vee\vee}$. 
Using a famous result called Farkas's lemma, we can also prove the reverse inclusion: $C^{\vee\vee} \subseteq C$. 


*

*Step 3: prove this. Hints: suppose $v \notin C$ and show $v \notin C^{\vee\vee}$. This translates to showing there exists $f \in C^\vee$ such that $i(v)(f) = f(v) < 0$, or that there exists $f \in M^\ast$ such that $f(u) \geq 0$ for all $u \in C$, but $f(v) < 0$. For this, use Farkas's lemma. 

*Step 4: conclude $C = C^{\vee\vee}$ for all cones $C$ in $M$. Of course the same applies to cones $D$ in $M^\ast$ as well. 

*Step 5: use steps 1 and 4 to conclude the bi-implication $C_1 \subseteq C_2 \Leftrightarrow C_2^\vee \subseteq C_1^\vee.$ The forward implication is just step 1; for the backward implication, use both steps 1 and 4. 
Finally, to prove $(C_1 \cap C_2)^\vee \subseteq C_1^\vee + C_2^\vee$, it is enough (by step 5) to prove 
$$(C_1^\vee + C_2^\vee)^\vee \subseteq (C_1 \cap C_2)^{\vee\vee} = C_1 \cap C_2$$ 
For this, we just prove $(C_1^\vee + C_2^\vee)^\vee \subseteq C_1$ and similarly $(C_1^\vee + C_2^\vee)^\vee \subseteq C_2$. By one of the steps above, this is equivalent to $C_1^\vee \subseteq (C_1^\vee + C_2^\vee)^{\vee\vee}$. But it is trivial that $C_1^\vee \subseteq C_1^\vee + C_2^\vee$. Similarly, $C_2^\vee \subseteq C_1^\vee + C_2^\vee$. 
A: This is my incomplete answer to get some advice.


*

*$X \subseteq Y \implies X^\vee \supseteq Y^\vee$: 
For $\eta \in Y^\vee$ and $x \in X \subseteq Y$, we get $\langle x, \eta \rangle \geq 0$ and conclude that $\eta \in X^\vee$.

*$C + D$ is a cone:
Since $C$ and $D$ are cones, there exist some finitely many elements $v_1, \dots, v_s, w_1 \dots, w_t \in \mathbb{Z}^n$ such that
$$ C = \Big\{ \sum p_i v_i \mid 0 \leq p_i \in \mathbb{R} \Big\},$$
$$ D = \Big\{ \sum q_j w_j \mid 0 \leq q_j \in \mathbb{R} \Big\}.$$
Hence, $C + D$ is
$$ \Big\{ \sum p_i v_i + \sum q_j w_j \mid 0 \leq p_i, q_j \in \mathbb{R} \Big\}$$
and is a cone, indeed.

*$(C + D)^\vee \subseteq C^\vee + D^\vee$:...

*$C + D \subseteq C \cap D$:
Let $x \in C + D$. From (2), there exist $p_i, q_j \geq 0$ such that
$$ x = \sum p_i v_i + \sum q_j w_j. $$
Assume that $x \not \in C$. By Farkas' lemma (and Riesz representation theorem), there exist $y \in M$ such that
$$ \langle y, v_i \rangle \geq 0, \quad \langle y, x \rangle < 0 $$
for all $i$. Since $\langle y, x \rangle = \sum p_i \langle y, v_i \rangle + \sum q_j \langle y, w_j \rangle$, there exist $\hat{\jmath}$ such that $\langle y, w_\hat{\jmath} \rangle < 0$...

*$(C \cap D)^\vee \subseteq C^\vee + D^\vee$:
From (1), (3), and (4), we get
$$ (C \cap D)^\vee \stackrel{(1), (4)}{\subseteq} (C + D)^\vee \stackrel{(3)}{\subseteq} C^\vee + D^\vee.$$

