How to prove a cone is convex and closed? 
Let $A$ be a $m\times n$ matrix and consider the cones $G_0=\{d\in\mathbb R^n:Ad<0\}$ and $G'=\{d\in\mathbb R^n:Ad\le0\}$
Prove that $G'$ is a convex closed cone.

Lets see that $G'=\overline{ G'}.$ Note that this contention $G'\subset\overline{ G'}$ is always true. Let's see the other contention.
Suppose $d\in\overline{G'}$ and $d\not\in G'. $ Thus For every open ball $B$ with $d\in B$ we have that $B\cap G'\neq\emptyset$ and $ Ad>0$
How can I reach the contradition? Is this a good way to prove it or is there a better way? I don't know
And how to prove convexity?
Please help me please thanks

And to show that $G'$ is convex just take two points $x,y \in G'$, the segment between these points must lie in $G'$. Let $0\leq \alpha\leq 1$, study the point $p = \alpha x + (1-\alpha)y$:
$$ Ap = A(\alpha x + (1-\alpha)y) = \alpha Ax + (1-\alpha)Ay \leq 0 $$
the last inequality is true because $\alpha, 1-\alpha \geq 0$ and since $x,y \in G'$ than $Ax \leq 0$ and $Ay \leq 0$.
