Let $X=l^2$ and $A$ be the set of all positve elements of $X$, Then show that $Cone(A-x)$ is not a subspace. Let $X=l^2 = \{x=\{x_i\}_{i=1}^{\infty}  ~ | \quad \sum_{i=1}^{\infty} |x_i|^2 < \infty   \}$ and $A = \{ x=\{x_i\}_{i=1}^{\infty} \in X  ~ | \quad x_i >0 ~ \forall i \}$ and take $a \in  A.$
My questions (I have conjectured that)  :  
$1-$ Is $ C:= \text{Cone}(A-a)$ a subspace of $X$? 
**This part was solved by Demophilus **
$2-$ $\bar{C} $ is indeed a subspace and it is indeed the whole space $X$. In other words, $ C:= \text{Cone}(A-a)$ is dense in $X$ for all $a \in A$
P.S.   $\text{Cone}(A-a) = \{\lambda (x-a ) ~ | \quad x \in A ~, \lambda \ge 0\}$

The second question is easy to answer. Let us define the finite sequences
$$c_c := \{v \in \ell^2 \mid \exists N > 0 : \forall n > N: v(n) = 0\}.$$
Then, it is easy to check $c_c \subset C \subset \ell^2$. Since $c_c$ is dense in $\ell^2$, you get
$\ell^2 = \overline{c_c} \subset \overline C \subset \ell^2$.
