If your set $Z$ is $\epsilon$-dense, then it is also dense!  This is in fact so for every normed space $X$, and for every  subset $Z\subseteq X$ which is positively homogeneous in the sense that $\alpha Z = Z$, for every scalar $\alpha>0$.  The reason is as follows, for any point $x$ in $X$, and any $\alpha>0$, one has that
  $$
  d(\alpha x,Z) = \alpha \,d(x,Z),
  $$
  where $d$ is the distance function.

Thus, if $Z$ is $\epsilon$-dense, one has for every $x$ in $X$ that
  $$
  d(x,Z) = (1/n)\,d(nx,Z) \leq (1/n)\varepsilon,
  $$
  from where you deduce that $d(x,Z)=0$