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 $d(x,Z)=0$

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 $$ d(\alpha x,Z) = d(\alpha x,\alpha Z) = \alpha \,d(x,Z), $$ where $d$ is the distance function.

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