**Definition** Doubling dimension ($\dim_D(M)=k$): A Metric space $M=(V,d)$ has doubling dimension at most $k$ if for any $x\in V$ & $r>0$, $B(x,r)\subseteq\bigcup^{2^k}_{i=1}B(x_i,r/2)$.

With this definition, a paper that i have read stated without proof $\dim_D(N)\leq2\dim_D(V)$ for any net $N\subseteq V$, where $N_\epsilon = \epsilon$-net means it is $\epsilon$-covering ($\forall v\in V, \exists n\in N_\epsilon$ s.t. $d(n,v)<\epsilon$) and $\epsilon$-separated (Take $n_1\neq n_2\in N_\epsilon$, $d(n_1,n_2)>\epsilon$)

I would like to know the reasoning for $\dim_D(N)\leq2\dim_D(V)$, many thanks. I have no idea how to construct $\dim_D(N)$ here.