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.

  • $\begingroup$ What's the relation between $N$ and $N_\epsilon$? $\endgroup$ – Jochen Wengenroth Jan 15 '16 at 8:17
  • $\begingroup$ THe title is quite misleading: not "half of" but "at most twice". $\endgroup$ – Ilya Bogdanov Jan 15 '16 at 10:33
  • $\begingroup$ @JochenWengenroth $N\in\{N_\epsilon\}$ $\endgroup$ – orb Jan 15 '16 at 12:27

This holds for every subset $N\subseteq V$, not only for nets.

Every ball $B(x,r)$ is covered by a union of $2^{2k}$ balls $B(x_i,r/4)$. Now, if for some $i$ there exists $n_i\in B(x_i,r/4)\cap N$, then $B(x_i,r/4)\subseteq B(n_i,r/2)$. So such balls $B(n_i,r/2)$ cover $B(x,r)\cap N$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.