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The Wikipedia article on doubling spaces gives a definition of doubling constant using open balls:

A metric space $X$ is said to be doubling if there exists some doubling constant $M>0$ such that for any $x\in X$ and $r>0$, it is possible to cover the ball $B(x,r) = \{y \mid d(x,y) < r\}$ with $M$ balls of radius $r/2$.

It then goes on to say that the doubling constant of $\mathbb R$ is $2$, but I don't see why:

Let $B=(-1,1)$ and let $C_1 = (-1,0)$ and $C_2 = (0,1)$. Then $C_1$ and $C_2$ are open balls of radius $1/2$, but they don't cover $B$, and I don't see how they could.

Am I missing something?

Edit: In line with the comment by Moishe Kohan, I had a look at the original paper that defined doubling dimension, and they never explicitly say open or closed balls (just "ball"). So I guess if we consider closed balls it makes sense, and the open ball definition in Wikipedia was an arbitrary choice by whoever wrote it.

Edit 2: The original paper actually does use open balls (didn't scroll far enough), so yeah, Wikipedia was (shockingly) just wrong! I suppose that if we use open balls then the correct answer is $3$, or $2$ if we allow closed balls instead.

pyridoxal_trigeminus
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    Have you tried other choices of $C_i$? It doesn't say that ALL pairs of balls must cover. (Though I also don't see that the constant should be $2$ immediately.) – Randall Feb 16 '23 at 22:29
  • I don't see how any two open balls of radius $1/2$ can cover $B$. – pyridoxal_trigeminus Feb 16 '23 at 22:31
  • I'm inclined to agree with you by the triangle inequality. – Randall Feb 16 '23 at 22:31
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    Keep in mind that anybody can edit a Wikipedia article regardless how little they inderstand the subject. Also, in metric geometry one frequen6uses closed balls, not open ones. This gives the doubling constant 2. – Moishe Kohan Feb 16 '23 at 22:34
  • I wonder if the article contains a typo. They call the doubling dimension $\log_2(M)$. I wonder if they meant that this is $2$, so that $M=4$. – Randall Feb 16 '23 at 22:36
  • @Randall I don't think so - see the picture for $d=2$ which clearly intends to demonstrate that the number $M$ itself is $7$, not that its logarithm is $7$. – Patrick Stevens Feb 16 '23 at 22:38
  • I suppose the other option is that it's just flat-out wrong. – Randall Feb 16 '23 at 22:40

1 Answers1

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I recommend you follow the references on the page. For instance, there is an article of Luukkainen and Saksman where the relevant terminology is defined. Their definition of doubling uses closed balls instead of open. In that case, the doubling constant of $\Bbb{R}$ is clearly $2$.

Note also that with the (open ball) definition given by wikipedia, the doubling constant is not $2$. Indeed, one can use an argument along the lines of: if $(-1,1)$ is covered by $(a_1,b_1)$ and $(a_2,b_2)$, then $b_1 > a_2$ (without loss of generality). If this is the case, then the triangle inequality tells you that if $b_1 - a_1 = b_2 - a_2 = 1$, then $$ \lvert b_2-a_1\rvert \le \lvert b_2-b_1\rvert + \lvert b_1-a_1\rvert < 2 $$

Alekos Robotis
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    Yes, you can check that they use closed balls, as in the Wikipedia link mentioned by the OP check their diagram illustration for M = 7 for $\mathbb{R}^2$. There is a point on the boundary of the original ball which are only covered if the covering balls are closed. – Anon Feb 16 '23 at 22:41
  • Thank you for the sanity check. – pyridoxal_trigeminus Feb 16 '23 at 22:42
  • In the original paper that OP posted in the edit, they clearly claim to use open balls (page 3, left column). However, the authors never make the claim on dimensions that is posted in the wiki article, so there are definitely conflicting definitions/results here. – Randall Feb 16 '23 at 22:50