# Why is $2$ the doubling constant of $\mathbb R$?

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.

• 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.) Commented Feb 16, 2023 at 22:29
• I don't see how any two open balls of radius $1/2$ can cover $B$. Commented Feb 16, 2023 at 22:31
• I'm inclined to agree with you by the triangle inequality. Commented Feb 16, 2023 at 22:31
• 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. Commented Feb 16, 2023 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 $M=4$. Commented Feb 16, 2023 at 22:36

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 $$\lvert b_2-a_1\rvert \le \lvert b_2-b_1\rvert + \lvert b_1-a_1\rvert < 2$$
• 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.