# How does a 0-ball have volume 1?

I've been reading about n-balls and I found in the Wikipedia article that in dimension 0, the volume of a 0-ball is 1. I have searched for more information but I can't find any resource explaining this.

So, how does a 0-ball have volume 1?

• There are currently two answers, which could each be an answer, depending on your point of view. Could you clarify, by explaining what you think the 0-ball is, and what you understand by 0-dimensional space? Mar 20, 2021 at 23:01
• I'm afraid I have a layman, if not less, understanding on this matter and I don't have a specific point of view to decide which answer is more correct. Hagen refers to the counting measure, while the other one just negates the premise of the existence of such a ball region in dimension 0. Answering your questions, my understanding of 0-dimensional space is just a point and I can just conceive a 0-ball in nildimensional space as being that point (if that makes sense at all)
– Jon
Mar 21, 2021 at 17:18

$$0$$-dimensional space is just a single point and every ball of positive radius contains that point. Moreover, the measure in this space is just the counting measure. So the volume of the ball is $$1$$ because it contains one point.

• What is the meaning of the words "positive radius" in this case?
– user
Mar 20, 2021 at 23:48
• @user $\mathbb{R}^0$ is a metric space in the same way as $\mathbb{R}^n$ for $n \geq 1$, and so it's fine to talk about balls of any positive radius. It just so happens that any such ball contains only the unique point of $\mathbb{R}^0$. Mar 21, 2021 at 8:04
• Thanks for your answer. Could you dive further into what do you mean by "the measure in this space is just the counting measure"? Is there some term I can refer to to search more about this?
– Jon
Mar 21, 2021 at 17:11

They don't. There are no zero-dimensional balls. the Wikipedia article is not precise. The reason for the imprecision can be filed under "semantic sugar": it so happens that if you plug $$n=0$$ in the formula for the volume of an $$n$$-dimensional ball, where $$n\geq 1$$, you get $$1$$.

• What do you mean with n >= 1?
– Jon
Mar 20, 2021 at 22:47
• I mean that the formula makes sense, geometrically, for $n\geq 1$. You could put in it $n=0$, but it does not mean that there exists a zero-dimensional ball whose volume is $1$. Mar 20, 2021 at 22:49
• The Wikipedia article also uses $A_n$ (a function of the radius $R$) to denote the surface area ($n$-volume) of the $n$-dimensional sphere, which can be though of as the boundary in $\mathbb{R}^{n+1}$ of the $(n+1)$-dimensianal solid ball. There is a formula for $A_n$, and if you naïvely plug in $n=0$ into it, you get $2$. Now, in the spirit of Hagen von Eitzen's answer, that is taken to mean that there are two points in $S^0$. What is your view on this? You say there are no 0-dimensional balls, but are there 0-dimensional spheres (like $\{\pm 1\}$ in $\mathbb{R}$), in your view? Mar 21, 2021 at 8:38
• $\mathbb{S}^{n-1}$ is an $n-1$-dimensional sphere. It is the boundary of an $n$-dimensional ball. The superscript $n-1$ is there for a reason: the dimension is reduced by $1$. Consequently, $\pm 1$ is the boundary of the $1$-dimensional ball $[-1,1]$. It is indeed a $0$-dimensional sphere. However, a sphere is not a ball. There aren't any zero-dimensional balls. Have you ever seen a minus-one dimensional sphere? With all due respect to Wikipedia, it is not the most reliable source for mathematics. Mar 21, 2021 at 8:42
• So do you agree that $\mathbb{S}^0$ has measure (generalized volume) $2$? Mar 25, 2021 at 12:36