Suppose $A\subset \mathbb{R}^n$ is a compact, convex and centrally symmetric set such that $(x_1,\ldots,x_n)\in A$ if $$ |x_1|+\ldots+|x_r|+2\left(\sqrt{x_{r+1}^2 + x_{r+2}^2} + \ldots + \sqrt{x_{n-1}^2 + x_{n}^2}\right) \leq n$$

If $n=r+2s$, I want to prove that (Lebesgue measure) $$ \mathrm{vol}(A) = \frac{n^n}{n!}2^r \left(\frac{\pi}{2}\right)^s$$

To prove this, I assumed that $V_{r,s}(t)$ denote the volume of the subset $\mathbb{R}^{r+2s}$ defined by $$|x_1|+\ldots+|x_r|+2\left(\sqrt{x_{r+1}^2 + x_{r+2}^2} + \ldots + \sqrt{x_{r+2s-1}^2 + x_{r+2s}^2}\right) \leq t$$

Therefore, $$V_{r,s}(t) = t^{r+2s}V_{r,s}(1)$$ I have so far proved that (following this):

$$V_{r,s}(1) =\frac{1}{n!}2^r \left(\frac{\pi}{2}\right)^sV_{0,0}(1)$$

But to conclude my proof I must show that


So, the question is :

Why $V_{0,0}=1$?

  • 1
    $\begingroup$ Maybe related to the fact that a product over the empty set is one? $\endgroup$
    – jdods
    Jul 27, 2016 at 11:25
  • $\begingroup$ @jdods and why is it so? $\endgroup$ Jul 27, 2016 at 11:30
  • 1
    $\begingroup$ Presumably you also have$$V_{r, s}(1) = \frac{1}{n!} 2^{r-1} \left(\frac{\pi}{2}\right)^{s} V_{1,0}(1),$$and $V_{1,0}(1) = 2$? $\endgroup$ Jul 27, 2016 at 11:31
  • 2
    $\begingroup$ Do the recursion $V_{r,s}\to V_{r,s-1}\to V_{r,0}\to V_{r-1,0}\to V_{1,0}$, and stop there. In this way no metaphysical questions arise. $\endgroup$ Jul 27, 2016 at 11:36
  • 1
    $\begingroup$ I found the text online and see this on p.140. If $V_{0,0}=1$ is necessary, then you've solved it I believe. It seems to be a convention then. However the empty product idea might be a reasonable philosophical justification for the sake of intuition. It also falls out of the second formula on p140 since $0!=1$. $\endgroup$
    – jdods
    Jul 27, 2016 at 13:14

1 Answer 1


There are different ways to define volume. I'd suppose that the formulas are only valid for positive dimension, but have a natural convention for zero dimension.

A quick search for Marcus Number Fields finds a pdf of the text. The formula $V_{r,s}(1)=\frac{1}{(r+2s)!}2^r \left(\frac{\pi}{2}\right)^s$ (which is being proved in the text) gives $V_{0,0}(1)=1.$ Intuitively, the author is only considering $n>0$ in the proof and calling attention to the fact that $V_{0,0}(1)=1$ is the natural way to extend it to zero dimension.

The question is:

How is $V_{0,0}(1)$ defined?

This seems to me to imply it is a definition and not something to be proved.

Consider an $n$-dimensional rectangle in $\mathbb R^n:$ $B_n=[a_1,b_1]\times\cdots\times[a_n,b_n].$ Assuming we have standard $n$-dimensional Lebesgue measure, the volume of this box is $\displaystyle \text{Vol}(B_n)=\prod_{k=1}^n\mu([a_k,b_k]).$

If $n=0$, then the formula becomes $\text{Vol}(B_0)=\displaystyle \prod_{k\in\emptyset}\mu([a_k,b_k])=1.$ Since an empty product is by convention $1$ since $1$ is the identity of multiplication. Similarly an empty sum is $0$ since that is the additive identity.

The Hausdorff dimension of a countable set is zero. Consider the origin as a zero dimensional space/set. How are we to define its volume? Let's consider it as a subset of $\mathbb R^n$ for some $n>0$ then cover it by a ball of radius $\epsilon$ so that the covering has volume proportional to $\epsilon^n.$ Obviously we can make $\epsilon$ as small as we like hence it has volume zero. However as we let $n$ decrease to zero, then $\epsilon^0=1$ for any $\epsilon>0.$ Hence in zero dimension, every ball covering the origin will have nonzero volume.

This may feel counter intuitive, but it still ties back to the idea that an empty product gives the multiplicative identity.


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