# Proof adding layers of constant width to a shape tends to an $d$-sphere as the number of layers tends to $\infty$

Good night,

I've recently seen one of Victoria Hart's videos on Youtube (it wasn't about this, it was about Fibonacci numbers, and I found it on a comment in this site), and in it she said that if you draw lines of constant width expanding any shape, like to this square you get closer and closer to a circle. That makes sense, even in higher dimensions (to $d$-spheres), because straight lines/surfaces/etc./$d-1$-things stay the same size and get away from each other, while sharp vertices/lines/etc./$d-2$-things become circle/$d$-sphere arcs.

Can any layer be decomposed into smaller others (like the blue-to-red layer can be made of the blue-to-green, and the green-to-red)? How can one prove this true for any dimension? Remembering the original shape may be non-convex, how do the layers evolve in it?

## Some of my ideas so far are:

Take the example of the blue-to-red layer: the red line is the furthest away from the center of the square at a value of $5+\sqrt2$, and the closest at $6$. Generalizing to a $d$-(hyper)cube, having a layer whose furthest limit (red line in the example) is $l$ units away from the original $d$-cube with an edge $e$, we have the closest point to the center at $l+\frac e2$ and the furthest at $l+\sqrt\frac {de}2$ (note these are the same for $0$ and $1$ dimensions).

As the edge and dimension are constant (because the shape is), and we are continuously expanding the layer (or adding more layers, which depending on the answer to the first question of the paragraph above, is the same), we have the limit (to infinity) of the distance from the center of the shape to the furthest points of the layer being equal to infinity, the same as the limit of the closest points,

$$\lim_{l\to\infty}l+\frac e2 = \infty = \lim_{l\to\infty} l+\sqrt\frac {de}2$$

and if the closest and furthest point approximate as $x$ goes to $\infty$, all others in between do as well (I think this is a bit of the squeeze theorem), and the definition of a $d$-(hyper)sphere is that exactly: all points are at the same distance to the center.

(end ideas so far)

However, these do not seem like proofs. There are things I can't say for sure because this is not a formal proof. How can one solidly prove this? Is this all right? How can I generalize to any shape of any dimension? Thank you in advance,

• Ravi Vakil gave a talk on this (among other things) over the summer at the Young Mathematicians Conference. I'll see if I can dig up some references. – Alex Becker Jan 2 '14 at 23:46
• @AlexBecker, I'm glad someone has explored this already, I'm not very comfortable in this area. Looking forward to those references. :) – JMCF125 Jan 2 '14 at 23:51
• $\lim_{n\to\infty} n = \infty = \lim_{n\to\infty} n^2$ but $n$ and $n^2$ certainly do not tend to something equal as $n\to\infty$. This mistake is similar to taking $\infty-\infty=0$, which is also incorrect. – user21820 Jan 18 '14 at 2:14
• @user21820, I see your point, but if you're talking about the limits I put above, note $\frac e2$ and $\sqrt\frac {de}2$ are constant expressions, they are not bound to $l$ in any way. Also, did you downvote this? Can you please explain if you did? – JMCF125 Jan 18 '14 at 11:20
• No, you don't see my point. $lim_{n\to\infty} n = \infty = lim_{n\to\infty} (n+1)$ but $n$ and $n+1$ never tend to something equal as $n \to \infty$. You should study the definition of a limit carefully. – user21820 Jan 23 '14 at 1:43

## 1 Answer

It takes a fair amount of work to rigorously define what it means to "tend to a $d$-sphere". Luckily this is done, and the answer is proven to be yes, in these notes on the question by Ravi Vakil.

Starting with some set $S$, Vakil defines the "doodle" of radius $R$ to be the boundary of the set $N_R(S)$ of points within $R$ of a point in $S$. Note that your "layers" are precisely the doodles of radius $n$ for $n$ an integer. He proves that there exist disks $d_R,D_R$ with the same center such that $d_R\subseteq N_R\subseteq D_R$ and that as $R\to \infty$, the difference in radius between $d_R$ and $D_R$ goes to $0$. He then goes on to list all sorts of awesome applications.

(This summary is from memory based on his talk, and may be slightly off)

• Seems interesting, and I'll definitely read it (not now, I'm at UTC+0), but can you sum it up here, for future reference? – JMCF125 Jan 3 '14 at 0:04
• If you don't want to I can do it after I read the paper, I already found the answer to one of my questions (as «Prior Question. How does our choice of r affect the problem?») – JMCF125 Jan 3 '14 at 0:12
• I've done my best to summarize it from my memory of his talk, and quick glances at the paper. – Alex Becker Jan 3 '14 at 0:13