# Set of diameter $\le 1$ contained in set of constant width $1$

I'm reading the paper Minimal universal covers in $$E^n$$ by H.G. Eggleston and they state that every set $$A\subseteq{\bf R}^2$$ of diameter at most $$1$$ (the diameter of $$A$$ is defined as $$\sup_{x,y\in A}|x-y|$$) is contained in a set which has width $$1$$ along any direction. I can't see how the proof of this should work, though.

EDIT: Maybe the proof is something along the lines of constructing a thing that has width $$1$$ along any rational direction, then it should has width $$1$$ along all directions. For this, it should suffice to show that for any $$A$$ of diameter $$\le 1$$ you can add a translate $$I$$ of the unit interval (embedded into $${\bf R}^2$$) such that $$A\cup I$$ has diameter $$1$$. Then you are able to add an infinite amount of intervals at different angles without the diameter exceeding $$1$$.

• Just a note that the "reflection" of this statement is false: if a convex set has width $\ge1$ in every direction, it does not necessarily contain a set of constant width. (Consider the equilateral triangle of height $1$ for a counterexample.) Feb 14, 2021 at 19:06

Let $$X\subseteq\mathbf{R}^2$$ be of diameter $$d(X)\le 1$$, w.l.o.g. we can assume $$X$$ to be closed. I attempt to show that there exists $$a\in\mathbf{R}^2$$ such that with $$b:=a+(1,0)$$ we have $$d(X\subseteq\{a,b\})\le 1$$. This way we can add pairs of points with distance $$1$$ along every rational angle, which should suffice to get a superset of $$X$$ with constant width $$1$$. In the following, $$b$$, $$b'$$, etc. always means $$a+(1,0)$$, $$a'+(1,0)$$, etc.

[EDIT] Look in the comments for a shorter approach that doesn't use the following section.

First choose $$a$$ such that we have $$d(X\cup\{a,b\})=\min_{a'\in\mathbf{R}}d(X\cup\{a',b'\}),$$ call this diameter $$D$$. Assume there is some $$a'\neq a$$ with the same property. Then all points on the line between $$a$$ and $$a'$$ must have this property as well, as when you travel in a direction and the distance to some point grows, it will not shrink again when you travel further in the same direction. Assume $$1. Then there exists $$x\in X$$ such that $$D$$ equals the distance from $$x$$ to $$(a+a')/2$$ (or just any other point strictly between $$a$$ and $$a'$$). But this implies $$D or $$D, a contradiction. So if $$1 then our $$a$$ must be unique.

Now assume w.l.o.g. that $$a=(-1/2,0)$$, $$b=(1/2,0)$$. Consider the set $$E$$ of points $$e$$ such that $$D=d(\{a,b,e\})$$, I called it $$E$$ because it has the shape of a vertical eye, note that $$X$$ must live in $$E$$ together with its enclosed area. For $$s,s'\in\{-,+,*\}$$ denote by $$E_{s,s'}$$ the set of points $$e\in E$$ such that the sign of the first component is $$s$$ or $$0$$ or anything if $$s=*$$ and similar for the sign of the second component with $$s'$$. When we shift $$a$$ to the right a little, the diameter of $$X\cup\{a,b\}$$ increases, so $$X\cap(E_{-,*})$$ cannot be empty. A similar argument can be made for $$E_{+,*}$$ and by shifting $$a$$ vertically also for $$E_{*,-}$$ and $$E_{*,+}$$. So there must be two points of $$X$$ in two opposite components of $$E$$ (i.e. $$E_{-,+}$$ / $$E_{+,-}$$ or $$E_{-,-}$$ / $$E_{+,+}$$), and one can easily see that those have distance greater than $$1$$, a contradiction.

• Note that the reason we can choose an $a$ which attains the minimum is that we can restrict our attention to a compact set of such $a$, and that the diameter as a function of $a$ is continuous. Feb 14, 2021 at 19:19
• Good point! Also, I just realized that I don't need the middle section, I can just use that if there are two neighbouring components of $E$ which contain no points of $X$, then the distance of those compenents to $X$ must be larger than some $\epsilon$, so by shifting $a$ away from this components by $\epsilon$, call the result $a'$, I get that $E':=\{e'\in\mathbf{R}^2:D=d(\{a',b',e'\})\}$ is empty hence $d(X\cup\{a',b'\})<D$, a contradiction. Feb 14, 2021 at 20:36

WLOG we can assume that $$X$$ is convex. $$p_1,\ p_2\in X$$ s.t. $$|p_1-p_2|=1$$. Then there is a circular sector $$V_1$$ at $$p_1$$ whose angle is $$\theta_1$$.

And $$V_1$$ contains arc $$p_2p_3$$. Here $$\theta_1$$ is the largest when $$V_1-X$$ is empty.

Now we do the same thing at $$p_2$$. Hence we have $$\sum_{i=1}^{2m+1} \ \theta_i =\pi,\ \theta_i>0$$ and $$\bigcup_{i=1}^{2m+1} \ V_i$$ contains $$X$$.

• Not here for this answer, but I left some comments under the question you bountied, over which we had a conversation regarding my wrong answer a few days back. You can take a look, thank you. Feb 23, 2021 at 4:46