Property of compact convex sets of width 1

[Mr Davis 97]

Homework Statement

A strip of width w is a part of the plane bounded by two parallel lines at distance w. The width of a set $X \subseteq \mathbb{R}^2$ is the smallest width of a strip containing $X$. Prove that a compact convex set of width $1$ contains a segment of length $1$ in every direction.

Homework Equations

The Attempt at a Solution

Let $K$ be a compact convex set in the plane. Suppose for contradiction that there exists a direction, call it $D$, for which there is no segment of length $1$ in $K$. This means that the length of all segments contained in $K$ in this direction are between $0$ and $1$. Let $v$ be the unit vector in direction $D$, and consider the translate $v+K$. $K$ and $v+K$ are disjoint, since if they weren't we could find a segment of length $1$ in the direction of $D$. $K$ and $v+K$ are also compact and convex, so we can use the strong separating theorem to find a line that separates them, and in particular a line that is the perpendicular of the two figures and has normal vector $v$. Note that for essentially the same reasons we can find a separating line between $K$ and $-v+K$ with the same properties.

Now. let $\epsilon$ be the orthogonal distance between the separating line of $K$ and $v+K$, and either $K$ or $v+K$ (the distance is the same by symmetry). For simplicity sake we will use the term "right" to mean in the direction of $v$, and the term "left" to mean in the direction of $-v$. Now, form a line segment from the leftmost point in $K$ to the leftmost point in $v+K$. This line segment must be of length $1$. Now, translate this line to the left by $\epsilon$. By symmetry, we have formed a line segment of length $1$ between the two separating lines, which means that $K$ is at most width $1$. However, remember that our separating lines are strict, meaning that if we translate the rightmost separating line to the left by $\epsilon$, and translate the leftmost separating line to the right by $\epsilon$, this shows that that the width of $K$ is at most $1-2\epsilon$, which contradicts the assumption that $K$ had width $1$.



I have two questions. Does this proof seem to work? And if it does, is there any wordiness I could cut down on?