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?

 

[mighty2000]

This proof seems to work, as it follows a logical and coherent argument. However, there are a few places where you can cut down on wordiness to make it more concise and easier to follow.

First, instead of saying "Suppose for contradiction that there exists a direction, call it D, for which there is no segment of length 1 in K," you can simply say "Suppose there exists a direction D where no segment of length 1 exists in K."

Similarly, instead of saying "Let v be the unit vector in direction D," you can say "Let v be the unit vector in direction D."

In the paragraph where you introduce the terms "right" and "left," you can use the phrases "in the direction of v" and "in the direction of -v" instead, as it is more concise and clear.

Other than that, the proof seems clear and concise. Good job!