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We say that a subset of $\mathbb R^2$ is convex if any line segment connecting two points in that subset is contained in it. If I'm missing some preconditions that are required so this definition makes sense, consider them valid here.

How would I prove that any tangent on the edge of that shape never crosses it's interior? Furthermore, is it true that any concave shape has a tangent that enters it's interior?

A tangent like that splits the shape into at least two parts. My intuition tells me that there are two points in those parts that can't be connected. I also have a feeling is should be pretty easy to prove but it's escaping me at the moment.

Any ideas?

Luka Horvat
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1 Answers1

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The definition of convex is slightly off, you mean that any line segment connecting two points in the subset is contained within it. Also, I'm not quite sure what you mean by tangent in this case, perhaps you mean that the extended line is tangent or that there is some point in the interior of the segment that is a boundary point.

Let $S$ be a line segment whose endpoints are in the set. If $S$ intersects the interior, then there is some point $s\in S$ and a radius $\delta$ such that the open ball centered at $s$ and of radius $\delta$ lies inside the convex set. In other words, $B(s,\delta)$ is contained within the convex set.

Since $B(s,\delta)$ is a open disk with $s$ its center, we can extend $S$ into a line $L$ whose intersection with the open disk is a diameter. Then, we can choose points $x$ and $y$ so that they are in the disk and to either side of the line $L$. These points are in the convex set.

Let $s_0$ and $s_1$ be the endpoints of $S$. Consider the line segments between $x$ and $s_0$ and $y$ and $s_0$. These segments must be part of the convex set. Let these segments be called $S_x$ and $S_y$, respectively. Finally, for any point $z$ in $S_x$ or $S_y$, consider the segment between $z$ and $s_1$. This segment is in the convex set.

Combining all of these, you end up showing that no point of the interior of $S$ can be a boundary point of the convex set.

Michael Burr
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  • This isn't really clear to me. What does "on either side" mean when talking about a line segment? Why is the convex hull a subset of the set? Why does that mean that s can't be a tangent? I've also edited the question as you suggested. – Luka Horvat Oct 03 '15 at 13:13
  • There are several unspoken assumptions that I'm assuming from your question. I am assuming that you are using standard Euclidean topology. What this means is that if the line segment intersects the interior, there is some point of the line segment for which there is an open ball around that point that is entirely contained within the convex set (this is a definition of the interior) Now, you can show that if you extend the line segment into a line, there are points in the open ball that are on either side of the line. – Michael Burr Oct 03 '15 at 13:19
  • I'm still having trouble understanding your argument. If I understand correctly you construct a quadrilateral from the two points on either side of $S$ and the endpoints of $S$. I understand why this shape must be contained in the original set. But how does that extend so the fact that the extension $L$ of $S$ is not a tangent? – Luka Horvat Oct 03 '15 at 13:32
  • You can then replace $S$ with the largest subsegment of $L$ that whose endpoints are in the set. – Michael Burr Oct 03 '15 at 13:34
  • Hmm... I'm starting to see problems with my question since a tangent line doesn't make sense for polygons. I assume my convex shape to be "smooth" and then a tangent to be a line that passes through a point on the edge of the shape such that there exists some small area around that point in which that point is the only intersection between the shape and the line. – Luka Horvat Oct 03 '15 at 13:35