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It's quite obvious to me that if I clip any convex polygon with a half-space, the resulting polygon will be convex. I explain it to myself by "a straight line can't introduce concavity", but is there a proper, mathematical proof to it?

I have used this property extensively in my thesis (computer science, not mathematics), but realized now I can't come up with any sound argument why my assumption holds. Proving it is out of scope of my work, but I would like to understand the proof myself. Points to any reading/papers that I could cite are also very welcome :)

elena
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  • A convex polygon is itself the intersection of half spaces. Adding one does not change matters. Likewise for the intersection of two convex polygons. –  Dec 17 '19 at 10:51

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The convex polygon and half-space are both convex sets, and the result of clipping the polygon by the half-space is merely their intersection, which will be convex from this result. So in case the intersection forms a polygon, it will be a convex one.

Shubham Johri
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  • thank you! Is it a well known fact that I can just state, or would you have any reference to cite? – elena Dec 17 '19 at 10:11
  • @elena This is an elementary result in the study of convex sets, so I don't think citation would be necessary. – Shubham Johri Dec 17 '19 at 10:12
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    @elena Try to prove it yourself from the definition of convexity: if $A$ and $B$ are convex sets, this means for any $p,q \in A$ the segment $[p,q]$ also lies in $A$, and similarly for $B$. Now, if $p,q \in A\cap B$, then $p,q \in A$ and $p,q \in B$, so... – lisyarus Dec 17 '19 at 10:39
  • @ShubhamJohri "So in case the intersection forms a polygon, it will be convex." - this is a bit misleading, "it" may be understood to refer to either "the intersection" or "a polygon". May I suggest rewriting the last part as "it will be a convex one"? – lisyarus Dec 17 '19 at 10:40
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    @lisyarus fixed. – Shubham Johri Dec 17 '19 at 10:42
  • @lisyarus I'm afraid I'm lacking some set theory knowledge. $p,q \in A \cap B$ is a subset of $A$ as well as $B$. Apparently that makes it convex, but I'm only guessing, it is not obvious for me. – elena Dec 17 '19 at 11:16
  • @elena You need to show $A\cap B$ is convex if $A,B$ are convex. Assume two points $p,q\in A\cap B$. Then since $A\cap B\subseteq A,p,q\in A$. Using the convexity of $A$, you get that the entire line joining $p,q$ lies in $A$. Similarly $p,q\in B$ and the said line also belongs to $B$. Hence the said line belongs to $A\cap B$ for every $p,q\in A\cap B$, giving that $A\cap B$ is convex. – Shubham Johri Dec 17 '19 at 11:21
  • For the general case, see the link I mentioned in my answer. – Shubham Johri Dec 17 '19 at 11:22
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    @ShubhamJohri I've been reading over it, and could kinda understand, but not really. Now since you've put it into words, it became really clear. "the line between $p$ and $q$" is what I was missing, I was thinking only in terms of points. Thank you both, ShubhamJohri for the answers and lisyarus for forcing me to think harder. – elena Dec 17 '19 at 11:27