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Is it true that there does not exist a closed convex plane curve containing an infinite number of segments, belonging to distinct lines each?

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Unless I am misunderstanding the question, the answer is NO. Consider a semicircle whose diameter is the segment $[0, 1]$ on the $x$-axis. Now, consider the curve composed of the diameter, and the chords joining points on the semicircle with arguments $\pi/n$ to $\pi/(n+1)$ (and their reflection in the $y$-axis). This seems to be an example of the convex curve of the type you claim does not exist.

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  • $\begingroup$ This is great, thanks! How would you show that the curve you described is closed? $\endgroup$ – Evgenii.Balai Sep 24 '17 at 19:06
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    $\begingroup$ It is the image of a circle under a surjective continuous map (since there is a point with every argument). $\endgroup$ – Igor Rivin Sep 24 '17 at 19:17
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    $\begingroup$ Alternately, the semicircle (half disk) is convex and this set is the intersection of that region and a bunch of (convex) half planes hence convex. (Note that any two of the planes intersect outside the semicircle.) For a similar piecewise linear example take the segment from $(0,0)$ to $(1,1)$ along with all the segments from $(1/k,1/k^2)$ to $(1/(k+1),1/(k+1)^2).$ $\endgroup$ – Aaron Meyerowitz Sep 24 '17 at 20:15
  • $\begingroup$ Wow, someone downvoted this! $\endgroup$ – Igor Rivin Sep 24 '17 at 20:32
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    $\begingroup$ Sorry, I think I messed up an upvote. I had to “edit” to reverse it. $\endgroup$ – Aaron Meyerowitz Sep 25 '17 at 0:01

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