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Some one asked me this question and I thought about it and I don't have any good idea to solve that. Can some one help me and give me an idea to start solve that?

Draw a Cantor set $C$ on the circle and consider the set $A$ of all chords between points of $C$. Is $A$ a convex set?

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  • $\begingroup$ Do you start with removing three arcs of $2\pi/9$ or one arc of $2\pi/3$? $\endgroup$
    – Fedor Petrov
    Commented Dec 7, 2015 at 20:15
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    $\begingroup$ Straight chords are the geodesics in the non-conformal Klein model of the hyperbolic plane, so one can use hyperbolic geodesics in the Poincare disk model instead of straight chords in case that helps. $\endgroup$
    – Douglas Zare
    Commented Dec 7, 2015 at 22:31
  • $\begingroup$ Note that this is an exercise in Pugh's analysis (2.42), so it seems there should be an answer. $\endgroup$
    – Brian Rushton
    Commented Dec 8, 2015 at 16:36

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Two easy observations: (1) If $C$ has (box counting) dimension $<1/2$, then the chord set doesn't even have positive area. This is mentioned in the linked question, but just to make the trivial argument explicit, here it is: cover $C$ by $N\gg 1$ intervals of length $\le \ell$ each. For a fixed pair of intervals, the corresponding chords have area $\lesssim \ell$, so the total area is $\lesssim N^2\ell\to 0$ (since $N\lesssim\ell^{-1/2+\epsilon}$ under our current assumptions).

(2) There are Cantor sets for which the chord set is convex. This follows from the observation that if $n$ open intervals have already been deleted and the current chord set is convex, then we can delete one more (small) interval at the center of the largest remaining arc in such a way that the chord set remains convex.

To see this, let's survey the damage done by deleting a small interval at the center of one of our arcs. Fix one of the other arcs and consider all the chords starting at that arc that are no longer available because they end in the deleted interval. When the distance from the deleted interval gets larger than a certain small critical distance $d$, these points are still all covered by using nearby endpoints from the remaining parts of our arc instead of the deleted ones. Moreover, $d\to 0$ as we shrink the deleted interval. We do slice off the region bounded by the deleted arc and its secant. Other than this, points that are close are still covered by going from one half of what remains of the original arc to the other. We need to take our interval so small that all $d$'s (corresponding to the finitely many other intervals) are smaller than this distance.

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  • $\begingroup$ I'm having a "slow" regarding your point (2). It would seem that points in the interior of the chord having the same endpoints as the deleted open interval are detached from the rest of the chords (except via the endpoints). I.e., there is no chord with an endpoint in the deleted arc, so there is no chord meeting the interior points of this chord. (... and/or I'm missing why the union of the triangular complements you mention don't leave an empty ellipse-like region near that chord.) $\endgroup$
    – Eric Towers
    Commented Dec 8, 2015 at 4:10
  • $\begingroup$ @EricTowers: Yes, this is right of course: deleting a small interval slices off a small piece from my chord set (the set bounded by the deleted arc and the secant). The point is that everything else stays in the chord set (the argument is clumsily described in my answer, but a picture should clarify), so it remains convex. $\endgroup$
    – Christian Remling
    Commented Dec 8, 2015 at 4:20
  • $\begingroup$ @EricTowers: I've reorganized, maybe it's clearer now. $\endgroup$
    – Christian Remling
    Commented Dec 8, 2015 at 4:44
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Start with a 3 point set {a,b,c} on the unit circle. The union of the 3 chords forms not a convex set. Blow up the points into 3 very small Cantor sets. The union of all the new chords still fails miserably to be a convex planar set. For example the origin is not contained in any chord, despite being in the convex hull of the set in question.

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It seems that the answer is sometimes the answer might be YES, as evidenced by Dennis Sullivan's commentary on this question: Measure of chords from a cantor set

(of course, "full measure" does not necessarily mean "convex").

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    $\begingroup$ If I understand correctly the statement of the OP in the link you provide, it seems that the set $A$ (which is $Chords(D)$ in the linked post) obtained from the standard Cantor set (the result of removing a fixed ratio of intervals) is never convex, although it happens that its measure be positive (that of $C$ is zero). $\endgroup$
    – Loïc Teyssier
    Commented Dec 7, 2015 at 19:45
  • $\begingroup$ @LoicTeyssier You are right! I did not read the question carefully. $\endgroup$
    – Igor Rivin
    Commented Dec 7, 2015 at 19:47
  • $\begingroup$ @LoïcTeyssier But, of course, it might be convex for other Cantor sets... $\endgroup$
    – Igor Rivin
    Commented Dec 7, 2015 at 19:47
  • $\begingroup$ Indeed. I'd be surprised to learn the situation is that different, though, but I already was proved wrong once on that topic today so I'm not going to make bets ;) $\endgroup$
    – Loïc Teyssier
    Commented Dec 7, 2015 at 19:48
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    $\begingroup$ @LoicTeyssier Actually, the argument for non-convexity uses very much that this is the $1/3$ Cantor set (you can show that there is a triangle in the middle of the circle which is not contained in the chordal set), but this argument disappears for $1/N$ Cantor sets for other values of $N.$ $\endgroup$
    – Igor Rivin
    Commented Dec 7, 2015 at 19:52

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