Is a certain subset of the disc a convex set? 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?
 A: 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.
A: 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. 
A: 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").
