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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, just delete the center itself and look at the (triangular) regions of chords that have now disappeared. We need alternative chords to still cover the same points. We're definitely good close to the deleted point, thanks to the remaining parts of the arc from which we deleted. At some distance from our point, we have some positive amount of space to maneuver because the partner point is from one of finitely many intervals of positive minimum length. So we see that we may delete a whole interval, as long as this interval stays smaller than the minimum optional wiggling length that is obtained in this way.

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.

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 cords 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, just delete the center itself and look at the (triangular) regions of chords that have now disappeared. We need alternative chords to still cover the same points. We're definitely good close to the deleted point, thanks to the remaining parts of the arc from which we deleted. At some distance from our point, we have some positive amount of space to maneuver because the partner point is from one of finitely many intervals of positive minimum length. So we see that we may delete a whole interval, as long as this interval stays smaller than the minimum optional wiggling length that is obtained in this way.

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, just delete the center itself and look at the (triangular) regions of chords that have now disappeared. We need alternative chords to still cover the same points. We're definitely good close to the deleted point, thanks to the remaining parts of the arc from which we deleted. At some distance from our point, we have some positive amount of space to maneuver because the partner point is from one of finitely many intervals of positive minimum length. So we see that we may delete a whole interval, as long as this interval stays smaller than the minimum optional wiggling length that is obtained in this way.

Two easy observations: (1) If $C$ has (box counting) dimension $<1/2$, then the cord 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 cords 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 cord 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, just delete the center itself and look at the (triangular) regions of chords that have now disappeared. We need alternative chords to still cover the same points. We're definitely good close to the deleted point, thanks to the remaining parts of the arc from which we deleted. At some distance from our point, we have some positive amount of space to maneuver because the partner point is from one of finitely many intervals of positive minimum length. So we see that we may delete a whole interval, as long as this interval stays smaller than the minimum optional wiggling length that is obtained in this way.

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 cords 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, just delete the center itself and look at the (triangular) regions of chords that have now disappeared. We need alternative chords to still cover the same points. We're definitely good close to the deleted point, thanks to the remaining parts of the arc from which we deleted. At some distance from our point, we have some positive amount of space to maneuver because the partner point is from one of finitely many intervals of positive minimum length. So we see that we may delete a whole interval, as long as this interval stays smaller than the minimum optional wiggling length that is obtained in this way.