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Is there a nonempty planar set that contains $0$ or $2$ vertices from each unit equilateral triangle?

I know that such a set cannot be measurable. In fact, my motivation is to extend a Falconer-Croft proof that works for measurable sets, see the details here. In more general, to make their proof work for other sets besides equilateral triangles, one can ask the following.

Suppose we are given two sets, $S$ and $A$ in the plane, such that $S$ is finite, with a special point, $s_0$, while neither $A$ nor its complement is a null-set, i.e., the outer Lebesgue measure of $A$ and $A^c=\mathbb R^2\setminus A$ are both non-zero. Can we find two congruent copies of $S$, $f_1(S)$ and $f_2(S)$, such that $f_1^{-1}(f_1(S)\cap A)\Delta f_2^{-1}(f_2(S)\cap A)=\{s_0\}$, i.e., $s_0$ is the only element of $S$ that goes in to/out of $A$ when we go from $S_1$ to $S_2$?

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  • $\begingroup$ If I understand this, you ask for a set with certain properties, one of which is that for every point in the set, the set has "half the points" of the unit circle with the first point as its center. I suspect the answer is that the empty set is such a set, and that there is no other. Gerhard "Of Course Suspicion Isn't Proof" Paseman, 2019.08.17. $\endgroup$
    – Gerhard Paseman
    Commented Aug 17, 2019 at 19:34

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I am probably misunderstanding something. Let me suppose a non empty planar set with the property, and that point A is in the set. Pick a unit equilateral triangle having A as a vertex, and also B and C.

Then exactly one of B and C is in the set. If we pick unit equilateral triangle BCD, then D (short for Different from A) must be in the set with A. So if A is in the set, then every point with distance r (where r^2 is 3) from A is in the set.

The contradiction comes from picking arbitrary point X and stepping to it from A using steps of length r, which I leave to you.

Afterthought For your more general question, I would consider studying the following scheme, which may be in the literature: Given S, consider A-colorings of S. These are labellings of S depending on your allowed coloring scheme. For the general problem above, let red correspond to "out of A", green to "not out of A", and you color according to how you transform S to a domain containing A. For the problem above, you color a triplet corresponding to the appearance of vertices of a unit equilateral triangle inside or outside the set A.

Now a challenging problem is to determine what A-colorings are interesting/feasible, when you put restrictions on the set of transforms of S. You want to find two colorings which differ precisely at the point $s_0$. You may be able to prove existence by cardinality considerations on the set of all allowed A-colorings.

Gerhard "Please Show Me What's Wrong" Paseman, 2019.08.18.

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  • $\begingroup$ Well, I better undelete my original question then! $\endgroup$
    – domotorp
    Commented Aug 17, 2019 at 19:59
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    $\begingroup$ We reach a contradiction if the domain in the plane is large enough to include a point of A and a disk of radius sqrt 3 around the point. It might be of interest to find the smallest region of the plane for which the above argument gives a contradiction. This might be of use in constrained versions of your general problem. Gerhard "Often Works In Tight Spaces" Paseman, 2019.08.17. $\endgroup$
    – Gerhard Paseman
    Commented Aug 18, 2019 at 5:58
  • $\begingroup$ That would be of interest regarding the original problem is instead of a set and its complement, we would partition the plane into three sets/colors, and require that only one of the triangle's vertices changes color. $\endgroup$
    – domotorp
    Commented Aug 18, 2019 at 6:12

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