Unbounded subsets $A$ of $\mathbb{R^2}$ such that $x,y \in A \implies \exists z \in A$ such that $x,y,z$ are the vertices of an equilateral triangle

Question

Let $A \subseteq \mathbb{R^2}$ be an unbounded set such that, for each (every) pair of distinct points $x,y \in A, \exists z \in A$ such that $x,y,z$ are the vertices of an equilateral triangle. I am trying to classify the types of set $A$ can be, and have discovered the following. $A$ could be:

• Any infinite hexagonal lattice of points.

• $\mathbb{R^2}$

• $\mathbb{R^2}$ with any one point removed.

• $\mathbb{R^2}$ with the vertices of any equilateral triangle triangle (3 points) removed. [Notice that $\mathbb{R^2}$ with any two points removed doesn't work. I also suspect: removing 3 or more finite points from $\mathbb{R^2}$ doesn't work].

• $\mathbb{R^2}$ with any infinite hexagonal lattice of points removed.

• Any infinite hexagonal lattice of points with any one point removed from the lattice.

• Any infinite hexagonal lattice of points with two or more points removed from the lattice that "works". [I think I can classify what doesn't work: If you can form two equilateral triangles using two remaining lattice points; one equilateral triangle uses a removed point as the 3rd vertex, and the other equilateral triangle uses another removed point as the 3rd vertex. Everything else should work. ]

• $\mathbb{R^2}$ with any infinite hexagonal lattice of points removed, bar any one point.

• $\mathbb{R^2}$ with any infinite hexagonal lattice of points removed bar any two or more points that "works".[It might require more thought to classify what works here and what doesn't.]

• $\mathbb{R^2}$ with a disc removed, $\mathbb{R^2}$ with a square and it's interior removed,but not $\mathbb{R^2}$ with any convex subset of $\mathbb{R^2}$ (and its interior) removed (e.g. some ellipses), and also not $\mathbb{R^2}$ with any rectangle (and its interior) removed e.g. a 3 unit x 1 unit rectangle.

• ??

I'm probably definitely missing some sets.

Feel free to edit the format of this list so that it is laid out clearer.

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This is an extension to the proposition that Dejan Govc proposed and proved in the link:

Is there such a function?

Dejan Govc gave a nice answer to the same question for bounded sets (as opposed to unbounded sets- which is what this question is about).