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The question of whether there exists an infinite subset of $\mathbb{R}^3$ such that every set of $3$ vectors in that set is linearly independent is answered in the affirmative here.

I wanted to know if there exists a Ramsey theoretic proof of the same, motivated by the fact that using infinitary Ramsey's theorem one can show that given a countably infinite set of points $S$ in the plane, there exists an infinite subset $A$ of $S$ such that either $A$ is contained in a line or no $3$ points of $A$ are collinear.

My attempt at the question at hand: Let $X=\mathbb{R}^3$ and let $X^{(3)}$ denote the set of all $3$-tuples of $X$. Colour $(v_1, v_2, v_3) \in X^{(3)}$ red if $\{v_i\}_{i=1}^3$ is linearly dependent and blue otherwise. This colouring is well-defined so by infinitary Ramsey's theorem, there exists a monochromatic subset of $X.$ That is, either there exists a subset of $\mathbb{R^3}$ such that every set of $3$ vectors are linearly dependent or there exists a subset of $\mathbb{R^3}$ such that every set of $3$ vectors is linearly independent. But the existence of the former clearly cannot be ruled out, so I don't know where to go with this.

So my question is, how would one construct a Ramsey theoretic proof of this assertion, if such a thing is possible here?

Michael Hardy
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Aryaman Jal
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1 Answers1

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I'll argue that this line of proof cannot work. Prove me wrong!

Any variant of Ramsey's theorem that relies only on the cardinality of $\mathbb R^3$ will also ensure a monochromatic subset of each plane, which makes the conclusion useless. In some applications there is a trick that will convert a red subset to a blue subset, so you always get a blue subset, but here that would really just mean solving the problem without Ramsey theory.

Ramsey theory is about finding structures. Here you really just want something unstructured; e.g. take countably many random vectors in the unit cube.

Dap
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