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I was just browsing through MSE and found this question. It is about finding an infinite set $S \subseteq \mathbb{R}^3$ such that for every three different vectors in $S$ are linearly independent.

Given the answer provided there, it is not hard to see that a similar result holds for any dimension $d\in \mathbb{N}$: For all $d\in \mathbb{N}$, there is an infinite subset $S\subseteq \mathbb{R}^d$ such that all $X \in \binom{S}{d} = \{X \subseteq S \space | \space |X| = d\}$ are linearly independent.

Now my question is: can we even choose $S$ in such a way that $|S| = |\mathbb{R}|$?

A few ideas I have so far:

  • For $d = 1$ we can just choose $S = \mathbb{R}\backslash\{0\}$.
  • For $d=2$ we can choose $S = \{(x,y) \in \mathbb{R}^2 \space | \space x,y > 0 \text{ and } x^2 + y^2 = 1\}$ enter image description here
  • For $d=3$ we can choose a suited curve on the sphere or on the plane defined by $z=1$, e.g. $S = \{ (x,x^2,1) \in \mathbb{R}^3 \space | \space x \in \mathbb{R} \}$ enter image description here
  • For $d \geq 4$, I have no idea. Maybe again some curve?

I accept any proof or disproof, however I would prefer an explicit example.

Jakob W.
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1 Answers1

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Just extrapolate Hagen von Eitzen's answer to your linked question: the vectors $(1,t,t^2,\ldots,t^{n-1})$ are linearly independent, because the determinant of the corresponding Vandermonde matrix is non-zero when all the rows are different.

TonyK
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