Existence of an infinite subset

Question

While going through some problems of Linear algebra, I found the statement: "There exists an infinite subset S of $\mathbb{R}^3$ such that any three vectors in S are linearly independent." What I know that not more than 3 vectors of the given vector space can be linearly independent and there exists infinite basis for this vector space, but the existence of an infinite subset in which any three vectors are linearly independent is not justified to me. Please help me what I am missing to reach the conclusion. Thanks in advance.

Answer 1

Let $S_2:=\{v_1,v_2\}$ be a set of two linearly independent vectors in $\mathbb{R}^3$, having defined $S_n$ we define $S_{n+1}$ in the following way:

Since $$\bigcup_{1\leq i<j \leq n} \text{Span}\{v_i,v_j\}$$ is a finite union of planes in $\mathbb{R}^3$ it can't be all of $\mathbb{R}^3$. Pick some vector $v_{n+1}$ not in this union and set

$$S_{n+1}:=S_n \cup \{v_{n+1}\}. $$

Finally, let $S:=\bigcup_{n=2}^\infty S_n$.

Answer 2

Hint, in $\Bbb{R}^2$ an example of an infinite subset, $S\subset \Bbb{R}^2$ such that any two vectors in it are linearly independent are:

$$S=\{ \left(\cos(\theta),\sin(\theta)\right)\colon \theta \in \left[0,\frac{\pi}{2}\right)\}$$

Can you carry this to $\Bbb{R}^3$?

Answer 3

Any three points on the curve $t\mapsto (1,t,t^2)$ are linearly independent. This is because $$\left|\begin{array}{lll}1&t_1&t_1^2\\1&t_2&t_2^2\\1&t_3&t_3^2\end{array}\right|=(t_3-t_2)(t_3-t_1)(t_2-t_1)$$ so the three vectors are linearly independent provided $t_1,t_2,t_3$ are pairwise distinct.