There exists an infinite set $S\subset \mathbb R^3$ such that any $3$ vectors in $S$ are linearly independent.

Question

There exists an infinite set $S\subset \mathbb R^3$ such that any $3$ vectors in $S$ are linearly independent.

Start with a basis $\beta=\{v_1,v_2,v_3\}$.Take a vector not in any coordinate plane,call it $v_4$.It is possible to choose a vector $v_5\notin span \{v_i,v_j\}$ for all $i,j=1,2,3,4$.Choose $v_6\notin span \{v_i,v_j\}$ for all $i,j=1,2,3,4,5$.Continue this process of choosing $v_{n+1}\notin span\{v_i,v_j\}$ (for each $i,j=1,2,...,n$).Thus we obtain an infinite set $S=\{v_n:n\in \mathbb N\}$ in which now three vectors are coplanar.So,any three vectors of $S$ are linearly independent in $\mathbb R^3$.

Is my solution correct.Can someone provide some alternative solution of this problem?

Answer 1

Your solution is correct. You might want to elaborate exactly why you can keep choosing these points (e.g. the set of points you cannot choose is a finite union of planes, and the finite union of sets with empty interior must have empty interior).

You can also take the (uncountable) set of vectors $(1, t, t^2)$ where $t \in \Bbb{R}$. Then any set of three points is linearly independent, using the Vandermonde determinant.

Answer 2

You also have an explicit solution : $v_n=(1,n,n^2)$.

Answer 3

Take any three distinct vectors whose line of support lies on a cone with vertex at origin and we are done.Clearly no three distinct vectors satisfying the above property are coplanar because any plane through origin can intersect the cone at $2$ lines at maximum.