Is there a set of $4$ vectors in $\mathbb{R}^3$, any $3$ of which form a linearly independent set? I have an exercise in my last assignment for linear algebra:

Is there a set of $4$ vectors in $\mathbb{R}^3$, any $3$ of which form
  a linearly independent set? Prove.

My answer intuitively is no for one reason. If we have 4 vectors in $\mathbb{R}^3$, then, if we consider then all together, one of them is a linear combination of another (or others), it's a multiple of another. From this set of $4$ vectors, we can pick $3$ vectors to check if they are linear independent, but we are going to have at least one group of $3$ vectors where we have a vector and its multiple.
Does my reasoning make some sense? How could I prove what they are asking?
 A: If you think of the standard orthonormal basis for $\mathbb{R}$, $\hat{i}, \hat{j}, \hat{k}$ and then also the vector $\overline{x} = (1, 1, 1)$, you know that the first three are independent and that you can only form $\overline{x}$ if you have all three. I.e., $\overline{x}$ is not coplanar with any pair of the standard basis vectors so any three of these four are linearly independent. 
A: Start with a non-singular matrix of size $3\times3$, Call the thee columns of this matrix $u,v$ and $w$. Now take $u+v+w$ as the fourth vector. These 4 vectors will always have the property that any 3 of them will be linearly independent.
A: Here is a way to construct an infinite set of such vectors.
Let $Y$ be any infinite set of points in $\mathbb{R}^2$ such that for any $3$ distinct elements in $Y$ there is no line through all these $3$ points (so for example taking $Y$ to be the unit circle works). Let $X$ the set of vectors of the form $(1,x,y)$ such that $(x,y)\in Y$. I claim that if we pick any $3$ vectors from $X$ then these will be linearly independent.
To see this note that if $a_1(1,x_1,y_1) + a_2(1,x_2,y_2) + a_3(1,x_3,y_3) = 0$ then $a_3 = -a_1 - a_2$ so we are looking at the two equations $$a_1x_1 + a_2x_2 -(a_1 + a_2)x_3 = 0$$ and $$a_1y_1 + a_2y_2 - (a_1 + a_2)y_3 = 0$$ which we can rewrite as $$a_1(x_1 - x_3) + a_2(x_2 - x_3) = 0$$ and $$a_1(y_1 - y_3) + a_2(y_2 - y_3) = 0$$ so if we define the vectors $v = (x_1,y_1)$, $u = (x_2,y_2)$ and $w = (x_3,y_3)$ then the claim is that the vectors $v-w$ and $u-w$ are linearly independent. But it is well-known (and a nice exercise to show) that the vectors $v-w$ and $u-w$ are linearly dependent if and only if there is a line through the points $v$, $u$ and $w$ in $\mathbb{R}^2$. So our choice of $Y$ guarantees that these vectors are indeed linearly independent as claimed.
