How many convexly independent vectors there are in $\mathbb{R}^n$ I know there are n linearly independent and n + 1 affinely independent vectors in $\mathbb{R}^n$. But how many convexly independent there are?
I think there are infinity number of them because if I have a convex polytope I can always add another point that is "outside" of said polytope. 
But I'm not sure if my reasoning is correct.
 A: Your reasoning is almost but not quite correct. It's not enough that you can always add another point that's outside of said polytope – the resulting polytope also has to remain convex, i.e. each of the other points also has to be outside the polytope formed by the others. You can prove that, too, but it's a bit more involved – an easier proof would be to exhibit the corners of a regular $k$-gon for arbitrary $k$ as an example. This of course only works for $n\ge2$, and indeed for $n=1$ there are at most $2$ convexly independent points.
A: No, your reasoning is not correct. If you add a new point outside the polytope, some of the vertices may become interior points of your new polytope. You have to ensure that all vertices remain vertices.
A: Consider the unit sphere in $\Bbb{R}^n$, $S^{n-1}=\{x : \|x\|=1\}$. I claim that the vectors in the unit sphere are convexly independent. To see this, it suffices to show that any vector is outside the convex hull of the others by symmetry. Choose the first basis vector, $e_1$. Suppose $e_1=\sum_{i=1}^r t_i v_i$ for $\sum t_i = 1$, and $v_i\in S^{n-1}$, $v_i\ne e_1$. We then look at the projection of this equation on the first coordinate. This gives $1=\sum_{i=1}^r t_i v_{i1}$. $v_{i1}< 1$ for all $i$ since $v_{i}\ne e_1$, and $\|v_{i}\|=1$. Then $1 < \sum_{i=1}^r t_i \cdot 1 =\sum_{i=1}^r t_i=1$. Contradiction.
