Arrangement of points on $S^d$ While reading Aigner and Ziegler's "proofs from the book", one finds the following phrase: 
"It suffices to take any arrangement of $2k+d$ points on $S^{d+1}$ in general position, meaning that no $d+2$ of the points lie on a hyperplane through the center of the sphere. Clearly, for $d \geq 0$ this can be done.''
With $S^{d+1}$ they mean the $d+1$ dimensional sphere (so $S^{d+1} \subset \mathbb{R}^{d+2}$). It is sort of intuitively clear to me why this can be done, since we have only finitely many points there is enough 'freedom' to choose an arrangement on $S^{d+1}$. In the cases $d=0$, $d=1$, I can somewhat more easily see that it is true. However, using the word 'clearly' in this context seems to imply there is a fast and simple proof for the general case.
My question is: could anyone explain why this is clear, and maybe give a proof? Thanks in advance.
EDIT: A reformulation into purely linear algebraic terms is as follows. Let $m>n$. Prove the existence $m$ vectors $\{v_1, .., v_m \}$ in $\mathbb{R}^n$ such that every subset of $n$ points $\{v_{i_1}, .., v_{i_n}  \}$ is linearly independent.
 A: Here's a sketch. Call a configuration "bad" if some set of $d+2$ points are co-hyperplanar (a "bad" set.) The gist is that the set of bad configurations has strictly smaller dimension than the set of all configurations, in this case because the bad configurations are defined by algebraic equations which hold for some, but not all, of the configurations.
First, fix a set $X$ of $d+2$ points (say, in the sphere $S^{d+1}$). These lie in a hyperplane through the origin iff $d(X):=\det (x_{ij})=0$, where $x_{ij}$ is the $j$th coordinate of the $i$th point. Note that we can get $d(X)$ equal to zero or one by taking $X$ to consist of equal vectors or the unit coordinate vectors respectively, so in particular $d$ is a non-constant (polynomial) function of $X$, and $d^{-1}(0)$ has strictly smaller dimension than the space of all $(d+2)$-subsets of $S^{d+1}$). 
It follows that (a) there's a "good" set of points arbitrarily close to any given set $X$, and (b) a randomly chosen set $X$ (e.g., i.i.d. w.r.t. Haar measure) is good. 
But a configuration is good iff all of its (finitely many) $(d+2)$-subsets are good, which they will be with probability one, so a random configuration is good with probability one. In particular, good configurations exist.
A: An easy answer is: Define
$p: (\mathbb{R}^n)^m \rightarrow \mathbb{R}$ by
$$p(v_1,..,v_m)= \prod_{i_1<..<i_n} \det (v_{i_i},..,v_{i_k}).$$
This defines a polynomial function on
$\mathbb{R}^{nm}$. Clearly, each of the factors in the above product is a non-vanishing polynomial on $\mathbb{R}^{nm}$, and since the polynomial ring in $nm$ variables is an integral domain the product is nonzero, and hence by a result from algebra it is also non-vanishing, and we get our point in general position. We can even say more: the set where the polynomial doesn’t vanish lies dense in $\mathbb{R}^{nm}$ as a nonzero polynomial cannot be identically $0$ on an open set.
