I have a problem while studying the proof of the following theorem (Thm 1.4 in Algebraic Geometry, A first course by Joe Harris):
If $\Gamma \subset \mathbb{P}^n$ is any collection of $d \leq 2n$ points in general position, then $\Gamma$ may be described as the zero locus of quadratic polynomials.
Here is a summary of the proof: Suppose $q$ is a point such that every quadratic polynomial vanishing on $\Gamma$ vanishes at $q$. We shall show that $q\in \Gamma$. Note that for any partition $\Gamma=\Gamma_1\cup\Gamma_2$ of $\Gamma$ into $n$-point sets, $q$ lies in the union of the hyperplanes spanned by points of $\Gamma_1$ or $\Gamma_2$. Now let $S=\{p_1,\ldots,p_k\}$ be any minimal subset of $\Gamma$ such that $q$ lies in their span. Then $q$ lies in the span of $p_1$ and any $n−1$ of the points in $\Gamma−S$, from which we deduce that $q=p_1$.
The proof deals with the case when $d=2n$, and says that the general case is easier. Why is that?
If $d\leq n$, we may restrict our attention to the $\Bbb P^{d-1}$ spanned by the $d$ points we have, which is cut out by linear equations. Squaring these equations, we see that if we can solve the problem for $d$ points in $\Bbb P^{d-1}$, we can solve the problem for our original $d$ points in $\Bbb P^n$. So we may assume that $n+1\leq d\leq 2n$, and in the case that $d<2n$ we may let $\Gamma = \Gamma_1\cup \Gamma_2$ be a non-disjoint union of sets of cardinality $n$. The rest of the proof goes through from there.