An Incidence problem

Question

Is this intuition true: If any $d+1$ points among $d+2$ given points in $\mathbb{R}^d$ lies on a hyperplane then all of them lies on a hyperplane?

Answer 1

Let $P = \{x_1,...,x_{d+2} \} \subset \mathbb{R}^{d}$.

Let $\delta = \dim \operatorname{aff} P$. (This is the dimension of the affine hull of $P$, which is the dimension of the smallest linear subspace that contains some translate of $P$.) Note that $\delta \le d$.

I claim that $\delta \le d-1$. Suppose $\delta = d$ to obtain a contradiction. Then there is an affine basis $B \subset P$ with $|B| = d+1$. However, by assumption, there is a hyperplane containing $B$, and hence $|B| $ is not affinely independent which is a contradiction.

Hence there is a hyperplane containing $P$.

Addendum:

This can all be expressed in the context of linear spaces. Note that $y_1,...,y_k \in \mathbb{R}^d$ lie on a hyperplane iff $\dim \operatorname{sp} \{y_i-y_1\}_{i=2}^k \le d-1$.

Now suppose $x_1,...,x_{d+2}$ are not contained in some hyperplane. Then we must have $\dim \operatorname{sp} \{x_i-x_1\}_{i=2}^{d+2} = d$. In particular, $x_2-x_1,...,x_{d+1}-x_1$ contains a basis and so there we can choose $d$ of them (let the indices be $I \subset \{2,...,d+2\}$, with $|I| = d$) such that $x_i-x_1$, $i \in I$ form a basis. Since they form a basis, the points $x_1, x_i$, $i \in I$ cannot be contained in a hyperplane, which contradicts the original assumption. Hence the points are contained in some subspace.