Set of points covered by subspaces of small dimensions

Question

Let $S \subset \mathbb R^d$ be finite set of points. We say that $S$ is $2$-covered if $S$ lies in a union $V_1\cup V_2$ of affine subspaces such that $\dim(V_1)+\dim (V_2)\leq d-1$. For example, if $d=2$, a set is $2$-covered iff all or all but one point are collinear.

Question: Can we show that for each $d$, there is a constant $C$ such that if the size of $S$ is $n\geq C$, and each subset of $n-1$ points in $S$ is $2$-covered, then so is $S$?

For $d=2$, it is easy to show that $C=5$ works, and that is the smallest value (any $3$ points is $2$-covered, but $4$ points in general position is not $2$-covered).

Obviously, one can ask variants of this question about collection of subspaces with similar constraints, or finite fields, projective spaces, etc, so please share if you know some references.

Answer 1

Yes. One can take $C=2d+2$.

I'll do the projective case, so my points will be nonzero points in a $d+1$-dimensional space $W$ and I'm looking for pairs of subspaces of total dimension $\leq d+1$ that cover them. My proof will work over any field, and probably for matroids too.

If $S$ is not $2$-covered, then clearly every set of $n$ points of $S$ spans $W$, as otherwise the space it spans and the line containing the missing point would give a $2$-covering. If this subset of $n-1$ points is $2$-covered, then since the two subspaces have total dimension $\leq d+1$ and together span $W$, they must be complements of each other.

If $S$ doesn't span $W$, then clearly it is $2$-covered. Let $v_1,\dots, v_{d+1}$ be elements of $S$ that form a basis of $W$ and let $w_1,\dots, w_{n-d-1}$ be the other elements. Form a graph with vertices $v_i, w_i$ and an edge connecting $w_i$ to $v_j$ if and only if $v_j$ appears in the expression of $w_i$ in the basis $v_1,\dots, v_{d+1}$. The graph must be connected, since a partition of the graph into two disconnected parts gives a 2-covering by the spaces spanned by the basis vectors in each part. However, removing any vertex $w_i$ must make it disconnected, since for a 2-covering, since the two subspaces are complementary, a vector in one subspace written as a combination of basis vectors must involve only basis vectors in that subspace.

Since every cycle involves a vector $w_i$, and removing a vertex in a cycle doesn't disconnect a graph, the assume the graph is a tree.

EDIT: this claim about graphs is not actually true, so the answer isn't currently correct.

Thus it contains exactly $n-1$ edges. But each $w_i$ has at least $2$ edges since it is not a basis vector (up to scaling), so the number of edges is at least $2 (n-d-1)$. Thus $n-1 \geq 2 (n-d-1)$ which gives $n \leq 2d+1$, contradicting $n\geq C=2d+2$.

This can probably be improved with a different idea...