Simple disproof of Danzer — Grünbaum conjecture

I asked this question on the MSE, but I did not get an answer. I hope that one of the experienced participants will check the correctness of the proof or the truth of the statement (and, perhaps, will offer a simpler and clearer proof).

A set of points in $\mathbb R^n$ is acute if any three points from this set form an acute triangle. In 1962 Danzer and Grünbaum conjectured that cardinality of acute set in $\mathbb R^d$ is $2d-1$, no more. In 1983 Erdős and Füredi disproved this conjecture in their well-known article. They used the probabilistic method and proved that the cardinality of the set grows exponentially with respect to the dimension of the space. The first proofs with explicit construction was given much later, as I know.

I found an explicit construction so simple, that the situation with this conjecture looks somewhat curious. I do not know if this proof was published somewhere, so I'm asking to check it (I'm only an amateur).

Let us consider unit cube $[0,1]^n$ with vertices $x^i = (x^i_1, x^i_2, ..., x^i_n)$, $x^i_j \in \{0,1 \}$, $i = 1, ..., 2^n, \ j = 1, ..., n.$ The coordinates $x^i_1, x^i_2, ..., x^i_n$ we will call the base coordinates. We will denote the modulo-2 addition, or XOR, operation by $\oplus$. To each $x^i$ we add $\frac {n(n-1)}{2}$ additional coordinates (all pairwise XORs of base coordinates): $$\tilde x^i = (x^i_1, ..., x^i_n, \ x^i_1\oplus x^i_2, \ x^i_1\oplus x^i_3, \ldots, x^i_ {n-1}\oplus x^i_ {n}), \quad i = 1, \ldots, 2^n. \quad (1)$$

Lemma 1. The set (1) is acute set.

Lemma 2. Cardinality of the set (1) grows exponentially with respect to $n$ and, roughly, grows exponentially with respect to $\sqrt{D}$. Here $D$ – dimension: $D=n + \frac{n (n-1)}{2} = \frac{n (n + 1)}{2}$ .

Proof of Lemma 1.

Below I use an obvious fact that the angle constructed on the vertices of the unit cube can not be obtuse and all the summands of the inner product are nonnegative. This is easy to prove.
Let us choose 3 arbitrary points $a, b, c$ from (1) and consider the angle with vertex in $a$. Let $\vec {ab}$ and $\vec {ac}$ be two vectors of this angle. Obviously, there are base coordinates $j_1, j_2 \in [1, n]$ such that $b_{j_1} \ne a_{j_1}$ and $c_ {j_2} \ne a_ {j_2}$. If $b_{j_1} = c_{j_1}$ or $b_{j_2} = c_{j_2}$ then the inner product $(\vec {ab}, \vec {ac})\ge 1$ (the summand $ab_ {j_1} \cdot ac_{j_1}=1$ or the summand $ab_ {j_2} \cdot ac_{j_2}=1$, respectively). If $j_1 \ne j_2$ then the summand $(ab_ {j_1} \oplus ab_{j_2}) \cdot (ac_{j_1} \oplus ac_{j_2})=1$ in the inner product. Therefore, the selected angle is acute. Q.E.D.

Proof of Lemma 2. Obvious.

In such a way, the Danzer — Grünbaum conjecture is disproved.

• I am not an expert in the topic but this site is not for discussing whether an argument is correct or not. It looks like you are promoting your own proof, which is also frowned upon here. – András Bátkai Sep 3 '18 at 6:23
• @AndrásBátkai I'm not a mathematician. Just accidentally found a statement that no one noticed before. The proof is so trivial that it would be ridiculous to claim its authorship. I can remove my own proof from my post and only ask the participants to give a simple proof. Will it comply with the rules? – grizzly Sep 3 '18 at 6:29
• Write the paper and submit it to an appropriate journal! – Watson Ladd Sep 3 '18 at 14:23
• The receptions here seem awfully un-welcoming. As far as I can tell, there is no attempt to make the results sound like more than they are; in fact, most of the post is dedicated to explaining how it must be simple, and might be wrong. Contributions from an amateur mathematician are great things and should be encouraged; they don't have to be at the very forefront of research still to be valuable. I second @WatsonLadd's recommendation (although unfortunately I don't know an appropriate journal). – LSpice Sep 3 '18 at 14:59
• Andras is right, technically: MathOverflow is not for checking results. However, this question can be easily turned into a reference request, which seems implicit in the post already; it just needs to be made explicit. If someone can find a reference for the poster, I think that would be a good question and answer for this forum. Gerhard "We Do Reference Requests, Right?" Paseman, 2018.09.03. – Gerhard Paseman Sep 3 '18 at 18:07